Mixtures At The Molecular Level | Exam Questions Chapter 12

Chemistry: Molecular Nature of Matter, 8e (Jespersen)

Chapter 12 Mixtures at the Molecular Level: Properties of Solutions

1) Wax is a solid mixture of hydrocarbon compounds consisting of molecules with long chains of carbon atoms. Which solvent would you expect to be most capable of dissolving wax?

A) CH3—C—CH3

║

O

B) CH3—O—H

C) CH3—C—H

║

O

D) H—O—CH2—CH2—O—H

E) CH3—CH2—CH2—CH2—CH2—CH2– CH2—CH3

Diff: 1

Section: 12.1

2) Wax is a solid mixture of hydrocarbon compounds consisting of molecules with long chains of carbon atoms. Which solvent would you expect to be most capable of dissolving wax?

A) H—O—H

B) CH3—O—H

C) CF3—O—H

D) H—O—CH2—CH2—O—H

E) CH3—CH2—CH2—CH2—CH2—H2—CH2—CH3

Diff: 1

Section: 12.1

3) Which response lists all the following pairs that are miscible liquids?

Pair # 1. octane (C8H18) and water

Pair # 2. ammonia (NH3) and water

Pair # 3. octane (C8H18) and carbon tetrachloride (CCl4)

A) 1, 3

B) 1, 2

C) 2

D) 2, 3

E) 3

Diff: 2

Section: 12.1

4) Concerning the process of separating of a solid substance into its component units (molecules or ions),

A) the process is exothermic and the potential energy increases.

B) the process is exothermic and the potential energy decreases.

C) the process is endothermic and the potential energy increases.

D) the process is endothermic and the potential energy decreases.

E) the process is endothermic and occurs with no change in potential energy.

Diff: 1

Section: 12.2

5) Concerning the solvation step of the solution process,

A) the process is exothermic and the potential energy increases.

B) the process is exothermic and the potential energy decreases.

C) the process is endothermic and the potential energy increases.

D) the process is endothermic and the potential energy decreases.

E) the process is endothermic and occurs with no change in potential energy.

Diff: 1

Section: 12.2

6) KBr has a lattice energy of −682 kJ/mol and a hydration energy of −657 kJ/mol. Using this

information, what is the heat of solution, ∆Hsol, for KBr?

Hint: What happens to the crystal lattice when an ionic compound dissolves?

A) −1339 kJ/mol

B) +1339 kJ/mol

C) +25 kJ/mol

D) −657 kJ/mol

E) +657 kJ/mol

Diff: 3

Section: 12.2

7) A solution in a beaker has some undissolved solute lying on the bottom of the beaker. If the rate of crystallization exceeds the rate of dissolution of the excess solute, the solution is described as

A) dilute.

B) concentrated.

C) unsaturated.

D) saturated.

E) supersaturated.

Diff: 2

Section: 12.3

8) A solution in a beaker has some un-dissolved solute lying on the bottom of the beaker. If the rate of crystallization is equal to the rate of dissolution of the excess solute, the solution is described as

A) dilute.

B) concentrated.

C) unsaturated.

D) saturated.

E) supersaturated.

Diff: 1

Section: 12.3

9) A solution in a beaker has some un-dissolved solute lying on the bottom of the beaker. If the rate of dissolution of the solute exceeds the rate of crystallization, the solution is described as

A) dilute.

B) concentrated.

C) unsaturated.

D) saturated.

E) supersaturated.

Diff: 1

Section: 12.3

10) A solution is sitting undisturbed on a side shelf in the laboratory. A small crystal of the same solute of which the solution is made was gently dropped into the solution. Suddenly, a mass of crystals forms and settles to the bottom of the container. The solution is, or must have been

A) dilute.

B) concentrated.

C) unsaturated.

D) saturated.

E) supersaturated.

Diff: 1

Section: 12.3

11) Which of the following will always cause an increase in the solubility of a gas in a solvent in which the gas does not react with the solvent to form a new substance?

A) increasing the temperature of the solvent and simultaneously decreasing the pressure of the gas in the space above the solvent

B) decreasing the temperature of the solvent and simultaneously increasing the pressure of the gas in the space above the solvent

C) increasing the temperature of the solvent and simultaneously increasing the pressure of the gas in the space above the solvent

D) decreasing the temperature of the solvent and simultaneously decreasing the pressure of the gas in the space above the solvent

E) increasing the temperature of the solvent while maintaining the pressure of the gas in the space above the solvent at a set value

Diff: 1

Section: 12.3

12) The solubility of O2 in water is approximately 0.00380 g L−1 when the temperature is 25.0°C and the partial pressure of gaseous oxygen is 760. torr. What will the solubility of oxygen be if the oxygen pressure is adjusted to 1000 torr?

A) 0.00289 g L−1

B) 0.00500 g L−1

C) 1.49 g L−1

D) 2.89 × 103 g L−1

E) 3.46 × 103 g L−1

Diff: 2

Section: 12.4

13) The solubility of O2 in water is approximately 0.00380 g L−1 of water when the temperature is 25.0°C and the partial pressure of gaseous oxygen is 760 torr. The oxygen gas above the water is replaced by air at the same temperature and pressure, in which the mole fraction of oxygen is 0.210. What will the solubility of oxygen in water be under these new conditions?

A) 1.05 × 10−6 g L−1

B) 7.98 × 10−4 g L−1

C) 1.33 × 10−3 g L−1

D) 0.606 g L−1

E) 1.01 g L−1

Diff: 2

Section: 12.4

14) A solution is made by mixing 138.2 grams of ethanol, C2H6O, (46.069 g mol−1); 103.6 grams of water (18.015 g mol−1), and 80.11 grams of methanol, CH4O, (32.042 g mol−1). What is the mole fraction of methanol in the mixture?

