Chapter 11
Properties of
Solutions
QUESTION
The influence of temperature and temperature changes on many
chemical processes and observations is quite significant. However,
of the four common ways of expressing solution concentration, only
one is influenced by temperature. Which of the following
expressions is that?
1.
2.
3.
4.
Mass percent
Molality
Molarity
Mole fraction
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CRS Question, 11–2
ANSWER
Choice 3 is the only one of the four to be affected by temperature
changes. This is because molarity depends on solution volume.
Volume may vary with temperature. The other three expressions
are mass based.
Section 11.1: Solution Composition
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CRS Question, 11–3
QUESTION
Household bleach is an aqueous solution of sodium hypochlorite. If
5.25 g of NaOCl (molar mass = 74.44 g/mol) were placed in 94.75 g
of water, what would you calculate as the molality? The density of
the solution is slightly greater than water. Would the molarity of the
solution be greater, less or the same as the molality?
1.
2.
3.
4.
0.0705 m; M would be greater
0.705 m; M would be the same
0.744 m; M would be greater
0.744 m; M would be less
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CRS Question, 11–4
ANSWER
Choice 4 provides the correct answer to both parts of the question.
The molality involves moles/kg of water, so the given mass of
solute can be converted to moles of solute and then divided by
0.09475 kg to obtain molality. The molarity of the solution will be
greater than the molality because the density of the solution shows
that one milliliter of solution has a mass greater than one. So one
liter of solution will contain more mass of solute than would be
found mixed with one kilogram of water.
Section 11.1: Solution Composition
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CRS Question, 11–5
QUESTION
A certain solution of ethanol and water produces a mole fraction
of ethanol (C2H5OH: molar mass = 46.07 g/mol) of 0.200.
What is the molality of the solution?
1.
2.
3.
4.
13.9 m
0.200 m
1.38 m
14.4 m
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CRS Question, 11–6
ANSWER
Choice 1 provides the correct molality. The mole fraction of
ethanol shows that in the solution, regardless of the size, there
will always be a ratio of 0.200 mole of ethanol per 0.800 mole
of water. For 0.200 mole of ethanol, there would be 0.800
mole of water (or 14.4 grams = 0.0144 kg)
0.200 mole/0.0144 kg = 13.9 m
Section 11.1: Solution Composition
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CRS Question, 11–7
QUESTION
A solution of ammonia (NH3 molar mass = 17.03 g/mol) that you
would likely encounter in a general chemistry lab would have a
concentration of approximately 15 M. If the mole fraction of NH3
in the solution was 0.29, what would you determine as the density
of the solution?
1.
2.
3.
4.
0.26 g/mL
0.92 g/mL
0.39 g/mL
I don’t see how to connect the mole fraction to the molarity and
density.
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CRS Question, 11–8
ANSWER
Choice 2 correctly relates mole fraction and molarity to density.
The molarity states that there are 15 moles of NH3 per liter of
solution. This equals 255 g of NH3 in the liter. To determine the
mass of water in the liter of solution, the mole fraction of NH3 could
be used to determine that there are 0.71 moles of water per 0.29
moles of NH3. That would indicate the presence of 36.7 moles of
water per 15 moles of NH3 in the solution. 36.7 moles of water =
661 g. The total mass of solution in 1.0 L would be 255 + 661 =
916 grams. Therefore, the density would be 916g/1000 mL.
Section 11.1: Solution Composition
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CRS Question, 11–9
QUESTION
Citric acid is found in typical fruit drinks. If the normality of
citric acid were determined to be 0.030 N, what would be reported
as the molarity? Note that citric acid has the potential to furnish
three moles of hydrogen ions per mole.
1.
2.
3.
4.
0.090 M
0.010 M
0.030 M
3.0 M
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CRS Question, 11–10
ANSWER
Choice 2 provides the correct answer for the relationship between
M and N in this acid. An acid that provides three moles of
hydrogen ions per mole will provide three “equivalents” of
hydrogen, so the molarity would be 1/3 of the normality.
Section 11.1: Solution Composition
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CRS Question, 11–11
QUESTION
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CRS Question, 11–12
QUESTION (continued)
A solvent in a beaker that you are holding becomes warmer as a
solute is introduced and begins to dissolve. Which of the
following conclusions could be drawn? Use the figure as a
reference for the following DH comparisons.
1.
2.
3.
4.
DH1 + DH2 must be greater than DH3.
DH3 has a larger value than DH1 but not DH2.
DH3 has a larger value than DH1 + DH2.
The combined total of DH1 + DH2 + DH3 will have to be > 0.
