CHAPTER 12
Solutions
CHAPTER TERMS AND DEFINITIONS
Numbers in parentheses after definitions give the text sections in which the terms are explained. Starred
terms are italicized in the text. Where a term does not fall directly under a text section heading,
additional information is given for you to locate it.
solution* homogeneous mixture of two or more substances, consisting of ions or molecules (12.1,
introductory section)
solute
solvent
gas or solid dissolved in a liquid or the solution component in smaller amount (12.1)
liquid that dissolves a gas or solid or the solution component in greater amount (12.1)
miscible fluids
immiscible*
fluids that mix with or dissolve in each other in all proportions (12.1)
said of two fluids that do not mix but form two layers (12.1)
dynamic equilibrium* equilibrium in which the forward and reverse processes, such as the
dissolving and depositing of crystals, are always occurring and at the same rate (12.2)
saturated solution
solution that is in equilibrium with respect to a given dissolved substance (12.2)
solubility amount that dissolves in a given quantity of liquid at a given temperature to give a
saturated solution (12.2)
unsaturated solution
solution not in equilibrium with respect to a given dissolved substance and in
which more of the substance can dissolve (12.2)
supersaturated solution
(12.2)
entropy*
solution that contains more dissolved substance than a saturated solution
measure of disorder (12.2, marginal note)
ion–dipole force*
attraction between an ion and a polar molecule (12.2)
hydration (of ions)
lattice energy*
heat of solution*
attraction of ions for water molecules (12.2)
energy holding ions together in a crystal lattice (12.2)
heat that is released or absorbed when a substance dissolves in a solvent (12.3)
Le Châtelier’s principle when a system in equilibrium is altered by a change of temperature,
pressure, or concentration variable, the system shifts in equilibrium composition in a way that tends to
counteract this change of variable (12.3)
Henry’s law the solubility of a gas is directly proportional to the partial pressure of the gas above
the solution; S = kHP (12.3)
colligative properties
properties of solutions that depend on the concentration of solute particles
(molecules or ions) in solution but not on the chemical identity of the solute (12.4, introductory section)
concentration*
molarity (M)*
amount of solute dissolved in a given quantity of solvent or solution (12.4)
number of moles of solute per liter of solution (12.4)
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Chapter 12: Solutions
mass percentage of solute
molality (m)
261
percentage by mass of solute contained in a solution (12.4)
number of moles of solute per kilogram of solvent (12.4)
mole fraction (X)
(12.4)
mole percent*
number of moles of a substance divided by the total number of moles of solution
mole fraction times 100 (12.4)
vapor-pressure lowering colligative property of a solution; equal to the vapor pressure of the pure
solvent minus the vapor pressure of the solution (12.5)
Raoult’s law the partial pressure of solvent PA over a solution is equal to the vapor pressure of the
pure solvent PAo multiplied by the mole fraction of solvent in the solution: PA = PAo XA (12.5)
ideal solution*
fractions (12.5)
solution in which both substances follow Raoult’s law for all values of mole
normal (boiling point of a liquid)*
atm (12.6)
temperature at which the vapor pressure of the liquid equals 1
boiling-point elevation (Tb)
colligative property of a solution; equal to the boiling point of the
solution minus the boiling point of the pure solvent (12.6)
boiling-point-elevation constant (Kb)*
and the molality of a solution (12.6)
proportionality constant between the boiling-point elevation
freezing-point depression (Tf)
colligative property of a solution; equal to the freezing point of the
pure solvent minus the freezing point of the solution (12.6)
freezing-point-depression constant (Kf)* proportionality constant between the freezing-point
lowering and the molality of a solution (12.6)
semipermeable*
describes a membrane that allows solvent molecules but not solute molecules to
pass through (12.7)
osmosis phenomenon of solvent flow through a semipermeable membrane from lower solute
concentration to higher concentration to equalize concentrations on both sides of the membrane (12.7)
osmotic pressure
colligative property of a solution; equal to the pressure that, when applied to the
solution, just stops osmosis (12.7)
reverse osmosis* reversing the osmosis process by applying greater pressure to the more
concentrated solution so that solvent flows from the concentrated solution through a membrane to the
more dilute solution (12.7)
desalinate*
activities*
to remove salts from seawater to make it usable for drinking or industrial uses (12.7)
effective concentrations of ions in solution (12.8)
Debye–Hückel theory* describes the distribution of ions in a salt solution and enables us to
calculate ionic activities (12.8)
colloid dispersion of particles of one substance (the dispersed phase) throughout another substance
or solution (the continuous phase) (12.9)
Tyndall effect
aerosols
emulsion
scattering of light by colloidal-sized particles (12.9)
colloidal dispersions of liquid droplets or solid particles throughout a gas (12.9)
colloidal dispersion of liquid droplets throughout another liquid (12.9)
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262
sol
Chapter 12: Solutions
colloidal dispersion of solid particles in a liquid (12.9)
hydrophilic colloid colloid in which there is a strong attraction between the dispersed phase and the
continuous phase (water) (12.9)
hydrophobic colloid colloid in which there is a lack of attraction between the dispersed phase and
the continuous phase (water) (12.9)
nuclei (in crystallization)*
centers about which crystallization occur (12.9)