A) 0.02504

B) 0.2222

C) 0.2493

D) 0.3333

E) 0.4490

Diff: 2

Section: 12.5

15) The CO2 gas sealed inside a carbonated beverage bottle has a pressure of 3.750 atm. At this pressure, the solubility of CO2 in water is 0.65 g CO2/100 g H2O. If the bottle is opened, as the gas in the space above the liquid escapes, the partial pressure of the CO2 falls to 0.30 torr, the value in the surrounding atmosphere of the room. What is the solubility of CO2 in the beverage at this new pressure?

Hint: Be sure check your pressure units.

A) 5.2 × 10−2 g/100 g H2O

B) 5.0 × 10−4 g/100 g H2O

C) 6.7 × 10−5 g/100 g H2O

D) 6.8 × 10−5 g/100 g H2O

E) 9.6 × 10−4 g/100 g H2O

Diff: 3

Section: 12.4

16) A solution of sodium nitrite is prepared by mixing 3.25 g of NaNO2 with 12.0 g of water. The percent, by mass, of NaNO2 is

A) 28.0%.

B) 23.3%.

C) 27.0%.

D) 21.3%.

E) 37.1%.

Diff: 1

Section: 12.5

17) A solution of lithium hydroxide is prepared by mixing 2.00 g of LiOH with 10.0 g of water. The percent, by mass, of LiOH is

A) 10.7%.

B) 12.0%.

C) 16.7%.

D) 20.0%.

E) 80.0%.

Diff: 1

Section: 12.5

18) How many grams of NaC2H3O2 should be dissolved in 400.0 g of water to prepare a solution that is 11.28% NaC2H3O2 by mass?

A) 3.146 g

B) 7.558 g

C) 21.17 g

D) 50.86 g

E) 127.15 g

Diff: 2

Section: 12.5

19) A solution of potassium nitrate is prepared by mixing 3.50 g of KNO3 with 12.0 g of water. The percent, by mass, of KNO3 is

A) 22.6%.

B) 23.3%.

C) 28.0%.

D) 29.2%.

E) 41.8%.

Diff: 1

Section: 12.5

20) A glucose solution is prepared by dissolving 5.10 g of glucose, C6H12O6, in 110.5 g of water. What is the molality of the glucose solution?

A) 0.283 m

B) 0.000256 m

C) 0.245 m

D) 0.256 m

E) 0.351 m

Diff: 2

Section: 12.5

21) An aqueous solution of orthophosphoric acid, H3PO4, has a measured density of 1.2089 g mL−1 and is 5.257 molal. How many moles of H3PO4 are there in one liter of this solution?

A) 0.4261 moles

B) 4.194 moles

C) 4.349 moles

D) 5.152 moles

E) 6.355 moles

Diff: 2

Section: 12.5

22) An aqueous solution of ethanol, C2H5OH, is 19.00% ethanol by mass and has a density of 0.9700 g mL−1. Calculate the molality of the ethanol solution.

A) 4.000 m

B) 4.124 m

C) 4.252 m

D) 5.092 m

E) 14.48 m

Diff: 2

Section: 12.5

23) An aqueous solution of glycerol, C3H8O3, is 48.0% glycerol by mass and has a density of Calculate the molality of the glycerol solution.

A) 11.2 m

B) 5.84 m

C) 0.584 m

D) 0.521 m

E) 10.0 m

Diff: 2

Section: 12.5

24) A solution is prepared by mixing 0.3355 moles of NaNO3 (84.995 g mol−1) with 235.0 g of water (18.015 g mol−1). Its density is 1.0733 g mL−1. What is the percent by mass of NaNO3 in the solution?

A) 10.16%

B) 10.82%

C) 11.61%

D) 14.19%

E) 26.56%

Diff: 2

Section: 12.5

25) A solution is prepared by mixing 0.3355 moles of NaNO3 (84.995 g mol−1) with 235.0 g of water (18.015 g mol−1). Its density is 1.0733 g mL−1. What is the molality of the solution?

A) 0.6474 m

B) 0.7004 m

C) 1.320 m

D) 1.428 m

E) 1.545 m

Diff: 2

Section: 12.5

26) An aqueous solution is prepared by mixing 0.2750 moles of NaOH (40.00 g mol−1) with 189.0 g of water. Its density is 1.065 g mL−1. What is the percent by weight of NaOH in the solution?

A) 0.001375%

B) 5.500%

C) 5.858%

D) 13.75%

E) 20.00%

Diff: 2

Section: 12.5

27) An aqueous solution of nitric acid has a density of 1.084 g mL−1 and a measured concentration of 2.580 molar. What is the percent by weight of nitric acid in the solution?

A) 2.380%

B) 13.82%

C) 15.00%

D) 22.29%

E) 44.38%

Diff: 2

Section: 12.5

28) Which is a concentration unit whose value changes if the temperature of an aqueous solution is changed?

A) mole fraction

B) molarity

C) molality

D) mass fraction

E) percent by weight

Diff: 2

Section: 12.5

29) Consider a 0.900 M Al(NO3)3 solution. This solution has a nitrate ion concentration of

A) 0.300 M.

B) 0.900 M.

C) 2.70 M.

D) 3.60 M.

E) 8.10 M.

Diff: 1

Section: 12.5

30) A solution of ethylene glycol (C2H6O2) in water is 3.981 molar and has a density of Calculate the percent, by weight, of ethylene glycol in the solution.

A) 3.867%

B) 4.099%

C) 15.14%

D) 24.00%

E) 25.45%

Diff: 1

Section: 12.5

31) A solution of sodium nitrate, NaNO3, in water is 5.181 molar and its density is 1.2680 g mL−1. Calculate the percent, by weight, of sodium nitrate in the solution.

A) 7.939%

B) 17.21%

C) 24.47%

D) 29.56%

E) 34.73%

Diff: 2

Section: 12.5

32) An aqueous solution of ethanol, C2H5OH, is 19.00% ethanol by mass and has a density of 0.9700 g mL−1. Calculate the molarity of the ethanol solution.

