Chemistry Reference
In-Depth Information
in it does not make any difference in the value. (However, for ionic solutes, the
total molality of cations
and
anions must be considered.)
EXAMPLE 15.19
Calculate the freezing point of a solution containing 0.135 mol of a nonionic
solute in 2.50 kg of benzene.
Solution
The freezing-point depression is proportional to the molality. The molality is
0.135 mol
2.50 kg
m
0.0540 m
The value of
k
f
is found in Table 15.2. Then,
¢
t
f
k
f
m
(5.12°C/m) (0.0540 m)
0.276
5
°C
The freezing point of the solution is equal to the freezing point of the solvent,
pure benzene, minus the freezing-point depression
(
¢
t
f
):
t
solution
t
solvent
¢
t
f
5.5°C
0.276
5
°C
5.2°C
Be sure to differentiate between the freezing point and the freezing-point depres-
sion, as well as between the freezing point of the solvent and the freezing point
of the solution.
Practice Problem 15.19
Calculate the freezing point of a 0.250 m
solution of benzene,
C
6
H
6
,
in naphthalene.
The molar mass of a solute may be determined from the number of grams
of solute dissolved in a given mass of solvent and the freezing-point depression
of the solution. The freezing-point depression yields the number of moles of
solute per kilogram of solvent, and the mass data yield the number of grams per
kilogram of solvent. Dividing the number of grams per kilogram by the number
of moles per kilogram yields the number of grams per mole—the molar mass.
EXAMPLE 15.20
A solution containing 1.66 g of nonionic solute in 171 g of water freezes at
Calculate the molar mass of the solute.
0.602°C.
Solution
The molality of the solution is determined from the freezing-point depression and
the value of
k
f
for water. The freezing-point depression for this solution is
0.602°C
¢
t
k
f
0.602°C
1.86°C/m
m
0.323
7
m