coagulation
process by which the dispersed phase of a colloid is made to aggregate and thereby
separate from the continuous phase (12.9)
micelle colloidal-sized particle formed in water by the association of molecules or ions that each
have a hydrophobic and a hydrophilic end (12.9)
association colloid
colloid in which the dispersed phase consists of micelles (12.9)
CHAPTER DIAGNOSTIC TEST
1.
When Solid A dissolves in water, there occurs a corresponding decrease in temperature.
Therefore, heating a mixture of Solid A and water should _______________________the
solubility of A in water.
(decrease/increase)
2.
The graph in Figure A plots solubility as a function of temperature.
Figure A
Which of the following statements concerning information available from the solubility curves in
Figure A is(are) incorrect?
a.
At all indicated temperatures, KBr is more soluble in water than is KCl.
b.
The solubility of KBr can be increased by warming the solution.
c.
At 35C, the order of solubilities is KBr > KNO3 > KCl > Ce2(SO4)3.
d.
The solubility of Ce2(SO4)3 is an endothermic process.
e.
The solubility of KNO3 is more affected by temperature than that of KCl.
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Chapter 12: Solutions
3.
4.
263
The energy of hydration increases with decreasing cation size, when cationic charge is constant,
because
a.
a larger cation can coordinate more water molecules about it.
b.
cationic charge has no effect on the coordination of water molecules.
c.
electrostatic attractive forces are inversely proportional to cation size.
d.
the cation size has no effect on hydration when the cationic charge changes.
e.
cations are attracted to water molecules and anions are repelled by water molecules.
For the following chemical reaction at equilibrium,
H3PO4 + KOH
KH2PO4 + H2O;
ΔH is negative
the addition of water to the reaction mixture will
5.
6.
a.
produce a larger ΔH per mole of H3PO4.
b.
increase the amount of KH2PO4 when the new equilibrium is established.
c.
have no effect.
d.
increase the amount of H3PO4 and KOH when equilibrium is reestablished.
e.
have no effect other than giving a more dilute solution of KH2PO4.
Which of the following carbonates would you expect to be insoluble in water?
a.
Na2CO3
b.
H2CO3
c.
(NH4)2CO3
d.
Rb2CO3
e.
BaCO3
A solution of 132.4 g Cu(NO3)2 per liter has a density of 1.116 g/mL. The mass percent of
Cu(NO3)2 in the solution is
a.
11.86.
b.
8.40.
c.
9.00.
d.
22.4.
e.
Insufficient data are given to calculate mass percent.
7.
Calculate the mole fraction of each component in a solution that contains 46.5 g of ethylene
glycol, CH2OHCH2OH, and 236 g of methanol, CH3OH.
8.
Calculate the molecular formula of a species that has the empirical formula CH3O if, when 25.1 g
of the compound was added to 0.150 kg of water, the solution froze at 5.0C. Kf = 1.86C/m.
9.
Which of the following should have the highest boiling point?
a.
Pure water
b.
A 0.2 m Ca(NO3)2 aqueous solution
c.
A 0.2 m (CH3)2CO aqueous solution
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264
Chapter 12: Solutions
d.
A 0.1 m KBr aqueous solution
e.
A 0.1 m Ba(NO3)2 aqueous solution
10. The graph in Figure B shows the effect of the addition of a nonvolatile solute on the vapor
pressure of water.
Figure B
Which of the following statements is a logical deduction from this graph?
a.
If a nonvolatile solute is added to water, the boiling point of the solution will be Tb.
b.
The addition of a volatile solute will change the boiling point of the water.
c.
Atmospheric pressure will affect the composition of the aqueous solution.
d.
If a nonvolatile solute is added to water, the freezing point of the solution will be lower than
that of water.
e.
If a nonvolatile solute is added to water, the boiling and melting points of the solution will
be greater than those of the water by Tb and Tf , respectively.
11. A liter of water at 25C and 1.0 atm dissolves 1.45 g of carbon dioxide. If the partial pressure of
CO2 is increased to 15 atm, what is its solubility in water?
12. An 8.0 M solution of KCl in water has a density of 1.218 g/mL. The density of pure water is 1.000
g/mL. Calculate the mole fraction of KCl in the solution.
13. Calculate the boiling point of a glucose solution consisting of 9.0 g of glucose dissolved in 100.0
g of water (glucose = C6H12O6, 180.1 g/mol, Kb for water is 0.512C/m).
14. A solution consists of 0.100 mol of naphthalene, C10H8, and 9.90 mol of benzene, C6H6, at 25C.
Calculate the vapor pressure and vapor-pressure lowering of the solution. (The vapor pressure of
pure benzene at 25C is 2.70  102 mmHg. Assume that naphthalene is a nonvolatile,
nonelectrolytic solute.)
15. Calculate the osmotic pressure of a solution that consists of 4.68 g of hemoglobin (molar mass =
6.83  104 g/mol) in 125 mL of water (molar mass = 18.0 g/mol) at 3.00  102 K.
16. Classify each of the following colloids as an aerosol, foam, emulsion, sol, or gel.
a.
Cigarette smoke
b.
Strawberry-flavored jello
c.
Muddy water
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Chapter 12: Solutions
d.
Milk
e.
Carbon ink
f.
Pearl
g.
Raised dough
265
ANSWERS TO CHAPTER DIAGNOSTIC TEST
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
Increase (12.2, 12.3)
2.
d (12.2, 12.3)
3.
c (12.2)
4.
d (12.3)
5.
e (12.2)
6.
a (12.3)
7.
Xglycol = 9.23  102; X CH3OH = 9.08  101 (12.4, PS Sk. 2)
8.
Molecular formula: C2H6O2 (12.6, PS Sk. 5, 6)
9.
b (12.6, 12.8)
10. d (12.5, 12.6)
11. 22 g CO2/L H2O (12.3, PS Sk. 1)
12. Mole fraction KCl = 0.19 (12.4, PS Sk. 3)
13. b.p. = 100.26C (12.4, 12.6, PS Sk. 2, 5)
14. Vapor pressure = 267 mmHg; vapor-pressure lowering = 2.70 mmHg (12.5, PS Sk. 4)
15. 0.0135 atm (12.7, PS Sk. 7)
16.
a.
Aerosol
b.
Gel
c.
Sol
d.
Emulsion
e.
Sol
f.
Gel
g.
Foam (12.9)
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266
Chapter 12: Solutions
SUMMARY OF CHAPTER TOPICS
12.1 Types of Solutions
Learning Objectives

