CHM 2046 Module 8II Chapter 14
Sample Multiple Choice Test#1 Answers
Based on Kotz 5th Edition
(pages
references to 5th Edition)
1. The
correct answer is:
moles of solute per kilogram of
solvent.
Explanation:
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Solution Reference:
Page 643, Molality
1. The molality of a solution is
|
a. |
moles of solute per liter of solution. |
|
b. |
moles of solute per mole of solvent. |
|
c. |
moles of solute per liter of solvent. |
|
d. |
moles of solute per kilogram of solvent. |
|
e. |
moles of solute per kilogram of solution. |
2. The
correct answer is:
1.00.
Explanation:
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Solution Reference:
Page 643, Molality
2. If 0.100 mole of naphthalene is dissolved in 100. g of
benzene (C6H6), the molality is
|
a. |
0.100. |
|
b. |
0.900. |
|
c. |
1.00. |
|
d. |
1.28. |
|
e. |
12.8. |
3. The
correct answer is:
0.49
Explanation:

Solution Reference:
Page 643 Mole Fraction
3. What is the mole fraction methanol, CH3OH, in
an ethanol, C2H5OH, solution that is 60.0% ethanol by
mass?
|
a. |
0.40 |
|
b. |
0.46 |
|
c. |
0.49 |
|
d. |
0.54 |
|
e. |
0.60 |
4. The
correct answer is:
0.500 m.
Explanation:

Solution Reference:
Page 643, Molality
4. The water of hydration becomes part of the solvent. In
terms of the anhydrous salt (Al2(SO4)3), the
solution has the concentration
|
a. |
0.500 m. |
|
b. |
0.600 m. |
|
c. |
0.500 M. |
|
d. |
0.600 M. |
|
e. |
0.400 m. |
5. The
correct answer is:
8.38%
Explanation:
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5. If
32.0 g of NaCl is dissolved in 350. g of H2O, what is the percent by
weight (by mass) of NaCl in the solution?
|
a. |
0.0838% |
|
b. |
0.0914% |
|
c. |
10.7% |
|
d. |
9.14% |
|
e. |
8.38% |
6. The
correct answer is:
1.3.
Explanation:
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Solution Reference:
Page 644, Parts Per Million
6. A
water sample contains 0.35 mg of arsenic in 275 g of water. The concentration
of arsenic in ppm is
|
a. |
1.3. |
|
b. |
2.3. |
|
c. |
3.5. |
|
d. |
7.6. |
|
e. |
23. |
7. The
correct answer is:
0.167
Explanation:

Solution Reference:
Page 643, Molality
7.
Calculate the mole fraction of urea (CO(NH2)2) that
results from mixing 12.0 g of urea with 32.0 g of methanol (CH3OH).
|
a. |
0.125 |
|
b. |
0.167 |
|
c. |
0.375 |
|
d. |
0.625 |
|
e. |
0.833 |
8. The
correct answer is:
dissolve in each other.
Explanation:
They mix with each other in all proportions.
Solution Reference:
Page 646, Liquids Dissolving in Liquids
8.
Miscible liquids
|
a. |
are polar. |
|
b. |
are nonpolar. |
|
c. |
are stable. |
|
d. |
are unstable. |
|
e. |
dissolve in each other. |
9. The
correct answer is:
3.56 M
Explanation:

