KEY
Practice Problems: Solutions
CHEM 1A
1. What is the boiling point and freezing point of a solution that is 3.725 m CaCl2 in ethanol?
Solvent
Normal Freezing
Point (ºC)
Normal Boiling
Point (ºC)
ethanol
–117.3
78.5
Tb = i Kb m =
3 mol particles
1 mol CaCl2
Kf
(ºC · kg solvent / mol
solute particles)
1.99
1.22 ºC · kg C2H5OH
mol particles
Kb
(ºC · kg solvent / mol
solute particles)
1.22
3.725 mol CaCl2
1 kg C2H5OH
= 13.6335ºC
BPsoln = BPnormal + Tb = 78.5ºC + 13.6335ºC = 92.1 335ºC
Tf = i Kf m =
3 mol particles
1 mol CaCl2
1.99 ºC · kg C2H5OH
mol particles
3.725 mol CaCl2
1 kg C2H5OH
= 22.23825ºC
FPsoln = FPnormal – Tf = –117.3ºC – 22.23825ºC = –139.5 3825ºC
92.1°C
BPsoln.: _________
–139.5°C
FPsoln.: _________
2. Use the vapor pressure curve for a solution of two volatile liquids below to complete the following
series of statements. (circle one of the underlined choices in each set of choices)

The solution is ideal / non-ideal, and exhibits
positive / negative / no deviation from Raoult’s law.

The intermolecular forces in the solution (attractions
to each other) are stronger / weaker than those in the
pure liquids (attractions to themselves).

Based on the relative strengths of the IMFs described
above, Hsoln would be exothermic / endothermic.

Liquid A is more / less volatile than Liquid B.

The solution with A = 0.6 will have a higher / lower
boiling point than either pure liquid
PºB
PºA
1
0
A
B
3. An aqueous solution of sulfuric acid has a density of 1.67 g/mL and is 75 %( ) H2SO4. How many
moles of H2SO4 are contained in 500. mL of this solution?
500. mL soln.
1.67 g soln.
1 mL soln.
6.4 mol H2SO4
Answer.: _____________
75 g H2SO4
100 g soln
1 mol H2SO4
= 6.38470322 mol H2SO4
98.086 g H2SO4
0
1
4. A solution is prepared by mixing 25 mL of pentane (C5H12, d = 0.63 g/cm3) with 45 mL of hexane
(C6H14, d = 0.66 g/cm3). Assuming that the volumes are additive upon mixing, calculate the
concentration of pentane in the following units:
a)
b)
c)
d)
e)
f)
M
m
%( )
ppm
%( )

a) molarity
25 mL C5H12
1 cm3
1 mL
0.63 g C5H12
1 cm3
1 mol C5H12
= 0.2183073213 mol C5H12
72.146 g C5H12
10–3 L
1 mL
( 25 mL C5H12 + 45 mL C6H14 )
molarity (M) =
b) molality
45 mL C6H14
= 0.070 L soln.
0.2183073213 mol C5H12
mol solute
= 3.118676018 M C5H12
=
0.070 L soln.
L solution
1 cm3
1 mL
0.66 g C6H14
1 cm3
1 kg
103 g
= 0.0297 kg C6H14
from part a)
molality (m) =
0.2183073213 mol C5H12
mol solute
= 7.350414857 m C5H12
=
0.0297 kg C6H14
kg solvent
c) mass %
25 mL C5H12
1 cm3
1 mL
0.63 g C5H12
1 cm3
= 15.75 g C5H12
45 mL C6H14
1 cm3
1 mL
0.66 g C6H14
1 cm3
= 29.7 g C6H14
15.75 g C5H12
g solute
(100%) =
%( ) =
(100%) = 34.65346535 %( ) C5H12
g solution
(15.75 g C5H12 + 29.7 g C6H14)
=
d) ppm
ppm =
from part c)
15.75 g C5H12
g solute
(106 ppm) =
(106 ppm) = 346534.6535 ppm C5H12
g solution
(15.75 g C5H12 + 29.7 g C6H14)
e) volume %
25 mL C5H12
mL solute
(100%) =
(100%) = 35.71428571 %( ) C5H12
%( ) =
mL solution
(25 mL C5H12 + 45 mL C6H14)
=
f) mole fraction
45 mL C6H14
1 cm3
1 mL
0.66 g C6H14
1 cm3
1 mol C6H14
= 0.3446595182 mol C6H14
86.172 g C6H14
from part a)
 =
0.2183073213 mol C5H12
mol solute
= 0.3877800715 ( =  C5H12)
=
(0.2183073213 mol C5H12 + 0.3446595182 mol C6 H14)
mol total
3.1 M C5H12
a) _______________
7.4 m C5H12
b) _______________
35 %( ) C5H12
c) _______________
350,000 ppm C5H12
d) _______________
36 %( ) C5H12
e) _______________
 C5H12 = 0.38
f) _______________