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CRS Question, 11–13
ANSWER
Choice 3 correctly utilizes the figure, with DH references, to point
out that when the solution process causes your hand to warm it
must be that the process in step 3 is providing excess heat beyond
what is needed for steps 1 and 2.
Section 11.2: The Energies of Solution Formation
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CRS Question, 11–14
QUESTION
Many people have had a blood analysis that included a test for
the concentration of cholesterol. In water, cholesterol has a
solubility of only around 0.2 mg/dL. However, in liver bile the
solubility increases to over 380 mg/dL. Which of the following
offers consistent, accurate comments about these observations?
1. Cholesterol is likely to be a polar molecule, and the bile
must also be polar.
2. Cholesterol is likely a non-polar molecule, and the bile
is likely to be polar.
3. Cholesterol is likely non-polar, and the bile is likely to
be non-polar.
4. Cholesterol is likely polar, and the bile is likely to be
non-polar.
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CRS Question, 11–15
ANSWER
Choice 3 relates the most likely evaluation of the polarities.
Water is known to be polar. As cholesterol has only slight
solubility it must be non-polar - in general “like dissolves like.”
The large increase in solubility in the liver bile indicates that
the bile is also likely non-polar.
Section 11.3: Factors Affecting Solubility
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CRS Question, 11–16
QUESTION
A minimum of 1.3  10–4 M O2 must be maintained in freshwater
supplies to sustain aquatic life. In the mountains of Montana, the
partial pressure of O2 may drop to 0.15 atm. What is the water
concentration of O2 there? Henry’s constant for O2 = 1.3  10–3
mol/L-atm. At the lower elevations at the base of those mountains,
would more or less O2 be dissolved in water?
1.
2.
3.
4.
M = 2.0  10–4; more dissolved
M = 8.7  10–4; more dissolved
M = 2.0  10–4; less dissolved
M = 8.7  10–4; less dissolved
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CRS Question, 11–17
ANSWER
Choice 1 provides the correct M and the correct change in
concentration. Henry’s Law relates pressure of a gas over a
solution to the concentration of the gas in the solution: C = k  P.
At lower altitudes, the partial pressure of O2 would be higher, thus
more O2 would dissolve. The huge fishing population of Montana
is very appreciative of Henry’s Law.
Section 11.3: Factors Affecting Solubility
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CRS Question, 11–18
QUESTION
In a warm room the compound beta-pinene could have a vapor
pressure of approximately 97 torr. The vapor of this compound
can be detected from solutions sometimes used in products where
a pine scent is desired. If you had a solution containing 0.10 moles
of beta-pinene and 0.050 moles of a non-volatile compound, what
would be the new vapor pressure of beta-pinene?
1.
2.
3.
4.
9.7 torr
65 torr
194 torr
32 torr
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CRS Question, 11–19
ANSWER
Choice 2 shows the correct value when using Raoult’s Law. First
calculate the mole fraction of beta-pinene and then
P soln = C (P solvent).
Section 11.4: The Vapor Pressures of Solutions
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CRS Question, 11–20
QUESTION
Because of the dissolved salts, ocean water will have a lower
vapor pressure than fresh water. If 5.00 grams of NaCl (molar
mass = 58.44g/mol) were dissolved in 100.0 mL of water, at
25.0°C, what would be the vapor pressure of salt water? The
vapor pressure of water at 25.0°C is 23.76 torr.
1.
2.
3.
4.
23.4 torr
0.360 torr
23.0 torr
2.03 torr
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CRS Question, 11–21
ANSWER
Choice 3 takes into account Raoult’s Law and the dissociation of
an ionic solute. The mole fraction of water must be calculated
using the moles of NaCl  2. Once that is accomplished the
mole fraction of water (0.970) is multiplied by the original vapor
pressure of water to obtain the salt water vapor pressure.
Section 11.4: The Vapor Pressures of Solutions
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CRS Question, 11–22
QUESTION
Mixing some solutes and solvents can lead to non-ideal solutions
where Raoult’s law does not predict the solution vapor pressure
accurately. Which of the following situations is most likely to
produce the listed result?
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CRS Question, 11–23
QUESTION (continued)
1. Two polar liquids when mixed have a good possibility of
producing negative deviations (actual vapor pressure is less
than expected) from Raoult’s Law.
2. Two polar liquids when mixed have a good possibility of
producing positive deviations (actual vapor pressure greater
than expected) from Raoult’s Law.
3. When a non-polar liquid and polar liquid are mixed they
have a good possibility of producing a negative deviation
from Raoult’s law. Mixing polar liquids seldom creates nonideal solutions.
