Physical Chemistry , 1st ed.

in temperature of the equilibrium melting or freezing process. Equation 7.44
becomes

xsolute

R

T

fu

2

M

sH

P

 Tf (7.45)

The relationship between molality and mole fraction is simple. IfMsolventis the
molecular weight of the solvent, then the molality of the solution is

msolute

xs

1

ol

0

ve

0

n

0

t



M

xso

so

lu

lv

te

ent

 (7.46)

The 1000 in the numerator of equation 7.46 represents a conversion from
grams to kilograms, so there is an implicit g/kg unit on it. Remember that the
mole fraction of the solvent is close to 1, so we further approximate by substi-
tuting 1 for xsolvent. We then rearrange equation 7.46 in terms ofxsolute, substi-
tute into equation 7.45, and then rearrange the equation to get an expression
for Tf, the amount that the freezing point is depressed. We get

TfM

1

s

0

o

0

lv

0

en



t

R

fu

T

sH

(^2) MP
msolute (7.47)
All of the terms relating to properties of the solvent have been grouped inside
parentheses, and the only term relating to the solute is its molal concentration.
Notice that all of the terms inside the parentheses are a constant for any par-
ticular solvent: its molecular weight Msolvent, its melting point TMP, and its heat
of fusion (^) fusH. (1000 and Rare also constants.) Therefore, this collection of
constants represents a constant value for any solvent. Equation 7.47 is more
commonly written as
TfKfmsolute (7.48)
where Kfis called the freezing point depression constantfor the solvent. It is also
called the cryoscopic constantfor the solvent.
Example 7.14
Calculate the cryoscopic constant for cyclohexane, C 6 H 12 , given that its heat
of fusion is 2630 J/mol and its melting point is 6.6°C. What are the units for
the constant?
Solution
The molecular weight of cyclohexane is 84.16 g/mol. The melting point,
which must be expressed in absolute temperature, is 6.6 273.15 279.8 K.
Comparing equations 7.47 and 7.48, we see that the expression for Kfis
Kf

M

10

so

0

lv

0

en

tR

fu

T

sH

(^2) MP

Substituting for the variables:
Kf
Working out the units, everything cancels but Kkg/mol
Kf20.83 

K

m

k

o

g

l



These units seem unusual until one remembers that the unit molality is

defined in terms of mol/kg. Since the above unit has the reciprocal of this

(84.16 mgol)(8.314 moJlK)(279.8 K)^2



1000 kgg 2630 mJol

7.8 Colligative Properties 195