A) 4.001 M

B) 4.124 M

C) 4.252 M

D) 5.092 M

E) 14.48 M

Diff: 2

Section: 12.5

33) An aqueous solution of glycerol, C3H8O3, is 48.0% glycerol by mass and

has a density of 1.120 g mL−1. Calculate the molarity of the glycerol solution.

A) 12.2 M

B) 5.84 M

C) 0.584 M

D) 0.521 M

E) 0.465 M

Diff: 2

Section: 12.5

34) A solution is prepared by mixing 0.3355 moles of NaNO3 (84.995 g mol−1) with 235.0 g of water (18.015 g mol−1). Its density is 1.0733 g mL−1. What is the molarity of the solution?

A) 1.186 M

B) 1.273 M

C) 1.350 M

D) 1.366 M

E) 1.428 M

Diff: 2

Section: 12.5

35) An aqueous solution of glycerol, C3H8O3, in which the mole fraction of C3H8O3 is 0.07070 was found to have a density of 1.0350 g mL−1. What is the molarity of this solution?

Hint: Assume one mole of glycerol and solve for moles of water. From there you can find the mass and then volume of the solution.

A) 3.147 molar

B) 4.371 molar

C) 4.223 molar

D) 2.938 molar

E) 3.651 molar

Diff: 3

Section: 12.5

36) An aqueous solution of glycerol, C3H8O3 is prepared by adding 150.0 g of glycerol to 200.0 g of water. What is the mole fraction of glycerol in the final solution?

A) 0.571

B) 0.429

C) 0.750

D) 0.872

E) 0.128

Diff: 2

Section: 12.5

37) An aqueous solution of ethanol, C2H5OH is prepared by adding 200. g of ethanol to 50.0 g of water. What is the mole fraction of ethanol in the final solution?

A) 0.200

B) 0.610

C) 0.390

D) 0.250

E) 0.800

Diff: 2

Section: 12.5

38) The vapor pressure of a solution containing a nonvolatile solute is directly proportional to the

A) mole fraction of the solvent.

B) mole fraction of the solute.

C) molality of the solvent.

D) molarity of the solvent.

E) osmotic pressure of the solute.

Diff: 2

Section: 12.6

39) When a nonvolatile solute such as ammonium sulfate is dissolved in a solvent like water, one of the observed effects is

A) a decrease in the vapor pressure of the solvent.

B) an increase in the vapor pressure of the solute.

C) an increase in the freezing point of the liquid.

D) a decrease in the boiling point of the liquid.

E) scattering of light beams by the solute particles in the solution.

Diff: 1

Section: 12.6

40) At 23.0°C, the vapor pressure of acetonitrile, CH3CN, is 81.0 torr while that of acetone, C3H6O, is 184.5 torr. What is the vapor pressure of a solution that contains 0.550 moles of acetonitrile and 0.350 moles of acetone? (Assume the mixture behaves as an ideal solution.)

A) 109 torr

B) 121 torr

C) 130 torr

D) 144 torr

E) 239 torr

Diff: 2

Section: 12.6

41) During osmosis,

A) pure solvent passes through a membrane but solutes do not.

B) pure solute passes through a membrane but solvent does not.

C) pure solvent moves in one direction through the membrane while the solution moves through the membrane in the other direction.

D) pure solvent moves in one direction through the membrane while the solute moves through the membrane in the other direction.

E) pure solute moves in one direction through the membrane while the solution moves through the membrane in the other direction.

Diff: 1

Section: 12.6

42) A very dilute solution contains 116 mg of fructose (molar mass = 180.16 g mol−1) in 1.000 liter of solution. It is placed in an osmotic membrane bladder, which is then suspended in pure water. What osmotic pressure would develop across the membrane if the temperature is 26.0°C?

A) 3.36 torr

B) 12.0 torr

C) 151 torr

D) 475 torr

E) 1217 torr

Diff: 1

Section: 12.6

43) Which aqueous solution will have the highest boiling point temperature?

A) 0.100 molal NiBr2(aq)

B) 0.100 molal MgSO4(aq)

C) 0.150 molal NH4NO3(aq)

D) 0.150 molal Na2SO4(aq)

E) 0.250 molal CH3OH(aq)

Diff: 1

Section: 12.6

44) Which aqueous solution will have the lowest boiling point temperature?

A) 0.100 molal NiBr2(aq)

B) 0.200 molal MgSO4(aq)

C) 0.150 molal NH4NO3(aq)

D) 0.150 molal Na2SO4(aq)

E) 0.250 molal CH3OH(aq)

Diff: 1

Section: 12.6

45) Which solution has the highest osmotic pressure at 25°C?

A) 0.200 M (NH4)2SO4

B) 0.200 M KClO3

C) 0.200 M C2H6O, ethanol

D) 0.200 M NiF2

E) 0.200 M Fe(NO3)3

Diff: 1

Section: 12.6

46) Arrange these aqueous solutions in order of increasing boiling points:

0.100 m Mg(NO3)2 0.200 m ethylene glycol, C2H6O2 0.175 m LiCl

A) C2H6O2 < Mg(NO3)2 < LiCl

B) Mg(NO3)2 < LiCl < C2H6O2

C) C2H6O2 < LiCl < Mg(NO3)2

D) LiCl < C2H6O2 < Mg(NO3)2

E) Mg(NO3)2 < C2H6O2 < LiCl

Diff: 1

Section: 12.6

47) A dilute aqueous solution of CaCl2 contains 0.159 grams of solute per liter of solution. It is fully dissociated. What is its osmotic pressure at 20.0°C?

Hint: CaCl2 is an ionic compound.

A) 1.79 torr

B) 3.49 torr

C) 26.2 torr

D) 78.6 torr

E) 2.65 × 103 torr

Diff: 1

Section: 12.6

48) At 28.0°C, the vapor pressure of n-propyl mercaptan, C3H7SH, is 175 torr, while that of acetonitrile, CH3CN, is 102 torr. What is the vapor pressure, at 28.0°C, of a solution made by mixing 100.0 g of C3H7SH and 100.0 g CH3CN, if Raoult's Law is obeyed?

