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16.2 Concentrations of 16.2 Concentrations of Solutions > Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties of Solutions 16.2 Concentrations of Soluti 16.3 Colligative Properties of Solutions 16.4 Calculations Involving Colligative Properties

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Page 1: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

1 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Chapter 16Solutions

161 Properties of Solutions

162 Concentrations of Solutions

163 Colligative Properties of Solutions164 Calculations Involving Colligative Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

2 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How can you describe the concentration of a solution

The federal government and state governments set standards limiting the amount of contaminants allowed in drinking water

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

3 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

Molarity

How do you calculate the molarity of a solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

4 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent

bull A solution that contains a relatively small amount of solute is a dilute solution

bull A concentrated solution contains a large amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

5 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

In chemistry the most important unit of concentration is molarity

bull Molarity (M) is the number of moles of solute dissolved in one liter of solution

bull Molarity is also known as molar concentration

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

6 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

The figure below illustrates the procedure for making a 05M or 05-molar solution

Add 05 mol of solute to a 1-L volumetric flask half filled with distilled water

Swirl the flask carefully to dissolve the solute

Fill the flask with water exactly to the 1-L mark

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

7 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

To calculate the molarity of a solution divide the number of moles of solute by the volume of the solution in liters

Molarity (M) = moles of soluteliters of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

8 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Intravenous (IV) saline solutions are often administered to patients in the hospital One saline solution contains 090 g NaCl in exactly 100 mL of solution What is the molarity of the solution

Sample Problem 162Sample Problem 162

Calculating Molarity

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

9 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

solution concentration = M

Analyze List the knowns and the unknown1

Sample Problem 162Sample Problem 162

solution concentration = 090 g NaCl100 mL

molar mass NaCl = 585 gmol

Convert the concentration from g100 mL to molL The sequence is

g100 mL rarr mol100 mL rarr molL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

10 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

11 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

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27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 2: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

2 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How can you describe the concentration of a solution

The federal government and state governments set standards limiting the amount of contaminants allowed in drinking water

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

3 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

Molarity

How do you calculate the molarity of a solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

4 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent

bull A solution that contains a relatively small amount of solute is a dilute solution

bull A concentrated solution contains a large amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

5 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

In chemistry the most important unit of concentration is molarity

bull Molarity (M) is the number of moles of solute dissolved in one liter of solution

bull Molarity is also known as molar concentration

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

6 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

The figure below illustrates the procedure for making a 05M or 05-molar solution

Add 05 mol of solute to a 1-L volumetric flask half filled with distilled water

Swirl the flask carefully to dissolve the solute

Fill the flask with water exactly to the 1-L mark

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

7 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

To calculate the molarity of a solution divide the number of moles of solute by the volume of the solution in liters

Molarity (M) = moles of soluteliters of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

8 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Intravenous (IV) saline solutions are often administered to patients in the hospital One saline solution contains 090 g NaCl in exactly 100 mL of solution What is the molarity of the solution

Sample Problem 162Sample Problem 162

Calculating Molarity

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

9 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

solution concentration = M

Analyze List the knowns and the unknown1

Sample Problem 162Sample Problem 162

solution concentration = 090 g NaCl100 mL

molar mass NaCl = 585 gmol

Convert the concentration from g100 mL to molL The sequence is

g100 mL rarr mol100 mL rarr molL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

10 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

11 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

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Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

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Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

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35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

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36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

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KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

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38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

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39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

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40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

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41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

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Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

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Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

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49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

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50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

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52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

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53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

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54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

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END OF 162END OF 162

Page 3: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

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MolarityMolarity

Molarity

How do you calculate the molarity of a solution

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MolarityMolarity

The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent

bull A solution that contains a relatively small amount of solute is a dilute solution

bull A concentrated solution contains a large amount of solute

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5 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

In chemistry the most important unit of concentration is molarity

bull Molarity (M) is the number of moles of solute dissolved in one liter of solution

bull Molarity is also known as molar concentration

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MolarityMolarity

The figure below illustrates the procedure for making a 05M or 05-molar solution

Add 05 mol of solute to a 1-L volumetric flask half filled with distilled water

Swirl the flask carefully to dissolve the solute

Fill the flask with water exactly to the 1-L mark

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7 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

