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Homework answers / question archive / Chapter 12  Properties of Solutions: Mixtures of Substances at the Molecular Level  Multiple Choice  1)Concerning the process of separating of a solid substance into its component units (molecules or ions),    the process is exothermic and the potential energy increases

Chapter 12  Properties of Solutions: Mixtures of Substances at the Molecular Level  Multiple Choice  1)Concerning the process of separating of a solid substance into its component units (molecules or ions),    the process is exothermic and the potential energy increases

Chemistry

Chapter 12  Properties of Solutions: Mixtures of Substances at the Molecular Level

 Multiple Choice

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

 

          1. the process is exothermic and the potential energy increases.
          2. the process is exothermic and the potential energy decreases.           c. the process is endothermic and the potential energy increases.
          1. the process is endothermic and the potential energy decreases.
          2. the process is endothermic and occurs with no change in potential energy. 

 

 

  1. 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.

          1. the process is endothermic and the potential energy increases.
          2. the process is endothermic and the potential energy decreases.
          3. the process is endothermic and occurs with no change in potential energy. 

 

 

  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?

 

          1. CH3?C?CH3

                                    ?

                                   O

          1. CH3?O?H
          2. CH3?C?H

                                    ?             

                                   O

          1. H?O?CH2?CH2?O?H

                       e. CH3?CH2?CH2?CH2?CH2?CH2?CH2?CH3

 

 

  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?

 

          1. H?O?H
          2. CH3?O?H
          3. CF3?O?H
          4. H?O?CH2?CH2?O?H

                       e. CH3?CH2?CH2?CH2?CH2?CH2?CH2?CH3

 

 

 

  1. 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

 

          1. dilute.
          2. concentrated.
          3. unsaturated.
          4. saturated.

                   e. supersaturated.

 

 

  1. A solution in a beaker has some undissolved 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

 

          1. dilute
          2. concentrated
          3. unsaturated

                  d. saturated

                    e. supersaturated

 

 

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

 

          1. dilute
          2. concentrated    c. unsaturated    d. saturated

                    e. supersaturated

 

 

  1. 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 

 

          1. dilute
          2. concentrated
          3. unsaturated
          4. saturated

                   e. supersaturated

 

  1. Which causes 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

          1. increasing the temperature of the solvent and simultaneously increasing the pressure of the gas in the space above the solvent
          2. decreasing the temperature of the solvent and simultaneously decreasing the pressure of the gas in the space above the solvent
          3. increasing the temperature of the solvent while maintaining the pressure of the gas in the space above the solvent at a set value

 

  1. 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 readjusted to 1000 torr?

 

   a. 0.00289 g L-1    b. 0.00500 g L-1    c. 1.49 g L-1

          1. 2.89 x 103 g L-1
          2. 3.46 x 103 g L-1

 

  1. 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 x 10-6 g L-1                     b. 7.98 x 10-4 g L-1                    c. 1.33 x 10-3 g L-1

          1. 0.606 g L-1
          2. 1.01 g L-1

 

  1. 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?

 

          1. 5.2 x 10-2 g/100 g H2O
          2. 5.0 x 10-4 g/100 g H2O
          3. 6.7 x 10-5 g/100 g H2O

 

                         d. 6.8 x 10-5 g/100 g H2O                                e. 9.6 x 10-4 g/100 g H2O

 

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

 

   a. mole fraction    b. molarity    c. molality

          1. mass fraction
          2. percent by weight

 

 

  1. 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

          1. 0.3333
          2. 0.4490

 

 

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

       

          1. 0.300 M
          2. 0.900 M

   c. 2.70 M    d. 3.60 M

                     e. 8.10 M

 

 

  1. 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 %

          1. 28.0 %
          2. 29.2 %
          3. 41.8 %

 

  1. 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

 

          1. 28.0 %
          2. 23.3 %
          3. 27.0 %    d. 21.3 %    e. 37.1 %

 

 

  1. A solution of sodium hydroxide is prepared by mixing 2.00 g of LiOH with 10.0 g of water.  The percent, by mass, of LiOH is

 

          1. 10.7 %
          2. 12.0 %    c. 16.7 %    d. 20.0 %

                    e. 80.0 %

 

 

  1. How many grams of NaC2H3O2  should be dissolved in 400.0 g of water to prepare a solution which is 11.28 % NaC2H3O2 by mass?