A) 35.7 torr

B) 128 torr

C) 139 torr

D) 149 torr

E) 277 torr

Diff: 2

Section: 12.6

49) A molecular solute with a molar mass of 50.0 g mol−1 is dissolved in 500 g of water and the resulting solution has a boiling point of 101.53°C. How many grams of solute were in the solution?

Kb = 0.512 °C m−1

A) 30.0 grams

B) 75.0 grams

C) 100 grams

D) 125 grams

E) 150 grams

Diff: 2

Section: 12.6

50) A solution contains 221 g of glycerol (C3H8O3) in 600 grams of water. For the solvent, the Kf is 1.86 °C m−1 and Kb is 0.512 °C m−1. What should the boiling point of the solution be?

A) 100.02°C

B) 2.04°C

C) 101.65°C

D) 102.04°C

E) 3.62°C

Diff: 2

Section: 12.6

51) A solution, which was made by dissolving 62.07 g of a nonelectrolyte in 500 g of water, exhibits a freezing point of −1.86°C. What is the molecular weight of this nonelectrolyte compound? For water, Kf is 1.86°C m−1 and Kb is 0.512°C m−1.

A) 57.7 g mol−1

B) 62.07 g mol−1

C) 115 g mol−1

D) 124 g mol−1

E) 231 g mol−1

Diff: 2

Section: 12.6

52) How many moles of the nonelectrolyte, propylene glycol (C3H8O2) should be dissolved in 800.0 g of water to prepare a solution whose freezing point is -3.72°C?

For water, Kf is 1.86°C m−1 and Kb is 0.512°C m−1.

A) 1.60 moles

B) 2.00 moles

C) 2.50 moles

D) 2.98 moles

E) 4.65 moles

Diff: 2

Section: 12.6

53) Which can be used to calculate the osmotic pressure of a solution?

A) de Broglie equation

B) Tyndall factor

C) Peters equation

D) van der Waals equation

E) van't Hoff equation

Diff: 2

Section: 12.6

54) How many grams of glycerol (C3H8O3, a nonelectrolyte) should be dissolved in 600 g of water to prepare a solution whose freezing point is -4.65°C? For water, Kf is 1.86°C m−1 and Kb is 0.512 °C m−1.

A) 22.1 grams

B) 93.6 grams

C) 138 grams

D) 384 grams

E) 478 grams

Diff: 2

Section: 12.6

55) Which aqueous solution has the highest boiling point?

A) 0.200 m KCl(aq)

B) 0.200 m Na2SO4(aq)

C) 0.200 m Ca(NO3)2(aq)

D) 0.200 m C3H8O3 (aq), glycerol

E) 0.200 m Na3PO4(aq)

Diff: 2

Section: 12.6

56) A solution is made by dissolving 0.840 moles of sodium hydroxide in 300.0 g of water. If the van't Hoff factor, i, for this particular concentration is 1.70, what is the expected freezing point of this solution? Kf = 1.86 °C m−1.

A) +0.80°C

B) -2.66°C

C) +3.06°C

D) -8.85°C

E) -9.97°C

Diff: 2

Section: 12.6

57) 9.92 grams of a compound with a molar mass of 124.0 grams/mole when dissolved in 150.0 grams of water gave a solution that had a freezing point of -2.69°C. Calculate the experimental value of the van't Hoff factor. Kf = 1.86 °C m−1.

A) 1.21

B) 1.45

C) 1.54

D) 2.69

E) 2.71

Diff: 2

Section: 12.6

58) A solution is made by dissolving 1.25 moles of magnesium nitrate in 300.0 g of water. If the van't Hoff factor, i, for this particular concentration is 2.25, what is the expected freezing point of this solution? Kf = 1.86 °C m−1.

A) −0.45°C

B) −1.57°C

C) −3.44°C

D) −5.23°C

E) −17.4°C

Diff: 2

Section: 12.6

59) An aqueous solution which is 12.00% sodium hydroxide by weight has a freezing point of −10.40°C. What is the observed value of the van't Hoff factor, i, in this solution? Kf = 1.86 °C m−1.

A) 1.59

B) 1.86

C) 1.64

D) 1.75

E) 1.70

Diff: 2

Section: 12.6

60) Which aqueous solution should have the highest osmotic pressure?

A) 0.100 molar Al(NO3)3

B) 0.150 molar Ba(NO3)2

C) 0.100 molar CaCl2

D) 0.150 molar NaCl

E) 0.200 molar NH3

Diff: 2

Section: 12.6

61) Below is a diagram of vapor pressure versus composition for a mixture of two liquids. The dark straight lines represent the vapor pressure of each pure liquid, while the gray curved line gives the vapor pressure of the mixture. Using this information, which of the following statements is true about the mixture?

A) The mixture is an ideal mixture that has no deviation from Raoult's Law.

B) The mixture is an ideal mixture with a positive deviation from Raoult's Law.

C) The mixture is not ideal and interactions between unlike molecules in solution are weaker than the interactions between like molecules in pure solution.

D) The mixture is not ideal and interactions between unlike molecules in solution are stronger than the interactions between like molecules in the pure solution.

E) The mixture is ideal and interactions between unlike molecules in solution are the same as the interactions between like molecules in the pure solution.

Diff: 2

Section: 12.6

62) At 28.0°C, the vapor pressure of npropyl mercaptan, C3H7SH, is 175 torr, while that of acetonitrile, CH3CN, is 102 torr. What is the vapor pressure, at 28.0°C, of a solution made by mixing 120.0 g of C3H7SH and 80.0 g CH3CN, if Raoult's Law is obeyed?

Hint: Organizing the given information into moles and mole fraction of each component will help in solving this problem.

