Physical Chemistry Third Edition

294 6 The Thermodynamics of Solutions

Exercise 6.31

a.Write the next two terms in the Taylor series for ln(1−x 2 ).

b.Evaluate the two terms of part a forx 2  0 .0100.

c.Find the percentage error for the approximation of Eq. (6.7-8) forx 2  0 .100, 0.0100,

0 .00100, and 0.000100.

Use of the approximation of Eq. (6.7-8) gives

x 2 ≈∆fusHm,1∗

Tm,1−T

RTm,1T

where we omit the prime superscript (′)onT. This equation is accurate only for dilute

solutions, in which caseTis approximately equal toTm,1. We replaceTbyTm,1in the

denominator of this expression:

x 2 ≈

(

∆fusHm,1∗

RTm,1^2

)

∆Tf (6.7-9)

where we use the symbol∆TfforTm,1−T, the freezing point depression.

Equation (6.7-9) is often rewritten in terms of the molality, using Eq. (6.2-10) to

relate the molality and the mole fraction for a dilute solution:

x 2 M 1 m 2

whereM 1 is the molar mass of the solvent (measured in kilograms). The result is

∆TfKf,1m 2 (6.7-10)

wherem 2 is the molality of the solute. The quantityKf,1is called thefreezing point

depression constantfor the specific solvent.

Kf,1

M 1 RTm,1^2

∆fusHm,1

(6.7-11)

The freezing point depression constantKf,1has a different value for each solvent, but

is independent of the identity of the solute. If there are several solutesm 2 is replaced

by the sum of the molalities of all solutes. If a solute dissociates or ionizes, the total

molality of all solute species must be used, although it might be necessary to account

for ion pairing. The freezing point depression can be used to determine the extent of

ionization of a weak electrolyte.

Exercise 6.32

The molar enthalpy change of fusion of water is equal to 6.01 kJ mol−^1. Show that the value of

the freezing point depression constant for water is equal to 1.86 K kg mol−^1.