 

          1. 3.146 g
          2. 7.558 g
          3. 21.17 g    d. 50.86 g

                    e. 127.15 g

 

 

  1. 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?

 

          1. 0.283 m
          2. 0.000256 m 
          3. 0.245 m      d. 0.256 m      e. 0.351 m 

 

 

 

  1. A solution of ethylene glycol (C2H6O2) in water is 3.981 molar and has a density of 1.0296 g mL-1.  Calculate the percent, by weight, of ethylene glycol in the solution.

 

          1. 3.867 %
          2. 4.099 %
          3. 15.14 %

   d. 24.00 %    e. 25.45 %

 

  1. 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.

 

          1. 7.939 %
          2. 17.21 %
          3. 24.47 %
          4. 29.56 %    e. 34.73 %

 

 

  1. 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

          1. 5.152 moles
          2. 6.355 moles

 

 

  1. 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.

 

          1. 4.000 m
          2. 4.124 m
          3. 4.252 m    d. 5.092 m    e. 14.48 m

 

 

  1. 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

          1. 4.252 M
          2. 5.092 M
          3. 14.48 M

 

  1. 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

          1. 0.584 M
          2. 0.521 M
          3. 0.465 M

 

 

  1. An aqueous solution of glycerol, C3H8O3, is 48.0% glycerol by mass and has a density of  1.120              g mL-1.  Calculate the molality of the glycerol solution.

 

          1. 11.2 m
          2. 5.84 m
          3. 0.584 m
          4. 0.521 m

                   e. 10.0 m

 

 

  1. 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 weight of NaNO3 in the solution?

 

   a. 10.16 %    b. 10.82 %    c. 11.61 %

          1. 14.19 %
          2. 26.56 %

 

 

  1. 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?

 

          1. 0.6474 m
          2. 0.7004 m
          3. 1.320 m    d. 1.428 m    e. 1.545 m

 

 

  1. 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?

 

          1. 1.186 M
          2. 1.273 M
          3. 1.350 M

   d. 1.366 M    e. 1.428 M

 

  1. 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%

                    d.  20.00 %

 

 

  1. 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?

 

          1. 2.380 %
          2. 13.82 %    c. 15.00 %    d. 22.29 %

                    e. 44.38 %

 

 

  1. 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.

          1. molality  of the solvent.
          2. molarity of the solvent.
          3. osmotic pressure of the solute.

 

 

  1. When a 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.

          1. an increase in the freezing point of the liquid.
          2. a decrease in the boiling point of the liquid.
          3. scattering of light beams by the solute particles in the solution.

 

 

  1. 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 which 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

          1. 144 torr
          2. 239 torr

 

 

  1. 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

          1. 149 torr
          2. 277 torr

 

 

  1. 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 120.0 g of C3H7SH and 80.0 g CH3CN, if Raoult's Law is obeyed?

 

   a. 131 torr    b. 135 torr    c. 142 torr

          1. 146 torr
          2. 277 torr

 

 

  1. 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?

 

   a. 121 torr    b. 131 torr

          1. 139 torr
          2. 146 torr
          3. 156 torr

 

 

 

  1. 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?  

 

   a. 203 torr    b. 243 torr    c. 262 torr

          1. 275 torr
          2. 610 torr

 

 

  1. 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?

 

          1. 198 torr
          2. 228 torr    c. 282 torr    d. 325 torr

                    e. 425 torr

 

 

  1. What is the expected freezing point of a solution that contains 25.0 g of fructose, C6H12O6, in 250.0 g of H2O?   Kf = 1.86 oC m-1.

 

          1. -0.10 oC
          2. +0.10 oC
          3. -0.186 oC
          4. +0.186 oC

                   e. -1.03 oC

 

 

  1. A nonionic solute with a molecular weight 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 = 0.51 °C  m-1

 

   a. 30. grams    b. 75. grams

          1. 100 grams
          2. 125 grams
          3. 150 grams

 

  1. 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 = 0.51 °C m-1.