A) 131 torr

B) 135 torr

C) 142 torr

D) 146 torr

E) 277 torr

Diff: 2

Section: 12.6

63) At 28.0°C, the vapor pressure of n-propyl mercaptan, C3H7SH, is 175 torr, while that of

acetonitrile, CH3CN, is 102 torr. What is the vapor pressure, at 28.0°C, of a solution made by

mixing 80.0 g of C3H7SH and 120.0 g CH3CN, if Raoult's Law is obeyed?

Hint: Organizing the given information into moles and mole fraction of each component will help in solving this problem.

A) 121 torr

B) 131 torr

C) 139 torr

D) 146 torr

E) 156 torr

Diff: 3

Section: 12.6

64) At 28.0°C, the vapor pressure of acetonitrile, CH3CN, is 102.0 torr while that of acetone, C3H6O, is 228.9 torr, and for CS2 the value is 378.7 torr. A three-component solution is made by adding 0.300 moles of CH3CN and 0.400 moles of C3H6O to 0.350 moles of CS2. The mixture behaves as an ideal mixture. What is the vapor pressure of the solution?

Hint: Organizing the given information into moles and mole fraction of each component will help in solving this problem.

A) 203 torr

B) 385 torr

C) 262 torr

D) 275 torr

E) 610 torr

Diff: 3

Section: 12.6

65) At 28.0°C, the vapor pressure of pure carbon disulfide, CS2, is 378.7 torr, while that of acetone, C3H6O, is 228.9 torr. What is the vapor pressure, at 28.0°C, of a solution made by mixing 0.250 moles of carbon disulfide and 0.450 moles of acetone, if Raoult's Law is obeyed?

Hint: Organizing the given information into moles and mole fraction of each component will help in solving this problem.

A) 198 torr

B) 228 torr

C) 282 torr

D) 325 torr

E) 425 torr

Diff: 3

Section: 12.6

66) What is the expected freezing point of a solution that contains 25.0 g of fructose, C6H12O6, in 250.0 g of H2O? For water, Kf = 1.86 °C m−1.

A) -0.10°C

B) +0.10°C

C) -0.186°C

D) +0.186°C

E) -1.03°C

Diff: 2

Section: 12.6

67) What is the freezing point of a solution that contains 20.0 g of glucose (C6H12O6) in 100.0 g of H2O? Kf for water is 1.86°C/m.

A) -0.206°C

B) -2.06°C

C) -1.86°C

D) +0.111°C

E) +1.11°C

Diff: 2

Section: 12.6

68) What is the normal boiling point of a solution that contains 30.0 g of glucose (C6H12O6) in 100.0 g of H2O? Kb for water is 0.512°C/m.

A) 0.17°C

B) 99.01°C

C) 101.67°C

D) 100.85°C

E) 101.53°C

Diff: 2

Section: 12.6

69) Pure benzene, C6H6, has a freezing point of 5.45°C. Its freezing point depression

constant is: Kf = 5.07 °C m−1. A solution was made by taking 24.20 g of an unknown nonelectrolyte and dissolving it in 125.0 g of benzene. The measured freezing point of the solution was -1.65°C. Calculate the molecular weight of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

A) 138 g mol−1

B) 145 g mol−1

C) 258 g mol−1

D) 272 g mol−1

E) 595 g mol−1

Diff: 3

Section: 12.6

70) Pure cyclohexane, C6H12, has a freezing point of 6.53°C. Its freezing point depression constant is: Kf = 20.0 °C m−1. A solution was made by taking 11.40 g of an unknown nonelectrolyte and dissolving it in 150.0 g of cyclohexane. The measured freezing point of the solution was -0.78°C. Calculate the molecular weight of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

A) 27.8 g mol−1

B) 46.8 g mol−1

C) 208 g mol−1

D) 264 g mol−1

E) 1949 g mol−1

Diff: 3

Section: 12.6

71) Pure cyclohexane, C6H12, has a freezing point of 6.53°C. Its freezing point depression constant is: Kf = 20.0 °C m−1. A solution was made by taking 18.55 g of an unknown nonelectrolyte and dissolving it in 150.0 g of cyclohexane. The measured freezing point of the solution was -4.28°C. Calculate the molecular weight of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

A) 61.8 g mol−1

B) 66.8 g mol−1

C) 229 g mol−1

D) 578 g mol−1

E) 1099 g mol−1

Diff: 3

Section: 12.6

72) Pure glacial acetic acid, HC2H3O2, has a freezing point of 16.62°C. Its freezing point depression constant is: Kf = 3.57 °C m−1. A solution was made by taking 9.755 g of an unknown nonelectrolyte and dissolving it in 90.50 g of glacial acetic acid. The measured freezing point of the solution was 8.64°C. Calculate the molecular weight of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

A) 24.4 g mol−1

B) 30.5 g mol−1

C) 45.3 g mol−1

D) 48.2 g mol−1

E) 174 g mol−1

Diff: 3

Section: 12.6

73) Pure benzene, C6H6, has a freezing point of 5.45°C. Its freezing point depression

constant is: Kf = 5.07 °C m−1. A solution was made by taking 10.00 g of an unknown nonelectrolyte and dissolving it in 105.0 g of benzene. The measured freezing point of the solution was -1.05°C. Calculate the molecular weight of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

A) 74.3 g mol−1

B) 82.7 g mol−1

C) 110 g mol−1

D) 122 g mol−1

E) 460 g mol−1

Diff: 3

Section: 12.6

74) An aqueous ethylene glycol solution being considered for use as a radiator coolant is 16.0% C2H6O2 by weight. What would be the expected boiling point of this solution? For water, Kf is 1.86 °C m−1 and Kb = 0.512 °C m−1.

Hint: convert percent by mass to molality, then find △ T.

