CHAPTER 12
Solutions
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
12.1. An example of a solid solution prepared from a liquid and a solid is a dental filling made of liquid
mercury and solid silver.
12.2. The C4H9OH molecules will be more soluble in water because their –OH ends can form hydrogen
bonds with water.
12.3. The Na+ ion has a larger energy of hydration because its ionic radius is smaller, giving Na+ a
more concentrated electric field than K+.
12.4. Write Henry's law (S = kHP) for 159 mmHg (P2), and divide it by Henry's law for 1 atm, or 760
mmHg (P1). Then substitute the experimental values of P1, P2, and S1 to solve for S2.
S2 k P P
= H 2 = 2
S1 kH P1 P1
Solving for S2 gives
P2 S1 (159 mmHg)(0.0404 g O2 /L)
S2 = = = 8.452 10−3 = 8.45 10−3 g O2/L
P1 760 mmHg
12.5. The mass of HCl in 20.2% HCl (0.202 = fraction of HCl) is
0.202 35.0 g = 7.070 = 7.07 g HCl
The mass of H2O in 20.2% HCl is
35.0 g solution − 7.07 g HCl = 27.93 = 27.9 g H2O
12.6. Calculate the moles of toluene using its molar mass of 92.14 g/mol:
1 mol toluene
35.6 g toluene = 0.3863 mol toluene
92.14 g toluene
To calculate molality, divide the moles of toluene by the mass in kg of the solvent (C6H6):
0.3863 mol toluene
Molality = 3.0904 = 3.09 m toluene
0.125 kg solvent
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, 414 Chapter 12: Solutions
12.7. The number of moles of toluene = 0.3863 (previous exercise); the number of moles of benzene is
1 mol benzene
125 g benzene = 1.6003 mol benzene
78.11 g benzene
The total number of moles is 1.6003 + 0.3863 = 1.9866, and the mole fractions are
1.6003 mol benzene
Mole fraction benzene = 0.80554 = 0.806
1.9866 mol
0.3863 mol toluene
Mole fraction toluene = 0.1944 = 0.194
1.9866 mol
The sum of the mole fractions = 1.000.
12.8. This solution contains 0.120 moles of methanol dissolved in 1.00 kg of ethanol. The number of
moles in 1.00 kg of ethanol is
1 mol C2 H5OH
1.00 103 g C2H5OH = 21.706 mol C2H5OH
46.07 g C2 H5OH
The total number of moles is 21.706 + 0.120 = 21.826, and the mole fractions are
21.706 mol C2 H5OH
Mole fraction C2H5OH = = 0.994501 = 0.995
21.826 mol
0.120 mol CH3OH
Mole fraction CH3OH = = 0.005498 = 0.00550
21.826 mol
The sum of the mole fractions is 1.000.
12.9. One mole of solution contains 0.250 mol methanol and 0.750 mol ethanol. The mass of this
amount of ethanol, the solvent, is
46.07 g C2 H5OH
0.750 mol C2H5OH = 34.55 g C2H5OH (0.03455 kg)
1 mol C2 H5OH
The molality of methanol in the ethanol solvent is
0.250 mol CH3OH
= 7.2358 = 7.24 m CH3OH
0.03455 kg C2 H5OH
12.10. Assume an amount of solution contains 1 kg of water. The mass of urea in this mass is
60.05 g urea
3.42 mol urea = 205.4 g urea
1 mol urea
The total mass of solution is 205.4 + 1000.0 g = 1205.4 g. The volume and molarity are
1 mL
Volume of solution = 1205.4 g = 1153.49 mL = 1.15349 L
1.045 g
3.42 mol urea
Molarity = = 2.9649 mol/L = 2.96 M
1.15349 L solution
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 12: Solutions 415
12.11. Assume a volume equal to 1.000 L of solution. Then
Mass of solution = 1.029 g/mL (1.000 103 mL) = 1029 g
60.05 g urea
Mass of urea = 2.00 mol urea = 120.1 g urea
1 mol urea
Mass of water = (1029 − 120.1) g = 908.9 g water (0.9089 kg)
2.00 mol urea
Molality = = 2.2004 = 2.20 m urea
0.9089 kg solvent
12.12. Calculate the moles of naphthalene and the moles of chloroform:
1 mol C10 H8
0.515 g C10H8 = 0.004018 mol C10H8
128.17 g C10 H8
1 mol CHCl3
60.8 g CHCl3 = 0.50929 mol CHCl3
119.38 g CHCl3
The total number of moles is 0.004018 + 0.50929 = 0.5133 mol, and the mole fraction of
chloroform is
0.50929 mol CHCl3
Mole fraction CHCl3 = = 0.9921
0.5133 mol
0.004018 mol C10 H8
Mole fraction C10H8 = = 0.007828
0.5133 mol
The vapor-pressure lowering is
P = P X C10 H8 = (156 mmHg)(0.007828) = 1.221 = 1.22 mmHg
Use Raoult's law to calculate the vapor pressure of chloroform:
P = P X CHCl3 = (156 mmHg)(0.9921) = 154.7 = 155 mmHg
12.13. Solve for cm in the freezing-point-depression equation (T = Kfcm; Kf in Table 12.3):
T 0.150C
cm = = = 0.8073 m
Kf 1.858C/m
Use the molal concentration to solve for the mass of ethylene glycol:
0.08073 mol glycol
0.0378 kg solvent = 0.003051 mol glycol
1 kg solvent
62.1 g glycol
0.003051 mol glycol = 1.894 10−1 = 1.89 10−1 g glycol
1 mol glycol
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 416 Chapter 12: Solutions
12.14. Calculate the moles of ascorbic acid (vitamin C) from the molality, and then divide the mass of
0.930 g by the number of moles to obtain the molar mass:
0.0555 mol vit. C
0.0950 kg H2O = 0.005272 mol vit C
1 kg H 2 O
0.930 g vit. C
= 176.4 = 176 g/mol
0.005272 mol vit. C
The molecular weight of ascorbic acid, or vitamin C, is 176 amu.
