Solutions and Colligative Properties 193
- D—This is a dilution problem. Vbefore=(Mafter)
(Vafter)/(Mbefore)
(6.0 M HNO 3 )(0.500 L)(1000 mL/1 L)/
(16.0 M HNO 3 ) =190 mL
- D—To calculate the molar mass, the mass of the
solute and the moles of the solute are needed.
The molality of the solution may be determined
from the freezing-point depression, and the freezing-
point depression constant (I and II). If the mass
of the solvent is known, the moles of the solute
may be calculated from the molality. These
moles, along with the mass of the solute, can be
used to determine the molar mass.
- A—If the mole fraction of chloroform is 0.20
then the solution has a 0.80 mol fraction of
carbon tetrachloride. The moles of chloroform
and the kilograms of carbon tetrachloride are
needed. If 0.20 mol of chloroform are present, the
number of kilograms of carbon tetrachloride is:
(0.80 mol CCl 4 )(153.8 g CCl 4 /l mol CCl 4 )
(1 kg/1000 g) =0.12 kg
m CHCl 3 =(0.20 mol CHCl 3 )/(0.12 kg) = 1.7 m
- A—To calculate the molarity, the moles of urea
and the volume of the solution are needed.
Density is an intensive property, so any arbitrary
volume of solution may be used. One liter is a
convenient volume. Using this volume and the
density of the solution, you can calculate the mass
of the solution. Ten percent of this is the mass of
urea. The mass of urea and the molecular weight
of urea give the moles of urea.
- A—Freezing-point depression is a colligative
property, which depends on the number of parti-
cles present. The solution with the greatest con-
centration of particles will have the greatest
depression. The concentration of particles in E
(a nonelectrolyte) is 0.10 m. All other answers are
strong electrolytes, and the concentration of par-
ticles in these may be calculated by multiplying
the concentration by the van’t Hoff factor.
A: 3 ×0.10 =0.30 B, C, and D: 2 ×0.10 =0.20
- E—The strong electrolyte with the greatest con-
centration of ions is the best conductor. D is a
weak electrolyte, not a strong electrolyte. The
number of ions for the strong electrolytes may be
found by simply counting the ions: A––2, B––2,
C––2, E––4. The best conductor has the greatest
value when the molarity is multiplied by the
number of ions.
- D—Cooling the solution will change the tem-
perature and the volume of the solution. Volume
is important in the calculation of molarity and
density. A volume change eliminates answers A
and C. The mass and the number of moles are
not affected by the temperature change. Mole
fraction and molality will not change. This elim-
inates B and E.
- E—This is a dilution problem. Vafter=(Mbefore
Vbefore)/(Mafter)
(10.0 M HNO 3 ×50.0 mL)/(4.0 M HNO 3 )
=125 mL
The final volume is 125 mL. Since the original
volume was 50.0 mL, an additional 75.0 mL
must be added.
- E—The two most similar substances will be most
likely to be ideal.
- A—Solutions cannot be separated by titrations or
filtering. Electrolysis of the solution would pro-
duce hydrogen and oxygen gas. Chromatography
might achieve a minimal separation.
- C—The solubility of a gas is increased by increas-
ing the partial pressure of the gas, and by lower-
ing the temperature.