A) 100.3°C

B) 101.6°C

C) 105.1°C

D) 106.0°C

E) 156.5°C

Diff: 3

Section: 12.6

75) Pure chloroform, CHCl3, has a boiling point of 61.23°C. Its boiling point elevation constant, Kb, is 3.63°C m−1. A solution was made by taking 11.25 g of an unknown nonelectrolyte and dissolving it in 115.5 g of chloroform. The measured boiling point of the solution was 65.46°C. Calculate the molecular weight of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

A) 83.6 g mol−1

B) 113 g mol−1

C) 114 g mol−1

D) 120 g mol−1

E) 158 g mol−1

Diff: 3

Section: 12.6

76) Which property of a solution is not a colligative property?

A) solubility of a solute

B) freezing point depression

C) boiling point elevation

D) osmotic pressure

E) vapor pressure lowering

Diff: 1

Section: 12.6

77) An aqueous solution made by dissolving 168 mg of an unknown compound (a nonelectrolyte) in enough water to make 500.0 mL of solution, develops an osmotic pressure of 5.22 torr at a temperature of 23.5°C. What is the molecular mass of this unknown compound?

Hint: Find molarity, then moles, then molar mass, being sure to keep track of units.

A) 94.3 g mol−1

B) 124 g mol−1

C) 943 g mol−1

D) 1.19 × 103 g mol−1

E) 1.57 × 103 g mol−1

Diff: 3

Section: 12.6

78) Which aqueous solution will have the lowest freezing point temperature?

A) 0.100 molal NaBr(aq)

B) 0.100 molal MgSO4(aq)

C) 0.150 molal KClO3(aq)

D) 0.150 molal MgCl2(aq)

E) 0.250 molal C6H12O6(aq)

Diff: 2

Section: 12.6

79) A solution is prepared by mixing 0.3355 moles of NaNO3 with 235.0 g of water. Its density is 1.0733 g mL−1. If Kf = 1.86 °C m−1 and the solute is completely ionized, what is the expected freezing point of the solution?

Hint: be sure to classify the solute as an electrolyte or non-electrolyte.

A) 2.65°C

B) 2.87°C

C) -5.31°C

D) 5.75°C

E) 7.97°C

Diff: 2

Section: 12.6

80) Which aqueous solution will have the highest freezing point temperature?

Hint: be sure to closely read the question.

A) 0.100 molal MgBr2(aq)

B) 0.100 molal MgSO4(aq)

C) 0.150 molal KClO3(aq)

D) 0.150 molal MgCl2(aq)

E) 0.250 molal C6H12O6(aq)

Diff: 2

Section: 12.6

81) Below is a diagram of vapor pressure versus composition for a mixture of two liquids. The dark straight lines represent the vapor pressure of each pure liquid, while the gray curved line gives the vapor pressure of the mixture. Using this information, which of the following statements is true about the mixture?

A) The mixture is an ideal mixture that has no deviation from Raoult's Law.

B) The mixture is an ideal mixture with a positive deviation from Raoult's Law.

C) The mixture is not ideal and interactions between unlike molecules in solution are weaker than the interactions between like molecules in pure solution.

D) The mixture is not ideal and interactions between unlike molecules in solution are stronger than the interactions between like molecules in the pure solution.

E) The mixture is ideal and interactions between unlike molecules in solution are the same as the interactions between like molecules in the pure solution.

Diff: 2

Section: 12.6

82) What is the osmotic pressure of a solution prepared from 13.7 g of HCl and water to make 0.500 L of solution at 18°C? Assume HCl completely ionizes.

Hint: be sure to keep track of units when working this problem.

A) 0.52 atm

B) 1.10 atm

C) 8.95 atm

D) 17.2 atm

E) 35.9 atm

Diff: 2

Section: 12.6

83) An aqueous solution that would cause red blood cells to burst from lowered osmotic pressure is called a(n) ________ solution.

A) hypertonic

B) hydrophobic

C) hypotonic

D) asmorphic

E) isotonic

Diff: 1

Section: 12.7

84) An aqueous solution that would cause red blood cells to shrink from increased osmotic pressure is called a(n) ________ solution.

A) hypertonic

B) hydrophobic

C) hypotonic

D) asmorphic

E) isotonic

Diff: 1

Section: 12.7

85) Which of the following would best be classified as a heterogeneous suspension?

A) milk

B) dust from a dirt field

C) salt water

D) gelatin with fruit in it

E) air bubbles in frozen ice cream

Diff: 1

Section: 12.7

86) The mixture of air bubbles in whipped cream would best be classified as

A) a foam.

B) an emulsion.

C) a suspension.

D) a supersaturated solution.

E) an ideal solution.

Diff: 1

Section: 12.7

87) A mixture of flour in water would best be classified as

A) a colloid.

B) an emulsion.

C) a suspension.

D) a supersaturated solution.

E) an ideal solution.

Diff: 1

Section: 12.7

88) Which of the following is most likely to be hydrophobic?

A) Na+

B) C6H14

C) NO3−

D) Br−

E) NH4+

Diff: 1

Section: 12.7

89) The solubility of gases in water usually ________ when the temperature increases.

Diff: 1

Section: 12.3

90) The solubility of CO2 gas in water would ________ if the partial pressure of CO2 was increased.

Diff: 1

Section: 12.4

91) Oxygen has a solubility in water of 0.0043 g/100 mL at 20°C, but CO2 has a solubility of 0.169 g/100 mL at 20°C. What is the main reason for this large difference in solubilities between these gases?

Diff: 1

Section: 12.4

92) A scientist looking to remove oxygen from above a liquid sample pumps nitrogen into a sealed compartment to displace the oxygen, then places his liquid sample into the compartment. If the nitrogen pressure (partial pressure) were 1.8 atm, what would be the concentration of dissolved nitrogen gas if the Henry's Law constant for nitrogen gas was 6.8 × 10−4 M atm−1 at this temperature?

Diff: 2

Section: 12.4

93) If the concentration of hydrogen gas in a liquid sample at 1.25 atm and 25°C is 9.75 × 10−4 M, the concentration of hydrogen gas at 3.41 atm and 25°C would be ________.

