Properties of Solutions
Lesson 2
Measuring concentration
 Concentration
= amount of solute/amount
of solution
 Weight/volume percent

Mass solute in g/volume of soln in mL x 100%
 Weight/weight

percent
Mass solute in g/mass solution in g x 100%
Concentration
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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17a–3
Concentration
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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17a–4
Concentration
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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17a–5
Molarity
moles of solute
Molarity (M ) =
liters of solution
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 :
1 mole NaNO3
8.5 g NaNO3
 0.1 mole NaNO3
85 g NaNO3
Convert given ml of solution to liters
1 liter
500 ml
 0.5 liter
1000 ml
Apply the definition for Molarity: Molarity = moles NaNO3 / volume
of the solution in liters
M = 0.1 mole / .500 liters = 0.200 Molar NaNO3
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17a–7
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
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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17a–11
Molality
moles of solute
Molality (m) =
kilogram of solvent
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
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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17a–14
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
1 mole KCl
37.3 g KCl
 0.5 mole KCl
74.6 g 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
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17a–15
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
 2962.7 grams water 
1 mol
 164.6 mols H 2O
18.0 grams
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
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17a–16
Mole Fraction
molesA
Mole fraction ( A ) =
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 solute
Mass (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%
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 g
1 ppm =
1g
1 g 1 g 1 g


1 g 1 ml 1 ml
 1 ppm = 1 μg/mL = 1 mg/L
 1 ppb = 1 ng/mL = 1 μg/L
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17a–21
Molarity
 Concentration
is usually expressed in
terms of molarity:
 Moles

of solute/liters of solution (M)
Moles of solute = molarity x volume of solution
Moles = M x V
Molarity and concentration
 Molarity:
M = moles solute/liter of solution
 Dilution
M1V1 = M2V2
Example
 What
is molarity of 50 ml solution containing
2.355 g H2SO4?




Molar mass H2SO4 = 98.1 g/mol
Moles H2SO4 = .0240 mol (2.355 g/98.1 g/mol)
Volume of solution = 50 mL/1000 mL/L = .050 L
Concentration = moles/volume
= .0240 mol/.050 L = 0.480 M
Solution stoichiometry

How much volume of one solution to react with
another solution




Volume A
Given volume of A with molarity MA
Determine moles A
Determine moles B
Find target volume of B with molarity MB
mol = MV
Moles A
Mole:mole ratio
Moles B
V = mol/M
Volume B