To calculate the molarity of a solution divide the number of moles of solute by the volume of the solution in liters

Molarity (M) = moles of soluteliters of solution

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Intravenous (IV) saline solutions are often administered to patients in the hospital One saline solution contains 090 g NaCl in exactly 100 mL of solution What is the molarity of the solution

Sample Problem 162Sample Problem 162

Calculating Molarity

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KNOWNS

UNKNOWN

solution concentration = M

Analyze List the knowns and the unknown1

Sample Problem 162Sample Problem 162

solution concentration = 090 g NaCl100 mL

molar mass NaCl = 585 gmol

Convert the concentration from g100 mL to molL The sequence is

g100 mL rarr mol100 mL rarr molL

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Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

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bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

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Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

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KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

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Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

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bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

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16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

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Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

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Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

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Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

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Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

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Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

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Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

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How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

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25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

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27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

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40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

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END OF 162END OF 162

Page 4: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

4 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent

bull A solution that contains a relatively small amount of solute is a dilute solution

bull A concentrated solution contains a large amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

5 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

In chemistry the most important unit of concentration is molarity

bull Molarity (M) is the number of moles of solute dissolved in one liter of solution

bull Molarity is also known as molar concentration

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

6 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

The figure below illustrates the procedure for making a 05M or 05-molar solution

Add 05 mol of solute to a 1-L volumetric flask half filled with distilled water

Swirl the flask carefully to dissolve the solute

Fill the flask with water exactly to the 1-L mark

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

7 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

To calculate the molarity of a solution divide the number of moles of solute by the volume of the solution in liters

Molarity (M) = moles of soluteliters of solution

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8 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Intravenous (IV) saline solutions are often administered to patients in the hospital One saline solution contains 090 g NaCl in exactly 100 mL of solution What is the molarity of the solution

Sample Problem 162Sample Problem 162

Calculating Molarity

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KNOWNS

UNKNOWN

solution concentration = M

Analyze List the knowns and the unknown1

Sample Problem 162Sample Problem 162

solution concentration = 090 g NaCl100 mL

molar mass NaCl = 585 gmol

Convert the concentration from g100 mL to molL The sequence is

g100 mL rarr mol100 mL rarr molL

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Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

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bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

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Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

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KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

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Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

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bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

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How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

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Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

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Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

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Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

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Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

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Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

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Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

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How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

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KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

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26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

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27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

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1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

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Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

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Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

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33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

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34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

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35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

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36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

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KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

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State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

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Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

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bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

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Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

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Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

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What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

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How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

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KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

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Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

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Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

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49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

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What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

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What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

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Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

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Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

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Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

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END OF 162END OF 162

Page 5: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

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MolarityMolarity

In chemistry the most important unit of concentration is molarity

bull Molarity (M) is the number of moles of solute dissolved in one liter of solution

bull Molarity is also known as molar concentration

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MolarityMolarity

The figure below illustrates the procedure for making a 05M or 05-molar solution

Add 05 mol of solute to a 1-L volumetric flask half filled with distilled water

Swirl the flask carefully to dissolve the solute

Fill the flask with water exactly to the 1-L mark

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MolarityMolarity

To calculate the molarity of a solution divide the number of moles of solute by the volume of the solution in liters

Molarity (M) = moles of soluteliters of solution

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Intravenous (IV) saline solutions are often administered to patients in the hospital One saline solution contains 090 g NaCl in exactly 100 mL of solution What is the molarity of the solution

Sample Problem 162Sample Problem 162

Calculating Molarity

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9 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

solution concentration = M

Analyze List the knowns and the unknown1

Sample Problem 162Sample Problem 162

solution concentration = 090 g NaCl100 mL

molar mass NaCl = 585 gmol

Convert the concentration from g100 mL to molL The sequence is

g100 mL rarr mol100 mL rarr molL

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10 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

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bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

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14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

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15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

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16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

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18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

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20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

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21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

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22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

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23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

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27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

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40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

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41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

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42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

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45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

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47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

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48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

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49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

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50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

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52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

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53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

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54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

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END OF 162END OF 162

Page 6: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

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MolarityMolarity

The figure below illustrates the procedure for making a 05M or 05-molar solution

Add 05 mol of solute to a 1-L volumetric flask half filled with distilled water

Swirl the flask carefully to dissolve the solute

Fill the flask with water exactly to the 1-L mark

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MolarityMolarity

To calculate the molarity of a solution divide the number of moles of solute by the volume of the solution in liters

