Physical Chemistry Third Edition

6.7 Colligative Properties 295

EXAMPLE6.18

Find the freezing point depression of a solution of 10.00 g of sucrose in 1.000 kg of water.

Solution

m 2 

(

10 .00 g sucrose

)

(

1 mol sucrose

342 .30 g sucrose

)

1 .000 kg

 0 .02921 mol kg−^1

∆T

(

1 .86 K kg mol−^1

)(

0 .02921 mol kg−^1

)

 0 .0543 K

Exercise 6.33

At 25◦C a solution of acetic acid with a stoichiometric molality of 0.100 mol kg−^1 is approxi-

mately 1.32% ionized. Assuming that this percentage ionization applies at the freezing temper-

ature, find the freezing temperature of this solution.

Boiling Point Elevation

Consider a volatile solvent (component 1) and a nonvolatile solute (component 2) in

a solution that is at equilibrium with the gaseous solvent at a constant pressure. We

assume that the gas phase is an ideal gas and that the solvent acts as though it were

ideal. Our development closely parallels the derivation of the freezing point depression

formula earlier in this section. The fundamental fact of phase equilibrium gives

μ∗ 1 (liq)

T

+Rln

(

x(liq) 1

)



μ(gas) 1

T

(6.7-12)

wherex 1 is the mole fraction of the solvent in the liquid phase. Use of Eq. (6.7-4) gives

R

⎛

⎝

ln

(

∂x(liq) 1

)

∂T

⎞

⎠

P

−

Hm,1(gas)

T^2

+

H

∗(liq)

m,1

T 2

−

∆vapHm,1∗

T^2

(6.7-13)

where∆vapHm,1∗ is the molar enthalpy change of vaporization of the pure solvent.

We now multiply bydTand integrate fromTb,1, the normal boiling temperature of

component 1, to a higher temperature,T′. Over a small interval of temperature, the

enthalpy change of vaporization is nearly constant, so that integration gives an equation

analogous to Eq. (6.7-7):

Rln

(

x(

liq)

1

)

∆vapHm,1∗

(

1

T′

−

1

Tb,1

)

(6.7-14)

Equation (6.7-14) can be simplified in the case of small boiling point elevations by

using the same approximations as were used in Eq. (6.7-9b), with the result

x 2 

(

∆vapHm,1∗

RTb,1^2

)

∆Tb (6.7-15)

where∆TbT−Tb.