CK-12-Chemistry Intermediate

16.3. Colligative Properties http://www.ck12.org

FIGURE 16.11

The vapor pressure of a solution (blue) is

lower than the vapor pressure of a pure

solvent (purple). As a result, the freezing

point of a solvent decreases when any

solute is dissolved into it.

with temperature. As a liquid is heated, its molecules move slightly farther apart from one another, and the volume
slightly increases.

The magnitude of the freezing point depression is directly proportional to the molality of the solution. The equation
is:

∆Tf= Kf×m

The proportionality constant, Kf, is called themolal freezing-point depression constant.It is a constant that is
equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute. For water, the
value of Kfis−1.86 °C/m. Consequently, the freezing temperature of a 1-molal aqueous solution of any nonvolatile
molecular solute is−1.86°C. Every solvent has a unique molal freezing-point depression constant. A few of these
are listed below (Table16.3), along with a related value for the boiling point called Kb, which will be discussed in
the next section.

TABLE16.3: Molal Freezing-Point and Boiling-Point Constants

Solvent Normal freezing

point (°C)

Molal freezing-

point depression

constant, Kf(°C/m)

Normal boiling

point (°C)

Molal boiling-point

elevation constant,

Kb(°C/m)

Acetic acid 16.6 −3.90 117.9 3.07

Camphor 178.8 −39.7 207.4 5.61

Naphthalene 80.2 −6.94 217.7 5.80

Phenol 40.9 −7.40 181.8 3.60

Water 0.00 −1.86 100.00 0.512

Sample Problem 16.6 shows how the freezing point of a solution can be calculated.

Sample Problem 16.6: Freezing Point Depression in a Solution of a Nonelectrolyte

Ethylene glycol (C 2 H 6 O 2 ) is a molecular compound that is used in many commercial anti-freezes. An aqueous
solution of ethylene glycol is used in vehicle radiators to prevent the water in the radiator from freezing by lowering
its freezing point. Calculate the freezing point of a solution of 400. g of ethylene glycol in 500. g of water.

Step 1: List the known quantities and plan the problem.

Known

• mass of C 2 H 6 O 2 = 400. g


• molar mass of C 2 H 6 O 2 = 62.08 g/mol


• mass of H 2 O = 500. g = 0.500 kg


• Kf(H 2 O) =−1.86 °C/m

Unknown

• Tfof solution =? °C

This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the
solution. Finally, calculate the freezing point depression.

Step 2: Solve.

400. g C 2 H 6 O 2 ×

1 mol C 2 H 6 O 2

62 .08 g C 2 H 6 O 2

= 6 .44 mol C 2 H 6 O 2

490