Diff: 2

Section: 12.4

94) Dry air is a mixture of gases in which the partial pressure of nitrogen gas at 1.000 atm is typically 0.78 atm while that of argon gas is 0.009 atm. A liquid sample was exposed to dry air at 1.000 atm and a given temperature and found to have a nitrogen gas concentration of 5.30 × 10−4 M. If argon were added to the dry air until the partial pressure of argon was now 0.400 atm, the concentration of nitrogen gas in the liquid sample would now be ________.

Hint: Think of Dalton's law of partial pressures.

Diff: 3

Section: 12.4

95) A solution contains 25.50 grams of NaNO3 (M = 84.99 g mol−1) in 250.0 grams of water. Its density is 1.0620 g mL−1. Calculate the molality of the solution.

Diff: 2

Section: 12.5

96) What is the mass percent of CdSO4 in a 2.00 molal aqueous CdSO4 solution? Assume that exactly 1 ml water = 1 g water at this temperature. Do these numbers 1 mL and 1 g only have 1 significant figure?

Diff: 1

Section: 12.5

97) How many grams of water are needed to dissolve 13.9 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?

Diff: 2

Section: 12.5

98) Calculate the molality of a 25.0% by mass solution of MgCl2 in H2O. The density of this solution is 1.127 g/mL.

Diff: 2

Section: 12.5

99) What is the mole fraction of ethylene glycol, C2H6O2, in an aqueous solution that is 50.0% ethylene glycol by mass?

Diff: 2

Section: 12.5

100) Consider a 0.80 M Al(NO3)3 solution. What would be the concentration of aluminum and nitrate ions in this solution? The aluminum concentration would be ________ M and the nitrate concentration would be ________ M.

Diff: 1

Section: 12.5

101) A solution of chloroform (CHCl3) and acetone ((CH3)2CO) exhibits a negative deviation from Raoult's law similar to that shown on the diagram below. What does this tell us about the solution with respect to it being an ideal solution and the strength of the interactions between acetone and chloroform? This result shows us that the solution is ________ ideal and the interactions between chloroform-chloroform and acetone-acetone are ________ than the interactions between chloroform/acetone.

Diff: 2

Section: 12.6

102) Arrange the following solutions in order of increasing freezing point depression.

0.05 m KCl; 0.5 m CH3CO2H; 0.15 m (NH4)2SO4.

Diff: 2

Section: 12.6

103) Calculate the freezing point of a solution made from 22.0 g of octane (C8H18) dissolved in 248.0 g of benzene. Benzene freezes at 5.50 °C and its Kf value is 5.07 °C/m.

Diff: 2

Section: 12.6

104) Describe what a colligative property is. Explain Raoult's Law in terms of colligative properties.

Diff: 1

Section: 12.6

105) Raoult's law expresses that the higher the mole fraction of a volatile solvent in an ideal solution the ________ its partial pressure above the solution.

Diff: 1

Section: 12.6

106) The van't Hoff factor describes to what degree an ionic substance ________.

Diff: 2

Section: 12.6

107) The higher the concentration of an ionic substance in solution, the ________ the van't Hoff factor deviates from its predicted value.

Hint: Dilute solutions behave closer to expectations than concentrated solutions.

Diff: 3

Section: 12.6

108) An unknown ionic substance has a van't Hoff factor of 1.89 with a concentration of 0.1 M. If the concentration of this ionic substance was increased to 0.5 M the value of the van't Hoff factor would ________.

Hint: Dilute solutions behave closer to expectations than concentrated solutions.

Diff: 3

Section: 12.6

109) The Tyndall effect is a light effect that is observed with what kind of solution or mixture?

Diff: 2

Section: 12.7

110) A mixture of fine sand in water would best be classified as what kind of solution or mixture?

Diff: 1

Section: 12.7

111) What properties does a molecule need to form a micelle?

Diff: 1

Section: 12.7

112) Substances that dissolve more readily in oils than in water is best classified as ________.

Diff: 2

Section: 12.7

113) Substances that have both a hydrophobic end and a hydrophilic end to their molecular structure can form a spherical structure called a ________.

Diff: 1

Section: 12.7

114) One driving force toward formation of homogeneous gas mixtures is the spontaneity of mixing through random motion of small molecules.

Diff: 1

Section: 12.1

115) Liquids that are mutually miscible possess intermolecular forces of similar type and magnitude.

Diff: 1

Section: 12.1

116) The process, MgSO4(s) → MgSO4(aq), is an endothermic process. The solubility of magnesium sulfate in water should therefore increase as the solvent temperature is increased.

Diff: 1

Section: 12.3

117) Colligative properties depend primarily on the concentration of solute particles in the solution.

Diff: 1

Section: 12.6

118) Hydrophobic substances mix easily with water.

Diff: 1

Section: 12.7

119) All substances can be either hydrophobic or hydrophilic, but not both.

Diff: 1

Section: 12.7

120) After spilling vegetable oil on your clothes, you need to remove the stain. You have the following items available to help with the stain: vinegar (CH3CO2H); water (H2O); and toluene (C6H6). Which would be the best choice of solvent to remove the oil, and why?

Diff: 1

Section: 12.1

121) Many marine organisms that require oxygen live at ocean depths where high pressures and low temperatures would not support humans. Based on what you have learned in this chapter, what is one reason that organisms can survive under these conditions?

Diff: 1

Section: 12.3

122) The solubility of oxygen in lakes high in mountains is affected by the altitude. If the solubility of O2 from the air is 2.67 × 10−4 M at sea level when the pressure is 1.000 atm and the temperature is 25°C, what is the solubility of O2 at an elevation of 12,000 ft. where the atmospheric pressure is 0.643 atm? Assume the temperature is 25°C, and that the mole fraction of O2 in air is 0.211 at both 12,000 ft. and at sea level.

Diff: 2

Section: 12.4

123) What is the freezing point of an aqueous solution of a nonvolatile solute that has a boiling point of 102.45 °C? Kf = 1.86 °C m−1 and Kb = 0.512 °C m−1.

