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COLLIGATIVE PROPERTY
Section III
COLLIGATIVE PROPERTY LAW
It was found that adding a solute to a solvent lowers the freezing point and the vapor pressure but
elevates the boiling point of the pure solvent. The lowering of the freezing point and vapor pressure and the
elevation of the boiling point of the solvent depend on the number of solute particles added and not on the
type of particles added. It was found that adding 1 mole of a non-electrolyte lowers the freezing point of
1000 g of water by 1.86C. 1.86C is known as the molal freezing point constant for water. Each solvent
has its own molal freezing point constant, given a symbol Kf, for example, for benzene Kf is 4.9C. It was
also found that 1 mole of a non-volatile non-electrolyte elevates the boiling point of 1000 g of water by
.52C. .52C is the Kb, known as the molal boiling point constant for water.
If the lowering of the freezing point or the elevation of the boiling point of a solution is known then it is
possible to calculate the molality of the solution using Eq. 7a or 7b.
Eq. 7a
m=
 f . p.
molal f. p. constant
Eq. 7b
m=
 b. p.
molal b. p. constant
For example, the freezing point of a sugar solution was -.93C. Calculate the molality of the solution.
SOLUTION
using Eq. 7a
m=
.93
= .5
186
.
It is possible to use the lowering of the freezing point or the elevation of the boiling point to calculate
the molecular weight of an unknown non-electrolyte by using Eq. 6 and 7a or 6 and 7b.
EXAMPLE 1
10 g of an unknown non-electrolyte dissolved in 150 g of water lowered the f.p. of water 1.2C.
Calculate the molecular weight of the substance.
Given Kf = 1.86
SOLUTION
Using Eq. 7, Calculate the m
m=
12
.
= .64
186
.
50
Substitute the value in Eq. 6
10 g =
(.64)(150g of solvent)(mol. wt.)
1000
Solving the equation for the molecular weight
m.w. =
10  1000
= 104
.64  150
EXAMPLE 2
The molecular weight of non-electrolyte is 58. What will be the boiling point of a solution containing
24 g of the non-electrolyte dissolved in 600 g of water. Given Kb = .52C
SOLUTION
24
1000  24
 .69
Molality of solution is equal to 58 
600
58  600
1000
We know that m =
bp
kb
Therefore, bp = .52  .69 = .35C
Therefore, the boiling point of the solution is 100C + .35C = 100.35C.