 

          1. -0.67 °C
          2. -0.99 °C
          3. -2.45 °C
          4. -2.99 °C    e. -8.94 °C

 

 

  1. 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.

 

   a. 138 g mol-1    b. 145 g mol-1

          1. 258 g mol-1
          2. 272 g mol-1
          3. 595 g mol-1

 

 

  1. 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.

 

          1. 27.8 g mol-1
          2. 46.8 g mol-1

   c. 208 g mol-1    d. 264 g mol-1

                     e. 1949 g mol-1

 

 

  1. 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.

 

          1. 61.8 g mol-1
          2. 66.8 g mol-1

   c. 229 g mol-1    d. 578 g mol-1

                     e. 1099 g mol-1

 

  1. 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 non-electrolyte 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.

 

          1. 24.4 g mol-1
          2. 30.5 g mol-1
          3. 45.3 g mol-1    d. 48.2 g mol-1    e. 174 g mol-1

 

 

  1. 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.

 

   a. 74.3 g mol-1    b. 82.7 g mol-1

          1. 110 g mol-1
          2. 122 g mol-1
          3. 460 g mol-1

 

 

  1. 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 = 0.51 °C m-1.  

 

   a. 101.3 oC    b. 101.6 oC    c. 105.1 oC

          1. 106.0 oC
          2. 156.5 oC

 

 

  1. 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.

 

   a. 83.6 g mol-1    b. 113 g mol-1

          1. 114 g mol-1
          2. 120 g mol-1   
          3. 158 g mol-1

 

  1. A solution contains 221 g of glycerol (C3H8O3) in 600 grams of water.  The Kf is 1.86 °C m-1 and Kb is 0.51 °C m-1.  What should the boiling point of the solution be?

 

          1. 100.02 °C
          2. 100.73 °C
          3. 101.65 °C    d. 102.04 °C    e. 103.62 °C

 

 

  1. A solution, which was made by dissolving 62.07 g of a non-electrolyte 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 0.51 °C m-1.  

 

          1. 57.7 g mol-1
          2. 62.07 g mol-1
          3. 115 g mol-1    d. 124 g mol-1    e. 231 g mol-1

 

 

  1. 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 0.51 °C m-1.  

 

   a. 1.60 moles    b. 2.00 moles

          1. 2.50 moles
          2. 2.98 moles
          3. 4.65 moles

 

 

  1. Which can be used to calculate the osmotic pressure of a solution?

 

          1. de Broglie equation
          2. Tyndall factor
          3. Peters equation
          4. van der Waals equation    e. van't Hoff equation
  1. 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 0.51   °C m-1.  

 

          1. 22.1 grams
          2. 93.6 grams

   c. 138 grams    d. 384 grams

                     e. 478 grams

 

 

  1. Which property of a solution is not a colligative property?

 

                   a. solubility of a solute

          1. freezing point depression
          2. boiling point elevation
          3. osmotic pressure
          4. vapor pressure lowering

 

 

  1. During osmosis:

       

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

          1. pure solute passes through a membrane but solvent does not
          2. pure solvent moves in one direction through the membrane while the solution moves through the membrane in the other direction
          3. pure solvent moves in one direction through the membrane while the solute moves through the membrane in the other direction
          4. pure solute moves in one direction through the membrane while the solution moves through the membrane in the other direction

 

 

  1. 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

          1. 475 torr
          2. 1217 torr

 

 

 

  1. 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 weight of this unknown compound?

 

          1. 94.3 g mol-1
          2. 124 g mol-1
          3. 943 g mol-1

                         d. 1.19 x 103 g mol-1                           e. 1.57 x 103 g mol-1

 

 

 

  1. Which aqueous solution will have the lowest freezing point temperature?

 

          1. 0.100 molal NaBr(aq)
          2. 0.100 molal MgSO4(aq)
          3. 0.150 molal KClO3(aq)

                   d. 0.150 molal MgCl2(aq)

                     e. 0.250 molal sucrose(aq)

 

 

  1. 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.  The solute is completely ionized. 

        If Kf = 1.86 °C m-1, what is the expected freezing point of the solution?

