Physical Chemistry Third Edition

296 6 The Thermodynamics of Solutions

When Eq. (6.7-15) is solved for the boiling point elevation and written in terms of

the molality, the result is analogous to Eq. (6.7-10):

∆TbKb,1m 2 (6.7-16)

Theboiling point elevation constantfor substance 1 is given by

Kb,1

M 1 RTb,1^2

∆vapHm,1∗

(6.7-17)

This quantity has a different value for each solvent, but does not depend on the identity

of the solute. If more than one solute is present, the molalitym 2 is replaced by the sum

of the molalities of all solutes.

EXAMPLE6.19

Show that the value of the boiling point elevation constant for water is equal to

0 .513 K kg mol-1. At 100◦C∆vapHm∗ 40 .67 kJ mol−^1.

Solution

Kb,1

(0.01801 kg mol-1)(8.3145 J K-1mol-1)(373.15 K)^2

40670 J mol-1

 0 .513 K kg mol-1

Exercise 6.34

Find the boiling temperature at 1.000 atm of a solution of 5.00 g of glucose in 1.000 kg of water.

Vapor Pressure Lowering

For a nonvolatile solute and a volatile solvent that obeys Raoult’s law, the total vapor

pressure is equal to the vapor pressure of the solvent:

Pvapx 1 P 1 ∗ (6.7-18)

whereP1,vap∗ is the vapor pressure of the pure solvent (component 1) and wherex 1 is

the mole fraction of the solvent in the liquid phase. The lowering of the vapor pressure

is given by

∆PvapP1,vap∗ −PvapP1,vap∗ −x 1 P∗1,vapP1,vap∗ (1−x 1 )P1,vap∗ x 2 (6.7-19)

Exercise 6.35

a.Calculate the vapor pressure at 100.0◦C of the solution in Exercise 6.34.

b.From Eq. (6.7-19), obtain an expression for the vapor pressure lowering of a dilute solution

in terms of the molality.