Define solute and solvent.

Define miscible fluid.

Provide examples of gaseous solutions, liquid solutions, and solid solutions.
Whether we call a mixture of one or more substances in another a suspension, a colloid, or a solution
depends on the size of the particles, particularly of those we are adding. To illustrate, assume that we
have water as the solvent. If the particles we add are large and heavy, they will sink to the bottom. If we
stir the mixture vigorously, we will have a suspension, wherein the added particles are momentarily
suspended in the continuous phase, the water. If the particles are small and light, they may stay
suspended; light passing through this suspension, or colloid, will be scattered. If the particles are the
size of molecules and ions, they are so small that we cannot discern them unless they color the solution,
as does the Cu2+ ion in CuSO4.
The definitions of the key terms in this chapter are of particular importance. Memorize them as soon as
possible. Your ability to do the exercises and problems will depend on knowing them.
Exercise 12.1
Give an example of a solid solution prepared from a liquid and a solid.
Solution: The dental-filling alloy mentioned in the text has liquid mercury in silver (and other
metals), which gives a solid solution.
12.2 Solubility and the Solution Process
Learning Objectives

List the conditions that must be present to have a saturated solution, to have an unsaturated
solution, and to have a supersaturated solution.

Describe the factors that make one substance soluble in another.

Determine when a molecular solution will form when substances are mixed.

Learn what conditions must be met in order to create an ionic solution.
The solubility of one substance in another at a given temperature and pressure is a function of two
factors: the attraction the substance has for its own species and the attraction the substance has for the
other species. When we dissolve a solid in water, we can measure these attractions in terms of lattice
energy and energy of hydration, respectively.
Exercise 12.2
Which of the following compounds is likely to be more soluble in water: C4H9OH or C4H9SH? Explain.
Known: The molecules differ only in S and O; oxygen can hydrogen bond, whereas sulfur cannot.
Solution: C4H9OH is the more soluble because it can hydrogen bond with water.
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Chapter 12: Solutions
267
Exercise 12.3
Which ion has the larger hydration energy, Na+ or K+?
Known: Hydration energy is inversely proportional to ionic radius and proportional to charge;
K+ has the same charge as Na+ and is below it on the periodic table, so it is larger.
Solution: Na+, being smaller, has the larger hydration energy.
12.3 Effects of Temperature and Pressure on Solubility
Learning Objectives

State the general trends of the solubility of gases and solids with temperature.

Explain how the solubility of a gas changes with temperature.

Apply Henry’s law. (Example 12.1)
Problem-Solving Skill
1.
Applying Henry’s law. Given the solubility of a gas at one pressure, find its solubility at another
pressure (Example 12.1).
According to the principle of Le Châtelier (pronounced “la shaught lee ay”), if a system in a state of
equilibrium is disturbed by a change of some variable, the system shifts in equilibrium composition in a
way that tends to counter that change of variable. This statement is an observation. It is not an
explanation of what happens on a molecular or ionic level. A state of equilibrium exists when the rates
of both the forward and reverse reactions are equal. On the molecular level, when a system at
equilibrium is disturbed, the rate of one of the reactions is increased. Thus we say the equilibrium is
shifted. The system will again reach equilibrium if left undisturbed, but the composition of the mixture
will not be the same. It appears as if the system has tried to negate the change.
Exercise 12.4
A liter of water at 25C dissolves 0.0404 g O2 when the partial pressure of the oxygen is 1.00 atm.
What is the solubility of oxygen from air, in which the partial pressure of O2 is 159 mmHg?
Wanted: solubility (S2) in water of O2 from air
Given:
t = 25C; S1 = 0.0404 g/L; P1 = 1 atm = 760 mmHg; P2 = 159 mmHg
Known:
Henry’s law: the ratio of solubilities is proportional to the ratio of pressures:
S2
P
= 2
S1
P1
Solution: Solving for S2 and substituting, we get
S2 =
P2 S 1
159 mmHg  0.0404 g
=
= 0.00845 g/L
P1
760 mmHg L
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268
Chapter 12: Solutions
12.4 Ways of Expressing Concentration
Learning Objectives

Define colligative property.

Define molarity.

Define mass percentage of solute.

Calculate mass percentage of solute. (Example 12.2)

Define molality.

Calculate the molality of solute. (Example 12.3)

Define mole fraction.