Solution Reference:
Page 642, Molarity and Page 644, Weight
Percent
9.
What is the molarity of 20.0% nitric acid solution, HNO3(aq) if the
density of the solution is 1.12 g/cm3?
|
a. |
3.56 M |
|
b. |
3.89 M |
|
c. |
4.21 M |
|
d. |
4.48 M |
|
e. |
4.84 M |
10. The
correct answer is:
the solubility of the solid will
increase with increasing temperature.
Explanation:
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From a consideration of LeChatelier's principle, if we exert a stress
(increasing the temperature) on the left hand side of the equation by adding
heat, it will shift to the right in order to remove the stress.
Solution Reference:
Page 649, Heat of Solution and Page 655,
LeChatelier's Principle
10.
If the dissolution of an ionic solid in water in endothermic, then it can be
concluded that
|
a. |
the solubility of the solid will
increase with increasing temperature. |
|
b. |
DH is negative. |
|
c. |
the solution process releases
heat. |
|
d. |
the hydration energy exceeds the
lattice energy. |
|
e. |
the solubility of the solid is
independent of the temperature. |
11. The
correct answer is:
1 only
Explanation:
Both CCl4 and C6H6
are nonpolar molecules.
Solution Reference:
Page 646, Liquids Dissolving in Liquids
11.
Which of the following pairs of liquids are miscible with each other?
|
a. |
1 only |
|
b. |
2 only |
|
c. |
3 only |
|
d. |
1 and 2 only |
|
e. |
1, 2, and 3 |
12. The
correct answer is:
atmospheric pressure.
Explanation:
Henry's law is S(g) = kPg where k
is the constant at a given temperature for the gas and P is the partial
pressure of the gas. The atmospheric pressure does not enter into this
relationship. The constant k depends upon both nature of the gas and of the
solvent.
Solution Reference:
Page 653, Henry's Law
12.
All of the following affect the solubility of gases in solvents EXCEPT
|
a. |
nature of the gas. |
|
b. |
nature of the solvent. |
|
c. |
pressure of the gas. |
|
d. |
temperature. |
|
e. |
atmospheric pressure. |
13. The
correct answer is:
decrease.
Explanation:
S(g) = kP(g)
Solution Reference:
Page 653, Henry's Law
13. If the pressure of a gas over a liquid decreases, the amount of gas dissolved in the liquid will
|
a. |
increase. |
|
b. |
decrease. |
|
c. |
remain the same. |
|
d. |
depend on the polarity of the solvent. |
|
e. |
depend on the polarity of the gas. |
14. The
correct answer is:
solute solubility.
Explanation:
The solubility of a solute depends upon the nature of
the solute. Every solute has its own characteristic solubility. A colligative
property is independent of the kind of solute.
Solution Reference:
Page 656, Colligative Properties
14. All of the following are colligative properties of a solution EXCEPT
|
a. |
freezing point depression. |
|
b. |
boiling point elevation. |
|
c. |
vapor pressure lowering. |
|
d. |
osmotic pressure. |
|
e. |
solute solubility. |
15. The
correct answer is:
mole fraction of solvent.
Explanation:
Psolvent = Xsolvent P°solvent
Since a solution consists of a solute and a solvent and the solute is nonvolatile, then the vapor pressure of the solution is simply the vapor pressure of the solvent. The change in vapor pressure, DPsolvent = Xsolute P°solvent is directly proportional to the mole fraction of solute.
15. The vapor pressure of a solution containing a nonvolatile solute is directly proportional to the
|
a. |
mole fraction of solvent. |
|
b. |
mole fraction of solute. |
|
c. |
molality of the solvent. |
|
d. |
molality of the solute. |
|
e. |
osmotic pressure of the solute. |
16. The
correct answer is:
0.091
Explanation:

Solution Reference:
Page 643, Mole Fraction
16. Consider
that an ideal solution is formed from a mixture of the nonvolatile solute, urea
(CO(NH2)2), and methanol (CH3OH). The vapor
pressure of pure methanol at 20°C is 89 mm Hg. Calculate the mole fraction of
urea that is obtained from mixing 6.0 g of urea with 32.0 g of methanol.
|
a. |
0.091 |
|
b. |
0.10 |
|
c. |
0.16 |
|
d. |
0.84 |
|
e. |
0.91 |
17. The
correct answer is:
8.9 mm Hg
Explanation:
Psolution = Xsolvent P°solvent
Psolution = (1.00 = 0.091)(89 mm Hg) = 81 mm Hg
Solution Reference:
Page 656, Raoult's Law
17.
Consider that an ideal solution is formed from a mixture of the nonvolatile
solute, urea (CO(NH2)2,), and methanol (CH3OH).
The vapor pressure of pure methanol at 20?C is 89 mm Hg. Calculate the vapor
pressure of the methanol solution in problem No. 16.
|
a. |
8.1 mm Hg |
|
b. |
8.9 mm Hg |
|
c. |
14 mm Hg |
|
d. |
75 mm Hg |
|
e. |
81 mm Hg |
18. The
correct answer is:
8.0 mm Hg
Explanation:
DP = Psolvent - Psolution = 89-81 = 8
mm Hg
Solution Reference:
Page 656, Raoult's Law
18.
Consider that an ideal solution is formed from a mixture of the nonvolatile
solute, urea (CO(NH2)2), and methanol (CH3OH).
The vapor pressure of pure methanol at 20°C is 89 mm Hg. Calculate the vapor
pressure-lowering for the methanol solution in problem No. 16.
|
a. |
6.0 mm Hg |
|
b. |
8.0 mm Hg |
|
c. |
14 mm Hg |
|
d. |
75 mm Hg |
|
e. |
81 mm Hg |
19. The
correct answer is:
-0.046°C
Explanation:
DT = Kfm = -1.86°C/m x 0.025 = -0.046°C Freezing
point = -0.046°C
Solution Reference:
Page 662, Colligative Properties and Molar
Mass Determination
19.
The freezing point of a 0.025 m solution of formaldehyde (CH2O) in
water would be (Kf for water is -1.86°C/m).
|
a. |
-1.86°C |
|
b. |
-0.046°C |
|
c. |
0.046°C |
|
d. |
0.093°C |
|
e. |
-0.093°C |
20. The
correct answer is:
0.12.
Explanation:
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Solution Reference:
Page 662, Colligative Properties and Molar
Mass Determination
20.
The freezing point of an aqueous solution of a nonelectrolyte is -0.23°C. The
molality of this solution is (Kf(H2O) = -1.86°C/m)
|
a. |
0.12. |
|
b. |
0.25. |
|
c. |
0.45. |
|
d. |
0.51. |
|
e. |
1.86. |
21. The
correct answer is:
26.8.
Explanation:

Solution Reference:
Page 129, Empirical Formulas
21. A
compound containing boron, nitrogen, and hydrogen analyzes 40.3% B, 52.2% N,
and 7.5% H. A 3.562 g sample of this compound was dissolved in 50.00 g of
benzene producing a solution freezing at 1.3°C. Kf for benzene is
-5.12°C/molal. Pure benzene has a freezing point of 5.48°C. Determine the
empirical formula of the compound from the percentage composition. The
empirical weight of the B, N, H compound is
|
a. |
25.8. |
|
b. |
26.8. |
|
c. |
50.6. |
|
d. |
51.6. |
|
e. |
106. |
22. The
correct answer is:
0.82.
Explanation:
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Solution Reference:
Page 643, Molality
22. A
compound containing boron, nitrogen, and hydrogen analyzes 40.3% B, 52.2% N,
and 7.5% H. A 3.562 g sample of this compound was dissolved in 50.00 g of
benzene producing a solution freezing at 1.3°C. Kf for benzene is
-5.12°C/molal. Pure benzene has a freezing point of 5.48°C. The molality of the
benzene solution is
|
a. |
0.071. |
|
b. |
0.24. |
|
c. |
0.70. |
|
d. |
0.82. |
|
e. |
1.07 |
23. The
correct answer is:
87.
Explanation:

Solution Reference:
Page 662, Molar Mass Determination
23. A
compound containing boron, nitrogen, and hydrogen analyzes 40.3% B, 52.2% N,
and 7.5% H. A 3.562 g sample of this compound was dissolved in 50.00 g of
benzene producing a solution freezing at 1.3°C. Kf for benzene is
-5.12°C/molal. Pure benzene has a freezing point of 5.48°C. The experimental
molecular weight for the compound is
|
a. |
67. |
|
b. |
87. |
|
c. |
102. |
|
d. |
106. |
|
e. |
297. |
24. The
correct answer is:
B3N3H6.
Explanation:

Solution Reference:
Page 129, Molecular Formula
24. A
compound containing boron, nitrogen, and hydrogen analyzes 40.3% B, 52.2% N,
and 7.5% H. A 3.562 g sample of this compound was dissolved in 50.00 g of
benzene producing a solution freezing at 1.3°C. Kf for benzene is
-5.12°C/molal. Pure benzene has a freezing point of 5.48°C. The molecular
formula for the compound is
|
a. |
B2N2H. |
|
b. |
B4N4H2. |
|
c. |
BNH2. |
|
d. |
B3N3H6. |
|
e. |
B4N4H8. |
25. The
correct answer is:
5
Explanation:
How many ions are there in one formula unit of
Al2(SO4)3?
Solution Reference:
Page 667, van't Hoff Factor
25.
For a 0.001 molal solution of Al2(SO4)3, the
van't Hoff factor, i, would be about
|
a. |
1 |
|
b. |
2 |
|
c. |
3 |
|
d. |
4 |
|
e. |
5 |