 

          1. -2.65 °C
          2. -2.87 °C    c. -5.31 °C    d. -5.75 °C

                     e. -7.97 °C

 

 

  1. Which aqueous solution will have the highest freezing point temperature?

 

   a. 0.100 molal MgBr2(aq)    b. 0.100 molal MgSO4(aq)     c. 0.150 molal KClO3(aq)

          1. 0.150 molal MgCl2(aq)
          2. 0.250 molal sucrose(aq)

 

  1. Which aqueous solution will have the highest boiling point temperature?

 

          1. 0.100 molal NiBr2(aq)
          2. 0.100 molal MgSO4(aq)
          3. 0.150 molal NH4NO3(aq)

                   d. 0.150 molal Na2SO4(aq)

                     e. 0.250 molal CH3OH(aq), methanol

 

 

  1. Which aqueous solution will have the lowest boiling point temperature?

 

          1. 0.100 molal NiBr2(aq)
          2. 0.200 molal MgSO4(aq)
          3. 0.150 molal NH4NO3(aq)
          4. 0.150 molal Na2SO4(aq)

                   e. 0.250 molal CH3OH(aq), methanol

 

 

  1. Which aqueous solution has the highest boiling point?  Kf = 0.51 °C m-1.

 

          1. 0.200 m KCl(aq)
          2. 0.200 m Na2SO4(aq)
          3. 0.200 m Ca(NO3)2(aq)
          4. 0.200 m C3H8O3 (aq), glycerol

                   e. 0.200 m Na3PO4(aq)

 

 

  1. Which solution has the highest osmotic pressure at 25 °C?

 

          1. 0.200 M (NH4)2SO4 
          2. 0.200 M KClO3
          3. 0.200 M C2H6O, ethanol
          4. 0.200 M NiF2

                   e. 0.200 M Fe(NO3)3

 

 

  1. 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. ethylene glycol < Mg(NO3)2  < LiCl

          1. Mg(NO3)2   < LiCl < ethylene glycol
          2. ethylene glycol < LiCl < Mg(NO3)2 
          3. LiCl < ethylene glycol < Mg(NO3)2 
          4. Mg(NO3)2  < ethylene glycol < LiCl

 

  1. A solution was made by mixing 24.40 g of NaCl, 19.77 g of MgCl2, and 1.665 kg of water. What is the colligative molality (moles of particles per kilogram of solvent) of the solution?

 

                     a. 0.5371 molal

   b. 1.360 molal    c. 1.418 molal

          1. 1.452 molal
          2. 4.297 molal

 

 

  1. A solution was made by mixing 24.40 g of NaCl, 19.77 g of MgCl2, and 1.665 kg of water.    What is the expected freezing point of the solution?

 

   a. -1.00 °C    b. -2.53 °C    c. -2.64 °C

          1. -2.70 °C
          2. -7.99 °C

 

 

  1. 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.

 

          1. -0.80 °C
          2. -2.66 °C
          3. -3.06 °C    d. -8.85 °C    e. -9.97 °C

 

 

  1. 9.92 grams of a compound with a molecular weight of 124.0 grams/mole when dissolved in 150.0 grams of water gave a solution which had a freezing point of -2.69 °C.  Calculate the experimental value of the van’t Hoff factor.

 

          1. 1.21
          2. 1.45
          3. 1.54
          4. 2.69    e. 2.71

 

  1. 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.

 

          1. -0.45 °C
          2. -1.57 °C
          3. -3.44 °C
          4. -5.23 °C    e. -17.4 °C

 

 

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

 

          1. 1.59
          2. 1.86    c. 1.64    d. 1.75

                    e. 1.70

 

 

  1. Which aqueous solution has the highest osmotic pressure?

 

   a. 0.100 molar Al(NO3)3    b. 0.150 molar Ba(NO3)2    c. 0.100 molar CaCl2

          1. 0.150 molar NaCl
          2. 0.200 molar NH3 

 

 

  1. 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, in torr, at 20.0 °C?