Calculate the mole fraction of components. (Example 12.4)

Convert molality to mole fractions. (Example 12.5)

Convert mole fractions to molality. (Example 12.6)

Convert molality to molarity. (Example 12.7)

Convert molarity to molality. (Example 12.8)
Problem-Solving Skills
2.
Calculating solution concentration. Given the mass percent of solute, state how to prepare a
given mass of solution (Example 12.2). Given the masses of solute and solvent, find the molality
(Example 12.3) and mole fractions (Example 12.4).
3.
Converting concentration units. Given the molality of a solution, calculate the mole fractions of
solute and solvent, and given the mole fractions, calculate the molality (Examples 12.5 and 12.6).
Given the density, calculate the molarity from the molality, and vice versa (Examples 12.7 and
12.8).
This and the remaining sections of the chapter deal with colligative properties of solutions, which are
properties that depend on the number of particles in a given amount of solution or solvent rather than on
the identity of the particles. The word colligative comes from the same Latin root that the word
collection comes from—“to gather together.” If you think of collection when you see colligative, it will
help you to remember these properties. Memorize the four colligative properties of solutions. They are
(1) vapor pressure; the two properties related to vapor pressure, that is, (2) boiling-point elevation and
(3) freezing-point depression; and (4) osmotic pressure.
Exercise 12.5
An experiment calls for 35.0 g of hydrochloric acid that is 20.2% HCl by mass. How many grams of
HCl is this? How many grams of water?
Wanted: g HCl, g H2O
Given:
Need 35.0 g of solution; solution is 20.2% HCl by mass.
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Chapter 12: Solutions
Known:
269
20.2 mass% means
20.2 g HCl
 100
100 g solution
Solution: Find g HCl by dimensional analysis. Begin with what you can hold in a container—the
35.0 g of solution:
35.0 g solution 
20.2 g HCl
= 7.07 g HCl
100 g solution
Find g H2O by subtracting g HCl from 35.0 g of solution:
35.0 g solution  7.07 g HCl = 27.9 g H2O
Exercise 12.6
Toluene, C6H5CH3, is a liquid compound similar to benzene, C6H6. It is the starting material for other
substances, including trinitrotoluene (TNT). Find the molality of toluene in a solution that contains 35.6
g of toluene and 125 g of benzene.
Wanted: molality (m) of toluene solution
Given:
Solution contains 35.6 g of toluene, 125 g of benzene; formulas are given.
Known:
Definition of molality =
moles solute
.
kg solvent
Toluene, C6H5CH3 = 92.1 g/mol.
Solution: Convert g toluene to moles and g benzene to kg:
35.6 g
= 0.3865 mol toluene
92.1 g/mol
125 g benzene = 0.125 kg
m=
0.3865 mol
= 3.09 m C6H5CH3
0.125 kg
Exercise 12.7
Calculate the mole fractions of toluene and benzene in the solution described in Exercise 12.6.
Wanted: Xtoluene, Xbenzene
Given:
0.3865 mol of toluene (calculated in Exercise 12.6); 125 g of benzene; formula of
benzene is C6H6.
Known:
Mole fraction =
mol substance
; C6H6 = 78.1 g/mol
mol total
Solution: Calculate moles benzene:
125 g benzene
= 1.601 mol benzene
78.1 g/mol
Calculate total moles:
0.3865 mol toluene + 1.601 mol benzene = 1.988 total mol
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270
Chapter 12: Solutions
Then
Mole fraction toluene =
0.3865 mol
= 0.194
1.988 mol
Mole fraction benzene =
1.601 mol
= 0.805
1.988 mol
Exercise 12.8
A solution is 0.120 m methanol dissolved in ethanol. Calculate the mole fractions of methanol, CH3OH,
and ethanol, C2H5OH, in the solution.
Known:
m=
mol methanol
, so, since the solution is 0.120 m, moles of methanol in 1 kg of
kg ethanol
solvent = 0.120 mol; ethanol, C2H5OH = 46.1 g/mol.
Solution: Assume 1 kg solvent and calculate moles ethanol:
1000 g
= 21.69 mol ethanol
46.1 g/mol
Calculate total moles:
0.120 mol methanol + 21.69 mol ethanol = 21.81 total mol
Then,
Mole fraction of methanol =
Mole fraction of ethanol =
0.120 mol
= 5.50  10–3
21.81 mol
21.69 mol
= 0.994
21.81 mol
Exercise 12.9
A solution is 0.250 mole fraction methanol, CH3OH, and 0.750 mole fraction ethanol, C2H5OH. What is
the molality of methanol in the solution?
Known:
m=
mol methanol
; C2H5OH = 46.1 g/mol
kg ethanol
Solution: Assume 1 mol of solution, and find the kilograms of ethanol. Since the mole fraction of
ethanol is 0.750, in 1 mol of solution there would be 0.750 mol of ethanol. The mass of this
ethanol is
0.750 mol  46.1 g/mol = 34.58 g ethanol = 0.03458 kg
m=
0.250 mol
= 7.23 m CH3OH
0.03458 kg
Exercise 12.10
Urea, (NH2)2CO, is used as a fertilizer. What is the molar concentration of an aqueous solution that is
3.42 m urea? The density of the solution is 1.045 g/mL.
Wanted: molarity (M)
Given:
Molality (m) is 3.42; solution density is 1.045 g/mL.
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Chapter 12: Solutions
Known:
m,
Definitions; urea = 60.0 g/mol. We must convert
3.42 mol urea
mol urea
, to M,
L solution
1 kg water
Solution: First, find the mass of a solution that contains 3.42 mol urea and 1 kg water:
The mass of urea is
3.42 mol urea  60.0 g/mol = 205.2 g urea
The total mass of solution is
1000 g + 205.2 g = 1205.2 g of solution
The volume of solution is its mass divided by its density:
1205.2 g
= 1153.3 mL = 1.1533 L
1.045 g/mL
Thus
M=
3.42mol
= 2.97 mol/L = 2.97 M
1.1533L
Exercise 12.11
An aqueous solution is 2.00 M urea. The density of the solution is 1.029 g/mL. What is the molal
concentration of urea in the solution?
Known:
M,
We must convert
mol urea
2.00 mol urea
to m,
; urea = 60.0 g/mol
L solution
kg water
Solution: Find the mass of water in 1 L of a 2.00 M solution:
Total mass of solution = 1000 mL  1.029 g/mL = 1029 g
Mass of solute = 2.00 mol urea  60.0 g/mol = 120.0 g urea
Mass of solvent = 1029 g soln – 120.0 g urea = 909 g water
= 0.909 kg water
Thus
m=
2.00 mol
= 2.20 m
0.909 kg
12.5 Vapor Pressure of a Solution
Learning Objectives

Explain vapor–pressure lowering of a solvent.

