51
COLLIGATIVE PROPERTY
Section IV
1.
60 g of a non-electrolyte dissolved in 500 g of water lowered the freezing point by 2.04C.
Calculate the molecular weight of the non-electrolyte. Given Kf 1.86C.
2.
6 g of a non-electrolyte were dissolved in 60 g of water. The boiling point of the solution is
100.52. Calculate the molecular weight of the compound. Given Kb .52C.
3.
The molecular weight of a non-electrolyte is 60. Calculate the freezing point of a solution
containing 30 g of the non-electrolyte dissolved in 400 g of water. Kf 1.86.
4.
6 g of a non-electrolyte dissolved in 400 g of benzene. The freezing point of the solution was 5C.
Calculate the molecular weight of the non-electrolyte. Given the freezing point of pure benzene
5.6C and the Kf for benzene is 4.9C.
Answers
1.
109 g
2.
100 g
3.
-2.33C
4.
122.5 g
52
MOLE FRACTION
Section V
The mole fraction of a component in a solution is the number of moles of that composed divided by the
total number of moles. For a system compound of two components, A and B, the mole fraction of A is given
in Equation 8.
Eq. 8
mole fraction A =
moles A
moles of A + moles of B
Example 1
Calculate the mole fraction of a 4 molal solution of NaOH.
Solution
4 molal solution of NaOH contains 4 moles of NaOH in 1000 g of water.
moles of water =
1000
= 55.5
18
Substitute the values in Equation 8.
m.f. of NaOH =
4
4

 .066
4  555
.
59.5
It is important to note that for a binary system the mole fraction of solvent + mole fraction of solute =
1. Therefore, the mole fraction of water is 1 - .066 = .934.
At this point turn off the tape recorder and work the problem in the space provided, then work all
problems in Section VI.
53
MOLE FRACTION
Section VI
1.
10 g of NaOH were dissolved in 50 g of water. Calculate the mole fraction of the solute and the
solvent.
2.
16 g of CH3OH and 23 g of C2H5OH were added to 500 g of water. Calculate the mole fraction of
CH3OH and of C2H5OH.
3.
9.8 g of H2SO4 were dissolved in 500 g of water to give 490 ml of solution. (assume 100%
ionization) Calculate:
(a)
the freezing point of the solution.
(b)
the normality of the solution.
(c)
the molarity of the solution.
(d)
the molality of the solution.
(e)
the mole fraction of H2SO4.
1.
.08 m.f. NaOH
2.
.017 m.f. CH3OH
.017 m.f. C2H5OH
3.
a.
-1.12 C
b.
.41 N
c.
.205 M
d.
.2 m
e.
.1
or .0036 m.f. H2SO4
27.88