16.2 concentrations of solutions > 1 copyright © pearson education, inc., or its affiliates. all...
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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
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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
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MolarityMolarity
Molarity
How do you calculate the molarity of a solution
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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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
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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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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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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
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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
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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
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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 solutionbull Isopropyl alcohol (2-propanol) is sold
as a 91-percent solution by volume
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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 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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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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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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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
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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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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
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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
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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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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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
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Making DilutionsMaking Dilutions
Making Dilutions
What effect does dilution have on the amount of solute
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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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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
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
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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 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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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)
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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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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
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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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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MolarityMolarity
Molarity
How do you calculate the molarity of a solution
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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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
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
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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
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
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
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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
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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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
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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
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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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
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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
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
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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)
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
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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
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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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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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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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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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))
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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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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
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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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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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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
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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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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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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
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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
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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
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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 solutionbull Isopropyl alcohol (2-propanol) is sold
as a 91-percent solution by volume
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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 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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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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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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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
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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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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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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)
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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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
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Making DilutionsMaking Dilutions
Making Dilutions
What effect does dilution have on the amount of solute
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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
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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
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END OF 162END OF 162
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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162 Concentrations of Solutions gt162 Concentrations of Solutions gt
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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
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
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
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
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
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