Diff: 2

Section: 12.6

124) Pure benzene, C6H6, has a molar mass of 78.114 g mol−1, a density of 0.8765 g mL−1, a freezing point of 5.45°C, and a boiling point of 80.2°C. Its freezing point depression and boiling point elevation constants are: Kf = 5.07°C m−1; Kb = 2.53°C m−1. A solution was made by taking 33.88 g of an unknown nonelectrolyte and dissolving it in 175.0 g of benzene. The measured freezing point of the solution was -1.65°C. Calculate the molar mass of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

Diff: 3

Section: 12.6

125) A solution is made by mixing acetone and methyl alcohol, whose structures are shown as I and II below. Which statement below best describes the characteristics of this mixture?

I. CH3–C–CH3 II. CH3–O–H

║

O

Hint: How does the strength of hydrogen bonds compare to other intermolecular forces?

A) Because the intermolecular forces for acetone-acetone, methanol-methanol, and acetone-methanol interactions are comparable, the mixture behaves as an ideal solution.

B) Because the intermolecular forces for acetone-methanol interactions are weaker than either the acetone-acetone or methanol-methanol interactions we should observe a positive deviation from ideal solution behavior.

C) Because the intermolecular forces for acetone-methanol and methanol-methanol interactions are stronger than the acetone-acetone interactions due to hydrogen bonding, the experimental vapor pressure would be lower than calculated by Raoult's Law.

D) Because the intermolecular forces for methanol-methanol interactions are weaker than either the acetone-acetone or acetone-methanol interactions due to hydrogen bonding, the experimental vapor pressure would be greater than calculated by Raoult's Law.

E) Because the intermolecular forces for methanol-methanol interactions are weaker than either the acetone-acetone or acetone-methanol interactions due to hydrogen bonding, the experimental vapor pressure would be lower than calculated by Raoult's Law.

Diff: 3

Section: 12.6

126) Pure cyclohexane, C6H12, has a molar mass of 84.161 g mol−1, a density of 0.7785 g mL−1, a freezing point of 6.53°C, and a boiling point of 80.72°C. Its freezing point depression and boiling point elevation constants are: Kf = 20.0 °C m−1; Kb = 2.69 °C m−1. A solution was made by taking 9.50 g of an unknown nonelectrolyte and dissolving it in 125.0 g of cyclohexane. The measured freezing point of the solution was -0.78°C. Calculate the molar mass of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

Diff: 3

Section: 12.6

127) Pure cyclohexane, C6H12, has a molar mass of 84.161 g mol−1 and a density of

0.7785 g mL−1, a freezing point of 6.53°C, and a boiling point of 80.72°C. Its freezing point depression and boiling point elevation constants are: Kf = 20.0 °C m−1; Kb = 2.69°C m−1. A solution was made by taking 15.46 g of an unknown nonelectrolyte and dissolving it in 125.0 g of cyclohexane. The measured freezing point of the solution was -4.28°C. Calculate the molar mass of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

Diff: 3

Section: 12.6

128) Pure glacial acetic acid, HC2H3O2, has a molar mass of 60.052 g mol−1 and a density of 1.0492 g mL−1, a freezing point of 16.62°C, and a boiling point of 118.3°C. Its freezing point depression and boiling point elevation constants are: Kf = 3.57 °C m−1; Kb = 3.07 °C m−1. A solution was made by taking 19.51 g of an unknown nonelectrolyte and dissolving it in 181.0 g of glacial acetic acid. The measured freezing point of the solution was 8.64°C. Calculate the molar mass of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

Diff: 3

Section: 12.6

129) Pure benzene, C6H6, has a molar mass of 78.114 g mol−1, a density of 0.8765 g mL−1, a freezing point of 5.45°C, and a boiling point of 80.2°C. Its freezing point depression and boiling point elevation constants are: Kf = 5.07 °C m−1; Kb = 2.53 °C m−1. A solution was made by taking 16.00 g of an unknown nonelectrolyte and dissolving it in 168.0 g of benzene. The measured freezing point of the solution was -1.05°C. Calculate the molar mass of the unknown substance.

Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

Diff: 3

Section: 12.6

130) An aqueous solution of fructose, C6H12O6, has an osmotic pressure of 10.50 atm at 25.0°C. What is the boiling point of this solution? Kf = 1.86 °C m−1 and Kb = 0.512 °C m−1. The density of the solution is 1.077 g mL−1.

Hint: You will need to convert molarity to molality before finding the boiling point elevation.

Diff: 3

Section: 12.6

131) A solution is made by dissolving 48.07 g of MgSO4∙7H2O in 250.0 grams of water. What is the expected freezing point of this solution if the van't Hoff factor is 1.90? Kf = 1.86 °C m−1.

Hint: Don't forget to include the 7H2O when calculating the molar mass of solute.

Diff: 3

Section: 12.6

132) How many liters of ethylene glycol antifreeze (C2H6O2), with a density of 1.100 g/mL, would you add to your car radiator containing 15.0 kg of water if you needed to protect your engine to -21.5°C? For water, Kf = 1.86°C m−1.

Hint: Find molality, moles, grams of solute, milliliters, then liters, making sure to pay attention to units.

Diff: 3

Section: 12.6

133) A chemist isolates a new compound with an empirical formula C3H3O2. 3.52 g of the unknown compound was dissolved in 100.0 g of water. The resulting solution had a freezing point of -0.307°C. What is the molecular formula for the unknown compound?

For water, Kf = 1.86°C m−1.

Hint: Use the freezing point depression to find the molar mass of the compound, then find the molecular formula as in Chapter 3.

Diff: 3

Section: 12.6

134) A phase diagram, like the one below, for an aqueous solution shows a shift in the solid/liquid and liquid/gas phase transitions but there is not a shift in the sublimation line between solids and gases. Explain what causes the shift in the other transitions and why this does not change a shift in the solid/gas transition.

Hint: Think about the difference in vapor pressure between a solution and a pure substance.

Diff: 3

Section: 12.6

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