 

          1. 1.79 torr
          2. 3.49 torr
          3. 26.2 torr    d. 78.6 torr

                     e. 2.65 x 103 torr

 

 

 

 

Fill in the Blanks

 

 

 

  1. 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.  ________

 

 

  1. What is the mole fraction of ethylene glycol, C2H6O2, in an aqueous solution which is 50.0% ethylene glycol by mass?  ________

 

 

True and False

 

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

 

  1. Liquids which are mutually miscible possess intermolecular forces of similar type and magnitude.   

 

 

  1. 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.   ___ 

 

 

  1. Colligative properties depend primarily on the concentration of solute particles in the solution.  ___

 

Critical Thinking Questions

 

 

  1. 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?

 

   a. 3.147 molar    b. 4.371 molar

      1. 4.223 molar
      2. 2.938 molar
      3. 3.651 molar

 

 

  1. Pure benzene, C6H6, has a molecular weight 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 oC 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 oC.  Calculate the molecular weight of the unknown substance.

 

  a. 138 g mol-1    b. 145 g mol-1

      1. 258 g mol-1
      2. 272 g mol-1
      3. 595 g mol-1

 

 

  1. A solution is made by mixing acetone and methyl alcohol, whose structures are shown as I and II below.   Which statement below describes the most probable state of affairs?

 

                   I.   CH3?C?CH3                                     II.    CH3?O?H

                                   ?             

                                   O

 

      1. Since the intermolecular forces for acetone?acetone, methanol?methanol, and acetone?methanol interactions are comparable, the mixture behaves as an ideal solution.
      2. Since 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. Since 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.

      1. Since 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.
      2. Since 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.

 

 

  1. Pure cyclohexane, C6H12, has a molecular weight 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 molecular weight of the unknown substance.

 

      1. 27.8 g mol-1
      2. 46.8 g mol-1

   c. 208 g mol-1    d. 264 g mol-1

                     e. 1949 g mol-1

 

 

  1. Pure cyclohexane, C6H12, has a molecular weight 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 molecular weight of the unknown substance.

 

      1. 61.8 g mol-1
      2. 66.8 g mol-1

   c. 229 g mol-1    d. 578 g mol-1

                     e. 1099 g mol-1

 

 

  1. Pure glacial acetic acid, HC2H3O2, has a molecular weight 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 oC 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 molecular weight of the unknown substance.

 

      1. 24.4 g mol-1
      2. 30.5 g mol-1
      3. 45.3 g mol-1    d. 48.2 g mol-1    e. 174 g mol-1

 

 

  1. Pure benzene, C6H6, has a molecular weight 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 molecular weight of the unknown substance.

 

   a. 74.3 g mol-1    b. 82.7 g mol-1

      1. 110 g mol-1
      2. 122 g mol-1
      3. 460 g mol-1

 

 

 

 

 

  1. 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 = 0.51 °C m-1.  The density of the solution is 1.077 g mL-1.  

 

   a. 100.22 °C    b. 100.41 °C

      1. 100.43 °C
      2. 100.52 °C
      3. 100.75 °C

 

 

  1. 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.  ________  

 

 

Short Answers

 

  1. 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? 

 

 

  1. Many marine organisms live at ocean depths where high pressures and low temperatures would not support humans. Based on what you've learned in this chapter, what is one reason for this? 

 

 

  1. Explain the significance of the van't Hoff factor. 

 

 

  1. You are given four unlabeled beakers, containing the following substances: vinegar (CH3COOH); water (H2O); toluene (C7H8); and ethanol (CH3CH2OH). Using what you know about miscibility, devise a way of figuring out the identity of the liquids in each beaker. Hint: Vinegar has a very characteristic odor! 

 

 

  1. Describe two everyday, common examples of boiling point elevation or freezing point depression. 

 

 

  1. Arrange the following solutions in order of increasing freezing point depression. Explain your reasoning! 0.05 m KCl; 0.5 m CH3CO2H; 0.15 m (NH3)2SO4

 

 

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

 

 

 

Questions 98-102 refer to the following diagram.

 

 

Pressure/atm

 

 

 

 

 

  1. Is this the phase diagram for water? Why or why not? 

 

 

  1. At what temperatures do the normal boiling and freezing points occur? 

 

 

  1. At what temperatures do the triple point and critical point occur? 

 

 

  1. At what point will the species stop subliming (going from a solid to a gas)? 

 

 

  1. What is true about the boiling point of this substance? 

 

 

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