4. I am mixed up myself…none of these seem logical.
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CRS Question, 11–24
ANSWER
Choice 1 correctly lists the likely outcome for mixing the
described liquids. When two polar liquids mix their interaction
could be so strong that the solvent evaporation is inhibited. This
would decrease the vapor pressure below (negative deviation)
that would be calculated using mole fraction.
Section 11.4: The Vapor Pressures of Solutions
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CRS Question, 11–25
QUESTION
Assuming ideal behavior which of the following statements
provides correct comparisons when adding a non-volatile solute
to a solvent?
1. Vapor pressure decreases; freezing point increases; boiling
point increases; osmotic pressure decreases.
2. Vapor pressure decreases; freezing point decreases; boiling
point decreases; osmotic pressure decreases.
3. Vapor pressure increases; freezing point decreases, boiling
point increases; osmotic pressure increases.
4. Vapor pressure decreases, freezing point decreases, boiling
point increases; osmotic pressure increases.
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CRS Question, 11–26
ANSWER
Choice 4 is the only option that provides four correct comparisons.
Adding a (non-volatile) solute lowers the vapor pressure as shown in
Raoult’s Law. The freezing point is depressed with more solute
(lowers the rate at which molecules return from the liquid to the solid
state) while elevating the boiling point (solution must gain more heat
to increase the lowered vapor pressure back up to the atmospheric
pressure to insure boiling). Osmotic pressure is related to solute
molarity, so higher concentrations raise the osmotic pressure.
Section 11.5: Boiling-Point Elevation and Freezing-Point Depression
Section 11.6: Osmotic Pressure
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CRS Question, 11–27
QUESTION
Suppose you want to keep the water in your car cooling system
from freezing during a cold Nebraska winter night. If you added
5.00 kg of ethylene glycol (C2H4(OH)2 mm = 62.07 g/mol) to
5.50 kg of water, what would be the freezing temperature of the
coolant/water mixture in your automobile?
Kf H2O = –1.86°C/kg mol
1.
2.
3.
4.
–0.0367°C
–7.90°C
–14.7°C
–27.3°C
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CRS Question, 11–28
ANSWER
Choice 4 provides the correct, although very cold, answer for that
Nebraska night. The molality of ethylene glycol must be
calculated from the mass (in grams) and molar mass. This is then
multiplied by the freezing point depression constant for the
solvent to obtain the drop in freezing point. Since water normally
freezes at zero degrees Celsius, the change is the actual new
freezing point.
Section 11.5: Boiling-Point Elevation and Freezing-Point
Depression
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CRS Question, 11–29
QUESTION
Most proteins have large molar masses. For example, a type of
hemoglobin may have a molar mass of approximately 68,000
g/mol. If a 0.42 gram sample of hemoglobin was dissolved in
20.0 mL of water (assume no volume change) what would be the
osmotic pressure, at 25.0°C, of the solution?
1.
2.
3.
4.
580 torr
5.7 torr
0.11 torr
0.0096 torr
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CRS Question, 11–30
ANSWER
Choice 2 provides the correct osmotic pressure for the
solution. The molarity of the solution must be determined
(g/mm)/L. This is then multiplied by the gas constant
0.08206 Latm/K mol and the temperature (in K). To convert
this answer from atm to torr multiply by 760.
Section 11.6: Osmotic Pressure
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CRS Question, 11–31
QUESTION
Assuming ideal behavior, which of the following solutions would
have the highest boiling point?
1.
2.
3.
4.
0.010 m NaCl
0.030 m C6H12O6
0.020 m AlCl3
0.010 m MgCl2
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CRS Question, 11–32
ANSWER
Choice 3 provides the dual consideration of molality and van’t
Hoff factor. Assuming ideal behavior, the molality  i (factor
considering ionic dissociation) would be used to determine the
resulting boiling points. Note that C6H12O6 is the only
nonelectrolyte.
Section 11.7: Colligative Properties of Electrolyte Solutions
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CRS Question, 11–33
QUESTION
Of the following statements, which would accurately apply to
colloids?
1. Colloid particles must be smaller than 1 nm.
2. The easiest colloid to form is between two gases.
3. Because of their dispersed nature, colloids create small or
minimal surface areas.
4. The effects of electrostatic repulsion keep colloids dispersed.
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CRS Question, 11–34
ANSWER
Choice 4 provides the only correct statement about colloids. The
molecules or particles can, to some extent, organize a charge
layer around their outer surface. Since this layer would be the
same for the dispersed particles, the layers of each particle now
will repel each other, thus preventing the particles from getting
close enough together to precipitate.
Section 11.8: Colloids
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CRS Question, 11–35