Molarity (M) = moles of soluteliters of solution

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8 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Intravenous (IV) saline solutions are often administered to patients in the hospital One saline solution contains 090 g NaCl in exactly 100 mL of solution What is the molarity of the solution

Sample Problem 162Sample Problem 162

Calculating Molarity

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9 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

solution concentration = M

Analyze List the knowns and the unknown1

Sample Problem 162Sample Problem 162

solution concentration = 090 g NaCl100 mL

molar mass NaCl = 585 gmol

Convert the concentration from g100 mL to molL The sequence is

g100 mL rarr mol100 mL rarr molL

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10 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

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11 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

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14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

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15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

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16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

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19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

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Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

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21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

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22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

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Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

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How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

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25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

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26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

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27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

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28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

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33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

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34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

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35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

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36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

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38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

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39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

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40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

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41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

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49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

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52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

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53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

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54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

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END OF 162END OF 162

Page 7: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

7 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

MolarityMolarity

To calculate the molarity of a solution divide the number of moles of solute by the volume of the solution in liters

Molarity (M) = moles of soluteliters of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

8 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Intravenous (IV) saline solutions are often administered to patients in the hospital One saline solution contains 090 g NaCl in exactly 100 mL of solution What is the molarity of the solution

Sample Problem 162Sample Problem 162

Calculating Molarity

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

9 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

solution concentration = M

Analyze List the knowns and the unknown1

Sample Problem 162Sample Problem 162

solution concentration = 090 g NaCl100 mL

molar mass NaCl = 585 gmol

Convert the concentration from g100 mL to molL The sequence is

g100 mL rarr mol100 mL rarr molL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

10 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

11 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

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27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

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40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

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49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

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52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

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53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

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54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

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END OF 162END OF 162

Page 8: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

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8 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Intravenous (IV) saline solutions are often administered to patients in the hospital One saline solution contains 090 g NaCl in exactly 100 mL of solution What is the molarity of the solution

Sample Problem 162Sample Problem 162

Calculating Molarity

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9 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

solution concentration = M

Analyze List the knowns and the unknown1

Sample Problem 162Sample Problem 162

solution concentration = 090 g NaCl100 mL

molar mass NaCl = 585 gmol

Convert the concentration from g100 mL to molL The sequence is

g100 mL rarr mol100 mL rarr molL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

10 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

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11 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

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12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

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14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

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15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

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16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

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18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

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19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

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20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

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21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

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55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 9: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

9 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

solution concentration = M

Analyze List the knowns and the unknown1

Sample Problem 162Sample Problem 162

solution concentration = 090 g NaCl100 mL

molar mass NaCl = 585 gmol

Convert the concentration from g100 mL to molL The sequence is

g100 mL rarr mol100 mL rarr molL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

10 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

11 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 10: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

10 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Use the molar mass to convert g NaCl100 mL to mol NaCl100 mL Then convert the volume units so that your answer is expressed in molL

Calculate Solve for the unknown2

Sample Problem 162Sample Problem 162

The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL1 L

Solutionconcentration =

= 015 molL

= 015M

090 g NaCl 1 mol NaCl 1000 mL 100 mL 585 g NaCl 1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

11 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 11: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

11 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be less than 1M because a concentration of 090 g100 mL is the same as 90 g1000 mL (90 g1 L) and 90 g is less than 1 mol NaCl

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 162Sample Problem 162

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 12: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

12 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO) How many moles of solute are present in 15 L of 070M NaClO

Sample Problem 163Sample Problem 163

Calculating the Moles of Solute in a Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 13: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

13 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

moles solute = mol

Analyze List the knowns and the unknown1

Sample Problem 163Sample Problem 163

volume of solution = 15 L

solution concentration = 070M NaClO

The conversion is volume of solution rarr moles of solute Molarity has the units molL so you can use it as a conversion factor between moles of solute and volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 14: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

14 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the given volume by the molarity expressed in molL