State Raoult’s law.

Calculate vapor-pressure lowering. (Example 12.9)

Describe an ideal solution.
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271
272
Chapter 12: Solutions
Problem-Solving Skill
4.
Calculating vapor-pressure lowering. Given the mole fraction of solute in a solution of
nonvolatile, undissociated solute and the vapor pressure of pure solvent, calculate the vaporpressure lowering and vapor pressure of the solution (Example 12.9).
Recall that the vapor pressure of a solution is the pressure of the vapor above the liquid when
equilibrium exists—when the rate of vaporization equals the rate of condensation. We consider cases
where only the solvent is appreciably volatile, so in the discussion of colligative properties we will be
comparing the vapor pressure of the pure solvent with its vapor pressure in solution.
Memorize Raoult’s law, PA = PAo XA, which states that the vapor pressure of the solvent in solution
equals the vapor pressure of the pure solvent times its mole fraction in the solution.
Exercise 12.12
Naphthalene, C10H8, is used to make mothballs. Suppose that a solution is made by dissolving 0.515 g
of naphthalene in 60.8 g of chloroform, CHCl3. Calculate the vapor-pressure lowering of chloroform at
20C from the naphthalene. The vapor pressure of chloroform at 20C is 156 mmHg. Naphthalene can
be assumed to be nonvolatile compared with chloroform. What is the vapor pressure of the solution?
Wanted:
PCHCl3
Given:
o
t = 20C; 0.515 g naphthalene, C10H8; 60.8 g chloroform, CHCl3; PCHCl
= 156 mmHg
3
Known:
o
o
 PCHCl3 = PCHCl
∙ X C10H8 ;  PCHCl3 = PCHCl
 PCHCl3 ; naphthalene = 128.1 g/mol
3
3
Solution: Calculate the mole fraction X of naphthalene by first calculating the moles of each
compound.
0.515 g C10 H 8
= 0.004020 mol C10H8
128.1g/mol
60.8 g CHCl 3
= 0.5088 mol CHCl3
119.5 g/mol
X C10H8 =
0.0040 20 mol
0.0040 20
=
= 0.007839
(0.0040 20  0.50 87) mol
0.51 28
Putting values in the expression for vapor-pressure lowering gives
o
 PCHCl3 = PCHCl
∙ X C10H8 = 156 mmHg  0.007839 = 1.223 mmHg
3
Assuming that naphthalene is nonvolatile, the vapor pressure of the solution is the vapor pressure
of the chloroform in solution:
o
 Δ PCHCl3 = (156  1.223) mmHg = 155 mmHg
PCHCl3 = PCHCl
3
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Chapter 12: Solutions
273
12.6 Boiling-Point Elevation and Freezing-Point Depression
Learning Objectives

Define boiling–point elevation and freezing-point depression.

Calculate boiling-point elevation and freezing-point depression. (Example 12.10)

Calculate the molecular mass of a solute from molality. (Example 12.11)

Calculate the molecular mass from freezing-point depression. (Example 12.12)
Problem-Solving Skills
5.
Calculating boiling-point elevation and freezing-point depression. Given the molality of a
solution of nonvolatile, undissociated solute, calculate the boiling-point elevation and freezingpoint depression (Example 12.10).
6.
Calculating molecular mass. Given the masses of solvent and solute and the molality of the
solution, find the molecular mass of the solute (Example 12.11). Given the masses of solvent and
solute, the freezing-point depression, and Kf, find the molecular mass of the solute (Example
12.12).
Boiling-point elevation is not too difficult a concept to grasp. Since the vapor pressure of a solvent is
decreased when a solute is added, we have to heat the solution to a higher temperature to make the
vapor pressure equal the atmospheric pressure and thus to make the solution boil.
Freezing-point depression is a more difficult concept. You may have to read your text and the following
explanation a few times. It also will be a good idea to take another look at phase diagrams, text Section
11.3, to be sure you understand all the information there.
The freezing point of a substance is the temperature at which the liquid and solid are in dynamic
equilibrium; melting and freezing occur at the same rate. At this temperature, the vapor pressures of the
liquid and solid phases are the same. If we decrease the vapor pressure of the liquid by adding solute
particles, the liquid in solution will have a lower vapor pressure than the solid at the freezing point of
the pure liquid, and freezing will not occur. As we continue to lower the temperature, the vapor
pressure of the solid decreases more rapidly than that of the liquid in solution, and we reach a
temperature at which both solid and liquid in solution have the same vapor pressure. (See Figure B in
the Chapter Diagnostic Test.) At this lower temperature, then, the solution will freeze.
Memorize the formulas for boiling-point elevation and freezing-point depression. Do not worry about
memorizing the constants (they are provided in text Table 12.3) unless your instructor asks you to do
so.
Exercise 12.13
How many grams of ethylene glycol, CH2OHCH2OH, must be added to 37.8 g of water to give a
freezing point of 0.150C?
Wanted: g ethylene glycol
Given:
formula; 37.8 g water; Tf = 0.150C.
Known:
Tf = Kf,c/m; Kf = 1.858C/m for water; m =
ethylene glycol = 62.1 g/mol
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moles solute
;
kg solvent
274
Chapter 12: Solutions
Solution: Rearrange Tf = Kf ,c/m to solve for molality:
m=
ΔTf
0.150 C
=
= 0.08073 m
1.858 C/ m
Kf
Solve for grams of ethylene glycol (e. gly.):
37.8 g H2O 
62.1 g e. gly.
0.08073 mol e. gly

