properties of solutions lesson 2. measuring concentration concentration = amount of solute/amount...

Properties of SolutionsProperties of Solutions

Lesson 2Lesson 2

Measuring concentration Measuring concentration

Concentration = amount of solute/amount Concentration = amount of solute/amount of solutionof solution

Weight/volume percentWeight/volume percent Mass solute in g/volume of soln in mL x 100%Mass solute in g/volume of soln in mL x 100%

Weight/weight percentWeight/weight percent Mass solute in g/mass solution in g x 100%Mass solute in g/mass solution in g x 100%

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Molarity = Moles of solute/Liters of Solution (M)

Molality = Moles of solute/Kg of Solvent (m)

Mole Fraction=Moles solute/total number of moles

Mass %=Mass solute/total mass x 100

Concentration

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Molarity = Moles of solute/Liters of Solution (M)

Molality = Moles of solute/Kg of Solvent (m)

Mole Fraction=Moles solute/total number of moles

Mass %=Mass solute/total mass x 100

Concentration

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Molarity = Moles of solute/Liters of Solution (M)

Molality = Moles of solute/Kg of Solvent (m)

Mole Fraction=Moles solute/total number of moles

Mass %=Mass solute/total mass x 100

Concentration

Molarity

moles of soluteMolarity ( ) =

liters of solution M

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A sample of NaNO3 weighing 8.5 grams is placed in a 500 ml volumetric flask and distilled water was added to the mark on the neck of the flask. Calculate the Molarity of the resulting solution.

Convert the given grams of solute to moles of solute :

Convert given ml of solution to liters

Apply the definition for Molarity: Molarity = moles NaNO3 / volume of the solution in liters

M = 0.1 mole / .500 liters = 0.200 Molar NaNO3

33

33

1 mole NaNO

858.5 g NaNO 0.1 mole

g N

NaNOaNO

1 liter

1000 ml500 ml 0.5 liter

Exercise #1

You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.

8.00 M

Exercise #2

You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?

0.200 L

Exercise #3

Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity.

10.0 M NaOH

5.37 M KCl

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Molarity = Moles of solute/Liters of Solution (M)

Molality = Moles of solute/Kg of Solvent (m)

Mole Fraction=Moles solute/total number of moles

Mass %=Mass solute/total mass x 100

Molality

moles of soluteMolality ( ) =

kilogram of solvent m

Exercise #4

A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)

0.816 m

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Molarity = Moles of solute/Liters of Solution (M)

Molality = Moles of solute/Kg of Solvent (m)

Mole Fraction=Moles solute/total number of moles

Mass %=Mass solute/total mass x 100

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Determine the mole fraction of KCl in 3000 grams of aqueous solution containing 37.3 grams of Potassium

Chloride KCl.

1. Convert grams KCl to moles KCl using the molecular weight of KCl

2. Determine the grams of pure solvent water from the given grams of solution and solute

Total grams = 3000 grams = Mass of solute + Mass of water Mass of pure solvent = (3000 - 37.3) gram

= 2962.7 gram

1 mole KCl

74.6 g K37.3 g KCl 0.5 mo

Clle KCl

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Determine the mole fraction of KCl in 3000 grams of aqueous solution containing 37.3 grams of Potassium

Chloride KCl.

3. Convert grams of solvent H2O to mols

4. Apply the definition for mole fraction mole fraction = moles of KCl / Total mols of KCl and water =

0.5 / (0.5 + 164.6) = 0.5 / 165.1 = 0.00303

2

1 mol

18.0 grams2962.7 grams water 164.6 mols H O

Mole Fraction

AA

molesMole fraction ( ) =

total moles of solution

Exercise #5

A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4.

(Assume water has a density of 1.00 g/mL.)

0.0145

Mass Percent

mass of soluteMass (weight) percent = 100%

mass of solution

Exercise #6

What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water?

6.6%

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Assuming the density of water to be 1 g/mL we approximate the density of a dilute aqueous solution to be 1 g/mL

1 ppm = 1 μg/mL = 1 mg/L

1 ppb = 1 ng/mL = 1 μg/L

1 g1 ppm =

1 g

1 g 1 g

1

1 g

1 ml g 1 ml

MolarityMolarity

Concentration is usually expressed in Concentration is usually expressed in terms of molarity:terms of molarity:

Moles of solute/liters of Moles of solute/liters of solutionsolution (M) (M)

Moles of solute = molarity x volume of solutionMoles of solute = molarity x volume of solution

Moles = M x VMoles = M x V

Molarity and concentrationMolarity and concentration

Molarity:Molarity:

M = moles solute/liter of solutionM = moles solute/liter of solution DilutionDilution

MM11VV11 = M = M22VV22

ExampleExample

What is molarity of 50 ml solution containing What is molarity of 50 ml solution containing 2.355 g H2.355 g H22SOSO44?? Molar mass HMolar mass H22SOSO44 = 98.1 g/mol = 98.1 g/mol

Moles HMoles H22SOSO44 = .0240 mol = .0240 mol (2.355 g/98.1 g/mol)(2.355 g/98.1 g/mol)

Volume of solution = 50 mL/1000 mL/L = .050 LVolume of solution = 50 mL/1000 mL/L = .050 L Concentration = moles/volumeConcentration = moles/volume

= .0240 mol/.050 L = 0.480 M= .0240 mol/.050 L = 0.480 M

Solution stoichiometrySolution stoichiometry How much volume of one solution to react with another How much volume of one solution to react with another

solutionsolution Given volume of A with molarity MGiven volume of A with molarity MAA

Determine moles ADetermine moles A Determine moles BDetermine moles B Find target volume of B with molarity MFind target volume of B with molarity MBB

Volume Bmol = MV Mole:mole ratio V = mol/MVolume A Moles A Moles B