Calculate Solve for the unknown2

Sample Problem 163Sample Problem 163

Make sure that your volume units cancel when you do these problems If they donrsquot then yoursquore probably missing a conversion factor in your calculations

15 L = 11 mol NaClO070 mol NaCl

1 L

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 15: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

15 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The answer should be greater than 1 mol but less than 15 mol because the solution concentration is less than 075 molL and the volume is less than 2 L

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 163Sample Problem 163

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 16: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

16 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 17: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

17 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How much water is required to make a 100M aqueous solution of NaCl if 584 g of NaCl are dissolved

A 100 liter of water

B enough water to make 100 liter of solution

C 100 kg of water

D 100 mL of water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 18: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

18 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Making Dilutions

What effect does dilution have on the amount of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 19: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

19 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Both of these solutions contain the same amount of solutebull You can tell by the color of solution (a) that it is

more concentrated than solution (b)

bull Solution (a) has the greater molarity

bull The more dilute solution (b) was made from solution (a) by adding more solvent

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 20: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

20 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 21: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

21 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The definition of molarity can be rearranged to solve for moles of solute

Molarity (M) = moles of solute

liters of solution (V)

Moles of solute = molarity (M) liters of solution (V)

Moles of solute before dilution =

Moles of solute after dilution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 22: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

22 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

The total number of moles of solute remains unchanged upon dilution

Moles of solute = M1 V1 = M2 V2

bull M1 and V1 are the molarity and the volume of the initial solution

bull M2 and V2 are the molarity and volume of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 23: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

23 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Making DilutionsMaking Dilutions

She measures 20 mL of the stock solution with a 20-mL pipet

She transfers the 20 mL to a 100-mL volumetric flask

She carefully adds water to the mark to make 100 mL of solution

The student is preparing 100 mL of 040M MgSO4 from a stock solution of 20M MgSO4

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 24: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

24 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many milliliters of aqueous 200M MgSO4 solution must be diluted with water to prepare 1000 mL of aqueous 0400M MgSO4

Sample Problem 164Sample Problem 164

Preparing a Dilute Solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 25: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

25 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

V1 = mL of 200M MgSO4

Analyze List the knowns and the unknown1

Sample Problem 164Sample Problem 164

M1 = 200M MgSO4

M2 = 0400M MgSO4

V2 = 1000 mL of 0400M MgSO4

Use the equation M1 V1 = M2 V2 to solve for the unknown initial volume of solution (V1) that is diluted with water

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 26: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

26 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Solve for V1 and substitute the known values into the equation

Calculate Solve for the unknown2

Sample Problem 164Sample Problem 164

V1 = = = 200 mLM2 V2

M1

0400M 1000 mL200M

Thus 200 mL of the initial solution must be diluted by adding enough water to increase the volume to 1000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 27: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

27 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The initial concentration is five times larger than the dilute concentration

bull Because the number of moles of solute does not change the initial volume of solution should be one-fifth the final volume of the diluted solution

Evaluate Does the result make sense3

Sample Problem 164Sample Problem 164

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 28: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

28 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 29: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

29 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

1000 mL of a 0300M CuSO45H2O solution is diluted to 5000 mL What is the concentration of the diluted solution

M1 V1 = M2 V2

M2 = 00600M

M2 = = M1 V1

V2

0300M 1000 mL

5000 mL

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 30: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

30 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent SolutionsHow do percent by volume and percent by mass differ

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 31: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

31 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 32: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

32 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 33: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

33 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 34: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

34 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solutionbull Isopropyl alcohol (2-propanol) is sold

as a 91-percent solution by volume

bull The concentration is written as 91 percent by volume 91 percent (volumevolume) or 91 (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 35: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

35 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by volume of a solution is the ratio of the volume of solute to the volume of solution

Percent by volume ((vv)) = 100 volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 36: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

36 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the percent by volume of ethanol (C2H6O or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water

Sample Problem 165Sample Problem 165

Calculating Percent by Volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 37: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

37 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

Percent by volume = ethanol (vv)

Analyze List the knowns and the unknown1

Sample Problem 165Sample Problem 165

volume of solute = 85 mL ethanol

volume of solution = 250 mL

Use the known values for the volume of solute and volume of solution to calculate percent by volume

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 38: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

38 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

State the equation for percent by volume

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100volume of solution

volume of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 39: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