= 0.190 g e. gly.
mol e. gly.
1000 g H 2 O
Exercise 12.14
A 0.930-g sample of ascorbic acid (vitamin C) was dissolved in 95.0 g of water. The concentration of
ascorbic acid, as determined by freezing-point depression, was 0.0555 m. What is the molecular mass of
ascorbic acid?
Wanted: molecular mass of ascorbic acid (amu)
Given:
0.930 g ascorbic acid; 95.0 g = 0.0950 kg H2O; m = 0.0555m.
Known:
m=
moles ascorbic acid
; molar mass = g/mol = same number as for molecular mass
kg H 2 O
Solution: Find moles of ascorbic acid using a rearranged definition of molality:
Moles = m  kg H2O =
0.0555 mol
 0.0950 kg H2O = 0.005272 mol
kg H 2 O
Thus
Molar mass =
0.930 g asc.acid
= 176 g/mol
0.005272 mol
Molecular mass = 176 amu
Exercise 12.15
A 0.205-g sample of white phosphorus was dissolved in 25.0 g of carbon disulfide, CS2. The boilingpoint elevation of the carbon disulfide solution was found to be 0.159C. What is the molecular mass of
the phosphorus in solution? What is the formula of molecular phosphorus?
Wanted: MW of P (amu); formula of molecular P
Given:
0.205 g P; 25.0 g CS2 (0.0250 kg); Tb = 0.159C.
Known: Tb = Kb ∙ m ; definition of molality; Kb of CS2 = 2.40C/m (text Table 12.3). To get
formula of elemental molecular substance, divide molecular mass by atomic mass.
Solution: Find molality from Tb formula:
m=
 Tb
0.159 C
=
= 0.06625 m
2.40 C/ m
Kb
Find moles P from the definition of molality:
mol P = m  kg solvent = 0.06625 mol/kg CS2  0.0250 kg CS2
= 0.001656 mol P
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Chapter 12: Solutions
275
Thus
MW =
0.205 g P
= 124 g/mol
0.0016 56 mol P
Atomic mass of P is 31 amu. Find the molecular formula
Molecular weight
124
=
= 4.0
31
Atomic weight
Formula is P4.
12.7 Osmosis
Learning Objectives

Describe a system where osmosis will take place.

Calculate osmotic pressure. (Example 12.13)
Problem-Solving Skill
7.
Calculating osmotic pressure. Given the molarity and the temperature of a solution, calculate the
osmotic pressure (Example 12.13).
Exercise 12.16
Calculate the osmotic pressure at 20C of an aqueous solution containing 5.0 g of sucrose, C12H22O11, in
100.0 mL of solution.
Wanted: osmotic pressure π
Given:
5.0 g C12H22O11; 20.0C (293 K); 100.0 mL solution
Known:
π = MRT; M =
mol solute
; C12H22O11 = 342.3 g/mol
L solution
Solution: Find moles of sucrose:
5. 0 g
= 0.0146 mol C12H22O11
342 .3 g/mol
The solution concentration is
M=
0.0146 mol
= 0.146 M
0.100 L
Find osmotic pressure:
π = MRT =
0.146 mol  0.0821 L • atm  293 K
(L)(K • mol)
= 3.5 atm
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276
Chapter 12: Solutions
12.8 Colligative Properties of Ionic Solutions
Learning Objective

Determine the colligative properties of ionic solutions. (Example 12.14)
Problem Solving Skill
8.
Determining colligative properties of ionic solutions. Given the concentration of ionic
compound in a solution, calculate the magnitude of a colligative property; if i is not given, assume
the value based on the formula of the ionic compound (Example 12.14).
Exercise 12.17
Estimate the boiling point of a 0.050 m aqueous MgCl2 solution. Assume a value of i based on the
formula.
Wanted: Tb (C)
Given:
0.050 m aqueous MgCl2 solution
Known: Tb = Tb pure solvent + Tb solution; Tb = iKb ∙ m; Kb for water = 0.521 C/m. MgCl2 is
ionic, and each formula unit produces 3 ions, so i = 3.
MgCl2(s) 
 Mg2+(aq) + 2Cl–(aq) (3 ions)
Solution: Tb = 3  0.521C/m  0.050 m = 0.0782C
Tb = Tb pure solvent + Tb solution
= 100C + 0.0782C = 100.078C
12.9 Colloids
Learning Objectives

Define colloid.

Explain the Tyndall effect.

Give examples of hydrophilic colloids and hydrophobic colloids.

Describe coagulation.

