Chapter 13
Sec. 2
Solutions and Their Colligative
Properties
Solutions have them but pure
substances do not have them….
Colligative Properties -
Properties of solutions which depend
on the number of solute particles in
the solution and not the nature of the
solute.
• Three Colligative Properties
1) Vapor pressure lowering
2) Boiling point elevation
3) Freezing point depression
II. Units of Concentration
Molarity (M) 
moles solute (mol)
liters solution (L)
mass of solute
% by mass 
100
mass of solution
• Molarity and % mass are not useful in colligative properties
because the exact amount of solvent is unknown.
• Only molality concentration units reflect the number of solute
particles per solvent molecules and are useful with colligative
properties.
Molality (m) 
moles solute (mol)
kilograms of solvent
III. Colligative Properties
Vapor pressure – the pressure above a liquid created by the
liquid releasing gas molecules as it evaporates. The
pressure is directly related to temperature.
A. Vapor Pressure Lowering
•
The vapor pressure of
the solution is lowered
because the solute
particles at the
liquid/vapor boundary
block the solvent particle
from jumping into the
vapor state.
Dynamic Equilibrium

When the rate of freezing is the same as the
rate of melting, the amount of ice and the
amount of water won't change on average
(although there are short-term fluctuations at
the surface of the ice). The ice and water are
said to be in dynamic equilibrium with each
other.
The balance between freezing and melting
processes can easily be upset. If the ice/water
mixture is cooled, the molecules move slower.
The slower-moving molecules are more easily
captured by the ice, and freezing occurs at a
greater rate than melting.

Molecules in Motion Flash plug-in
IV. Colligative Properties
B. Boiling Point Elevation
• For nonvolatile, nonelectrolyte solvents the change in
boiling point (DTbp) is:
DTbp = Kbpmsolute
Kbp = boiling point elevation
constant
msolute = molality of solute
The Boiling point
Curve is shifted to
the right. Raising
the boiling Pt.
IV. Colligative Properties
C. Freezing Point Depression
• For nonvolatile, nonelectrolyte solvents the change in
freezing point (DTfp) is:
DTfp = Kfpmsolute
Kfp = freezing point depression
constant
msolute = molality of solute
The Freezing point
curve shifts to the
left. Lowering the
F.P. of the solvent.
Practice Problems
1. Determine the freezing point of a solution of 60.0
g of glucose, C6H12O6, dissolved in 80.0 g of
water.
DT fp  K fp m
m= mol/kg
m = 60.0g / 180g/mol / 0.080kg = 4.167m
∆Tfp = (-1.86°C/m) (4.167 m)
ans: -7.75°C
Practice Problems
2. What is the boiling point of a solution of 645 g of
urea, CON2H4, dissolved in 980. g of water?
.
DTbp  K bp m
m= mol/kg
m = 645g / 60g/mol / .980 kg = 10.97m
∆Tbp = (0.512°C/m) (10.97 m)
∆Tbp = 5.62°C
BP = 100° + 5.62°C = 105.6°C
IV. Colligative Properties
D. Colligative Properties of Solutions Containing Ions
• The change in VP, BP or FP is greater than expected for
electrolyte (ionic salt) solutions.
Predicted BP elevation of an aqueous 0.100 m NaCl solution
DTbp = Kbp • msolute
DTbp, calculated = (0.512 °C/m)(0.100 m) = 0.0512 °C
Actual BP elevation of an aqueous 0.100 m NaCl solution
DTbp, measured = 0.09470 °C
(Almost double the DTbp calculated)
• Colligative properties depend on the total number of solute
particles in solution. Ionic compounds form ions in solution
so the total number of solute particles in solution is equal to
the total ions in solution.
IV. Colligative Properties
D. Colligative Properties of Solutions Containing Ions
NaCl(s)  Na+(aq) + Cl-(aq)
0.100 m
0.100 m
0.100 m
0.200 m total
THE MULTIPLIER -van’t Hoff factor (i)
We will assume 100% ionization
So for electrolytes or ionic solutions:
DTbp  i K bp m
DT fp  i K fp m
IV. Colligative Properties
D. Colligative Properties of Solutions Containing Ions
Predicting van’t Hoff factors
NaCl(s)  Na+(aq) + Cl-(aq)
1 particle + 1 particle = 2 particles
ipredicted = 2
Na2SO4(s)  2 Na+(aq) + SO42-(aq)
2 particles
ipredicted = 3
+ 1 particle = 3 particles
Practice Problems
3. What is the expected boiling point of a brine
solution containing 30.00 g of KBr dissolved in
100.00 g of water?
DTbp  iK bp m
m= mol/kg
KBr  two ions , i = 2
m = 30.00g / 119.0g/mol / .100 kg = 2.52 m
∆Tbp = 2 (0.512°C/m) (2.52 m) = 2.6°C
B.P. of water = 100.0°C + 2.6°C
ans: 102.6°C
Practice Problems
4. What is the expected boiling point of a CaCl2
solution containing 385 g of CaCl2 dissolved in
1.230 kg of water?
DTbp  iK bp m
m= mol/kg
CaCl2  three ions , i = 3
m = 385 g / 111.1 g/mol / 1.230 kg = 2.82 m
∆Tbp = 3 (0.512°C/m) (2.82 m) = 4.3°C
B.P. of water = 100.0°C + 4.3°C
ans: 104.3°C
IV. Colligative Properties
E. Colligative Properties and Molar Mass Determination
• Colligative properties can be used to determine the molar
mass of a solute when it is dissolved in a solvent of
appreciable vapor pressure and a known Kbp or Kfp.
To Measure a
change in VP,
BP elevation or
FP depression.
Solution
Conc.
Use mass
of solvent
Moles of
Solute
g solute
mol solute
Molar
Mass
Example: Butylated hydroxyanisole (BHA) is used as an antioxidant
in margarine and other fats and oils; it prevents oxidation
and prolongs the shelf life of food. What is the molar mass
of BHA if 0.640 g of the compound, dissolved in 25.0 g of
chloroform, produces a solution whose boiling point is
62.22 °C. (Chloroform BP = 61.70 °C, Kbp = 3.63 °C/m)
Practice Problems
5. A solution of 0.827 g of an unknown nonelectrolyte compound in 2.500 g of water has a
freezing point of -10.18°C. Calculate the molar
mass of the compound.
DT fp  K fp m
Molar mass = g/mol solve for moles thru m
-10.18 = (-1.86°C/m) (m) ; m = 5.47 mol/kg
m = 5.47 mol/kg = mol/0.0025kg mol =.01367
Molar mass = 0.827g /.01367 mol
ans: 60.5 g/mol