39 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Substitute the known values into the equation and solve

Calculate Solve for the unknown2

Sample Problem 165Sample Problem 165

Percent by volume ((vv)) = 100 250 mL

85 mL ethanol

= 34 ethanol (vv)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 40: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

40 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The volume of the solute is about one-third the volume of the solution so the answer is reasonable

bull The answer is correctly expressed to two significant figures

Evaluate Does the result make sense3

Sample Problem 165Sample Problem 165

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 41: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

41 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution

Percent by mass ((mm)) = 100mass of solution

mass of solute

Another way to express the concentration of a solution is as a percent by mass or percent (massmass)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 42: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

42 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Percent SolutionsPercent Solutions

bull Percent by mass is sometimes a convenient measure of concentration when the solute is a solid

bull You have probably seen information on food labels expressed as a percent composition

Percent by mass ((mm)) = 100mass of solution

mass of solute

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 43: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

43 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 44: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

44 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What are three ways to calculate the concentration of a solution

CHEMISTRY amp YOUCHEMISTRY amp YOU

The concentration of a solution can be calculated in moles solute per liter of solvent or molarity (M) percent by volume ((vv)) or percent by mass ((mm))

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 45: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

45 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

How many grams of glucose (C6H12O6) are needed to make 2000 g of a 28 glucose (mm) solution

Sample Problem 166Sample Problem 166

Using Percent by Mass as a Conversion Factor

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 46: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

46 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

KNOWNS

UNKNOWN

mass of solute = g C6H12O6

Analyze List the knowns and the unknown1

Sample Problem 166Sample Problem 166

mass of solution = 2000 g

percent by mass = 28 C6H12O6(mm)

The conversion is mass of solution rarr mass of solute In a 28 C6H12O6 (mm) solution each 100 g of solution contains 28 g of glucose Used as a conversion factor the concentration allows you to convert g of solution to g of C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 47: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

47 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Write percent by mass as a conversion factor with g C6H12O6 in the numerator

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

100 g solution

28 g C6H12O6

You can solve this problem by using either dimensional analysis or algebra

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 48: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

48 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Multiply the mass of the solution by the conversion factor

Calculate Solve for the unknown2

Sample Problem 166Sample Problem 166

2000 g solution = 56 g C6H12O6100 g solution

28 g C6H12O6

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 49: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

49 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

bull The prepared mass of the solution is 20 100 g

bull Since a 100-g sample of 28 (mm) solution contains 28 g of solute you need 20 28 g = 56 g of solute

bull To make the solution mix 56 g of C6H12O6 with

1944 g of solvent

bull 56 g of solute + 1944 g solvent = 2000 g of solution

Evaluate Does the result make sense3

Sample Problem 166Sample Problem 166

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 50: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

50 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 51: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

51 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 50 (mm)

(mm) = 100

mass of glucosemass of solution

mass of glucose =

mass of glucose = 2000 g 0050 = 100 g C6H12O6

mass of water = 2000 g ndash 100 g = 1900 g H2O

( (mm)) mass of solution100

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 52: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

52 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key ConceptsKey Concepts

To calculate the molarity of a solution divide the moles of solute by the volume of the solution in liters

Diluting a solution reduces the number of moles of solute per unit volume but the total number of moles of solute in solution does not change

Percent by volume is the ratio of the volume of solute to the volume of solution Percent by mass is the ratio of the mass of the solute to the mass of the solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 53: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

53 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Key EquationsKey Equations

Percent by mass = 100 mass of solution

mass of solute

Percent by volume = 100volume of solution

volume of solute

Molarity (M) = moles of solute

liters of solution

M1 V1 = M2 V2

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 54: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

54 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

Glossary TermsGlossary Terms

bull concentration a measurement of the amount of solute that is dissolved in a given quantity of solvent usually expressed as molL

bull dilute solution a solution that contains a small amount of solute

bull concentrated solution a solution containing a large amount of solute

bull molarity (M) the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162

Page 55: 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 16 Solutions 16.1 Properties

162 Concentrations of Solutions gt162 Concentrations of Solutions gt

55 Copyright copy Pearson Education Inc or its affiliates All Rights Reserved

END OF 162END OF 162