Explain how micelles can form an association colloid.
Exercise 12.18
Colloidal sulfur particles are negatively charged with thiosulfate ions, S2O32–, and other ions on the
surface of the sulfur. Indicate which of the following would be most effective in coagulating colloidal
sulfur: NaCl, MgCl2, or AlCl3.
Known: Coagulation is caused by charge neutralization; the more compact the layer of charge,
the more effective is the coagulation; the higher the ionic charge, the more compact is the layer.
Solution: The positive ions will cause the coagulation as they interact with S2O32. AlCl3 would
be most effective because it has the most highly charged cation, Al3+.
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Chapter 12: Solutions
277
ADDITIONAL PROBLEMS
1.
List the letters of the following compounds in order of their increasing solubility in water. Explain
your answer.
a.
C5H11OH, 1-pentanol
b.
C5H12, pentane
c.
C5H11SH, pentanethiol
2.
The solubility of carbon dioxide gas, CO2, in water at 20C and 5.00 atm pressure is 8.05 g/L.
What mass of CO2 would dissolve in exactly 250 mL of water at 1 atm pressure?
3.
A 35.5% by mass hydrochloric acid solution has a density of 1.18 g/mL. What are the molality,
mole fraction, and molarity of HCl in the solution?
4.
State how you would prepare each of the following solutions.
a.
255 g of an aqueous solution containing 5.00% by mass BaCl2, starting with BaCl2 • 2H2O
b.
755 g of a 1.40 m solution of naphthalene, C10H8, in benzene, C6H6
c.
455 g of a solution of glycerin, C3H5(OH)3, in water in which the mole fraction of glycerin is
0.250
5.
The vapor pressure of benzene at 25C is 57.4 mmHg. Calculate the vapor-pressure lowering
when 160.0 g of naphthalene (MW = 128.19 g/mol) is added to 1.50  103 g of benzene (MW =
78.12 g/mol) at 25C.
6.
A solution of 2.24  101 g of an organic compound of unknown molecular mass in 1.664 g of
camphor gave a freezing point of 176.44C. The freezing point of pure camphor is 179.80C, and
the freezing-point-depression constant Kf is 39.7C/m. Calculate the molecular mass of the
unknown compound.
7.
Methanol and ethanol form a nearly ideal solution over a wide concentration range. What is the
vapor pressure of a 33.4% by weight solution of methanol in ethanol at 40.0C? The vapor
pressures of the pure alcohols at 40.0C are
o
o
= 236 mmHg Pethanol
= 119 mmHg
Pmethanol
8.
The freezing point of a 0.0766 m H3PO4 solution is observed to be 0.218C. Does the H3PO4
dissociate primarily to H2PO4–, HPO42–, or PO43–? The freezing point of pure water is 0C, and the
freezing-point-depression constant Kf is 1.86C/m.
9.
An aqueous solution of urea, a molecular compound, has a freezing point of 0.0631C. Calculate
the osmotic pressure of the solution at body temperature (37.0C).
10. Calcium chloride is used to salt icy roads in many communities in winter. Sodium chloride is also
used and is much cheaper. Compare the freezing-point lowering of a ton of each of the salts. If
you were the city manager, what other factors would you consider in choosing which to use?
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278
Chapter 12: Solutions
ANSWERS TO ADDITIONAL PROBLEMS
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
b, c, a. The more similar the attractions between the molecules of a substance are to the attractions
between water molecules, the more they can interact with water, and the more soluble the
compound is. Water molecules are bound by strong hydrogen bonds. Pentane is bound by only
London forces and pentanethiol by dipole–dipole interactions, but 1-pentanol forms hydrogen
bonds. (12.2)
2.
0.402 g (12.3, PS Sk. 1)
3.
Molality: 15.1 m; mole fraction HCl: 0.214; molarity: 11.5 M (12.4, PS Sk. 3)
4.
a.
Dissolve 15.0 g BaCl2 ∙ 2H2O in 240.0 g of water.
b.
Dissolve 115 g of naphthalene in exactly 640 g of benzene.
c.
Add 287 g of glycerin to 168 g of water. (12.4, PS Sk. 2)
5.
3.50 mmHg (12.5, PS Sk. 4)
6.
1590 g/mol (12.6, PS Sk. 6)
7.
168 mmHg (12.5)
8.
i = 1.53; dissociates primarily to H2PO4– (12.6, 12.8)
9.
0.864 atm (12.7, PS Sk. 7)
10. A ton of calcium chloride, CaCl2, is theoretically only ~79% as effective as a ton of sodium
chloride, NaCl. (The theoretical value of i for CaCl2 is 3 and for NaCl is 2, but the formula mass
of NaCl is only 53% of that of CaCl2.) Other factors would include supply, storage, and
transportation costs, as well as effect on the road surface and on the environment. (12.8)
CHAPTER POST-TEST
1.
From a molecular viewpoint, briefly discuss the meaning of the phrase “like dissolves like.”
2.
Which of the following statements is incorrect with respect to the reaction
Sn(s) + 2Cl2(g)
3.
SnCl4(l)
ΔH = 130 kcal
a.
Heating the reaction will decrease the yield of SnCl4.
b.
Increasing the pressure will decrease the yield of SnCl4.
c.
Lowering the temperature of the reaction will shift the equilibrium to the right.
d.
Light should have little effect on the overall Keq.
For each pair of ions below, choose the ion that should have the larger hydration energy.
a.
Ca2+, Ba2+
b.
Na+, Sr2+
c.
Al3+, Mg2+
d.
Fe3+, Fe2+
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Chapter 12: Solutions
4.
279
List the following compounds in order of decreasing solubility in C2H5OH:
ClCH2CH2CH2Cl,
H2O,
(CH3)3CCH2Cl,
[CH3(CH2)16CH2]3N,
Cs2SO4.
5.
Calculate the minimum mass of NaCl that would have to be added to 1.200  103 g H2O so that
the resulting solution would not freeze outside on a cold day (10.0F). (Kf = 1.858C/m.)
6.
The solubility of K2Pt(NO2)4 is endothermic and is measured to be 38.0 g per liter of water at
15C. If 14.6 g of the compound is dissolved in 412 mL of water after considerable stirring and
