Concise Physical Chemistry

c12 JWBS043-Rogers September 13, 2010 11:27 Printer Name: Yet to Come

190 SOLUTION CHEMISTRY

dT≈
Tbto restore the system to its original vapor pressure. These small changes

are related by the equation

p

p

=

vapH

R

(

1

T^2

)

Tb

where we have already shown that
p=p◦ 1 X 2 so
p/p=X 2 p◦ 1 /p. For experiments

carried out at atmospheric pressure,p◦ 1 is the vapor pressure at the normal boiling point

of the pure solvent. The difference
p=p−p◦ 1 is thevapor pressure depression.

The temperature rise
Tbnecessary to restorepto its original value is theboiling

point elevation.

In the limit of small amounts of solute, the vapor pressure change from that of the

pure solvent is small,p 1 ◦≈p, so they cancel to a good approximation and

X 2 p◦A

p

≈X 2 ≈

vapH

R

(

1

T^2

)

Tb

or

Tb≈

RTb^2 X 2


vapH 1

Writing outX 2 intermsofgramsandmolarmassofthesolventM 1 (problems) leads

to the boiling point elevation in practical laboratory terms as

Tb≈

RTb^2 M 1


vapH 1 · 1000

m=Kbm

whereKbis called theboiling point constantor sometimes theebullioscopicconstant.

The molality,m, is the number of moles per 1000 g of solvent, and M 1 is the molar

mass of the solvent; that is, we are expressing the amount of solute in units of

molality. Typical values forKbrange from about 0.5 to 5 K kg mol−^1 ; for example,

Kb(water)= 0 .51 K kg mol−^1 andKb(benzene)= 2 .53 K kg mol−^1.

As we have seen,Kb is derived through Gibbs chemical potentials. A simi-

lar treatment leads to thefreezing point depression
Tf and the freezing point

constantKf:

Tf≈

RTf^2 M 1


freezHA· 1000

m=Kfm

where
freezHAis the enthalpy of freezing of the pure solvent at 1 atm. Typical

values areKf(water)= 1 .86 K kg mol−^1 (that famous number we memorized in

elementary chemistry) andKf(benzene)= 5 .07 K kg mol−^1. If we measure the