Solutions

Colligative Property
 property that depends on the
concentration of solute particles, not
their identity

Freezing Point Depression (tf)


f.p. of a solution is lower than f.p. of the pure solvent
Boiling Point Elevation (tb)

b.p. of a solution is higher than b.p. of the pure
solvent
Freezing Point Depression
View Flash animation.
Boiling Point Elevation
Solute particles weaken IMF in the solvent.

Applications
 salting icy roads
 making ice cream
 antifreeze
 cars (-64°C to 136°C)
 fish & insects
t = k · m · n
t:change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
m: molality (m)
n: # of particles

# of Particles

Nonelectrolytes (covalent)
 remain intact when dissolved
 1 particle

Electrolytes (ionic)
 dissociate into ions when dissolved
 2 or more particles

At what temperature will a solution that is composed of
0.73 moles of glucose in 225 g of phenol boil?
GIVEN:
WORK:
b.p. = ?
m = 0.73mol ÷ 0.225kg
tb = ?
tb = (3.60°C·kg/mol)(3.2m)(1)
kb = 3.60°C·kg/mol
tb = 12°C
m = 3.2m
b.p. = 181.8°C + 12°C
n=1
b.p. = 194°C
tb = kb · m · n

Find the freezing point of a saturated solution of
NaCl containing 28 g NaCl in 100. mL water.
GIVEN:
f.p. = ?
tf = ?
kf = 1.86°C·kg/mol
WORK:
m = 0.48mol ÷ 0.100kg
tf = (1.86°C·kg/mol)(4.8m)(2)
m = 4.8m
n=2
tf = kf · m · n
f.p. = 0.00°C - 18°C
tf = 18°C
f.p. = -18°C
Percent Solutions

If both solute & solvent are liquids
Must be the same
unit: mL or L
Percent by volume (% v/v) = volume of solute × 100%
solution volume
If a solid is dissolved in a liquid
Percent (mass/volume) (%(m/v)) = mass of solute (g)

× 100%
solution volume (mL)
Must be this unit
Example 1
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?
% (v/v) = volume of solute × 100%
Volume of solute = 85 mL
volume of solution
Volume of solution = 250 mL
% (v/v) = 85 mL ethanol × 100%
250 mL solution
= 34% ethanol
Example 2
How many grams of glucose (C6H12O6) would
you need to prepare 2.0 L of 2.8% glucose
(m/v) solution?
Solution volume = 2.0 L → change to mL
Percent by mass = 2.8%
Percent (mass/volume) (%(m/v) = mass of solute (g) × 100%
solution volume (mL)
2.0L 1000mL= 2,000 mL
1L
2.8% = mass of solute (g) × 100%
2,000 mL
100%
100%
0.028 =
X
X = 56 g of solute
2,000 mL
Percent Solution Problems
You do not have to write the problem. You MUST show your work.
1. What is the concentration, in percent (m/v), of a
solution with 75g K2SO4 in 1500mL of solution?
2. A bottle of hydrogen peroxide antiseptic is labeled
3.0% (v/v). How many mL H2O2 are in a 400.0 mL
bottle of this solution?
3. Calculate the grams of solute required to make 250
mL of 0.10% MgSO4 (m/v).