heating and remains dissolved after the solution is cooled to 15C, which of the following
statements is a logical conclusion?
7.
a.
K2Pt(NO2)4 is relatively insoluble in water.
b.
The final solution is supersaturated.
c.
Heating the K2Pt(NO2)4 solution decreases the solute solubility.
d.
The final solution is not saturated.
e.
A change in temperature has no effect on solute solubility.
Which of the following is generally true for solutions that obey Raoult’s law?
a.
The vapor phase is richer in the more volatile component than is the liquid phase.
b.
The component richer in the liquid phase will be the first to distill from the solution.
c.
The mole fraction of the most volatile component will be less than that of the less volatile
components.
d.
The components in the solutions are usually not chemically similar.
e.
All of the above are generally correct.
8.
What is the molality of a sulfuric acid solution that contains 30.0% by weight H2SO4 and has a
density of 1.21 g/mL?
9.
Calculate the quantities of water and acetone, C3H6O, necessary for the preparation of 150.0 g of
solution that is 2.00 m in acetone.
10. Indicate whether each of the following statements is true or false. If a statement is false, change it
so that it is true.
a.
The spontaneous flow of solute from high to low concentration is known as osmosis.
True/False: ________________________________________________________________
__________________________________________________________________________
b.
The addition of a nonvolatile solute to a solvent produces a vapor-pressure lowering of the
solvent. True/False:__________________________________________________________
__________________________________________________________________________
c.
The molality of a solution can be calculated if only the percentage composition by weight
and the identity of the solute are known. True/False: _______________________________
__________________________________________________________________________
d.
The specific gravity of a solution is a colligative property of the solution. True/False:
__________________________________________________________________________
__________________________________________________________________________
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280
Chapter 12: Solutions
e.
When NaCl is sprinkled on grass, the grass withers and dies. This occurs because of the
poisonous nature of NaCl. True/False: ___________________________________________
__________________________________________________________________________
f.
The freezing-point depression of a 0.25 m aqueous solution of Na2CO3 will be three times
that of a 0.25 m CH3OH aqueous solution. True/False: ______________________________
__________________________________________________________________________
g.
Raoult’s law relates the vapor pressure of solvent to the mole fraction of solvent in the
solution. True/False: _________________________________________________________
__________________________________________________________________________
11. A solution is made of two volatile components, toluene and benzene. At 60C, the vapor pressure
of benzene, C6H6, is 396 mmHg and the vapor pressure of toluene, C7H8, is 1.40  102 mmHg.
The solution is made of equimolar amounts of these two substances. Calculate the total vapor
pressure above the solution. (Assume that Raoult’s law is valid for volatile solutes.)
12. Calculate the osmotic pressure of a solution that consists of 43.5 g of an enzyme with a molar
mass of 9.80  104 g/mol dissolved in benzene to make 1.500 L of solution at 30.0C. The density
of benzene is 0.879 g/mL, and Kb = 2.61C/m. Pure benzene boils at 80.2C.
13. A liter of water dissolves 0.0045 g of oxygen at 25C and 1.0 atm. If the partial pressure of O2 is
increased to 5.6 atm, what is its solubility in water?
14. A 4.28-g sample of a nonionizing solute is dissolved in exactly 250 mL of water (molar mass =
18.02 g/mol) at 25C. The molality of the solution is 0.121. The vapor pressure of water at 25C is
23.8 mmHg. Calculate (a) the molecular mass of the solute, (b) the mole fraction of the solute, (c)
the vapor-pressure lowering due to the solute, and (d) the vapor pressure of the solution.
ANSWERS TO CHAPTER POST-TEST
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
Solutes and solvents that have similar forces of interaction, functional groups, and/or structural
features tend to be completely miscible. (12.2)
2.
b (12.3)
3.
a.
Ca2+
b.
Sr2+
c.
Al3+
d.
Fe3+ (12.2)
4.
H2O > ClCH2CH2CH2Cl > (CH3)3CCH2Cl > [CH3(CH2)16CH2]3N > Cs2SO4 (12.2)
5.
440 g NaCl (12.4, 12.6, 12.8, PS Sk. 2, 5)
6.
d (12.2, 12.3)
7.
a (12.5)
8.
4.37 mole H2SO4/1 kg solvent, or 4.37 m (12.4, PS Sk. 2)
9.
134.4 g of water, 15.6 g of acetone (12.4, PS Sk. 2)
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Chapter 12: Solutions
281
10.
a.
False. The spontaneous flow of solvent from low to high concentration of solute is known as
osmosis. (12.7)
b.
True. (12.5)
c.
True. (12.4)
d.
False. The specific gravity of a solution is not a colligative property of the solution because
it is not a property that depends on the number of particles of substance. (12.4)
e.
False. This occurs because of the phenomenon of osmotic pressure. (12.7)
f.
True. (12.6, 12.8)
g.
True. (12.5)
11. 268 mmHg (12.5)
12. π = 7.36  103 atm (12.7, PS Sk. 7)
13. 0.025 g O2/L (12.3, PS Sk. 1)
14.
a.
141 g/mol (12.4, 12.6, PS Sk. 6)
b.
0.00218 (12.4, PS Sk. 3)
c.
0.0519 mmHg (12.5, PS Sk. 4)
d.
23.7 mmHg (12.5, PS Sk. 4)
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