12.15. The molal concentration of white phosphorus is
T 0.159°C
cm = = = 0.06625 m
Kb 2.40C/m
The number of moles of white phosphorus (Px) present in this solution is
0.06625 mol Px
0.0250 kg CS2 = 0.001656 mol Px
1 kg CS2
The molar mass of white phosphorus equals the mass divided by moles:
0.205 g 0.001656 mol = 123.77 = 124 g/mol
Thus, the molecular weight of Px is 124 amu. The number of P atoms in the molecule of white
phosphorus is obtained by dividing the molecular weight by the atomic weight of P:
123.77 amu Px
= 3.9965 = 4.00
30.97 amu P
Hence, the molecular formula is P4 (x = 4).
12.16. The number of moles of sucrose is
1 mol sucrose
5.0 g sucrose = 0.0146 mol sucrose
342.3 g sucrose
The molarity of the solution is
0.0146 mol sucrose
= 0.146 M sucrose
0.100 L
The osmotic pressure, , is equal to MRT and is calculated as follows:
0.146 mol sucrose 0.0821 L atm
293 K = 3.51 = 3.5 atm
1L K mol
12.17. The number of ions from each formula unit is i. Here,
i=1+2=3
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solutions
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
12.1. An example of a solid solution prepared from a liquid and a solid is a dental filling made of liquid
mercury and solid silver.
12.2. The C4H9OH molecules will be more soluble in water because their –OH ends can form hydrogen
bonds with water.
12.3. The Na+ ion has a larger energy of hydration because its ionic radius is smaller, giving Na+ a
more concentrated electric field than K+.
12.4. Write Henry's law (S = kHP) for 159 mmHg (P2), and divide it by Henry's law for 1 atm, or 760
mmHg (P1). Then substitute the experimental values of P1, P2, and S1 to solve for S2.
S2 k P P
= H 2 = 2
S1 kH P1 P1
Solving for S2 gives
P2 S1 (159 mmHg)(0.0404 g O2 /L)
S2 = = = 8.452 10−3 = 8.45 10−3 g O2/L
P1 760 mmHg
12.5. The mass of HCl in 20.2% HCl (0.202 = fraction of HCl) is
0.202 35.0 g = 7.070 = 7.07 g HCl
The mass of H2O in 20.2% HCl is
35.0 g solution − 7.07 g HCl = 27.93 = 27.9 g H2O
12.6. Calculate the moles of toluene using its molar mass of 92.14 g/mol:
1 mol toluene
35.6 g toluene = 0.3863 mol toluene
92.14 g toluene
To calculate molality, divide the moles of toluene by the mass in kg of the solvent (C6H6):
0.3863 mol toluene
Molality = 3.0904 = 3.09 m toluene
0.125 kg solvent
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 414 Chapter 12: Solutions
12.7. The number of moles of toluene = 0.3863 (previous exercise); the number of moles of benzene is
1 mol benzene
125 g benzene = 1.6003 mol benzene
78.11 g benzene
The total number of moles is 1.6003 + 0.3863 = 1.9866, and the mole fractions are
1.6003 mol benzene
Mole fraction benzene = 0.80554 = 0.806
1.9866 mol
0.3863 mol toluene
Mole fraction toluene = 0.1944 = 0.194
1.9866 mol
The sum of the mole fractions = 1.000.
12.8. This solution contains 0.120 moles of methanol dissolved in 1.00 kg of ethanol. The number of
moles in 1.00 kg of ethanol is
1 mol C2 H5OH
1.00 103 g C2H5OH = 21.706 mol C2H5OH
46.07 g C2 H5OH
The total number of moles is 21.706 + 0.120 = 21.826, and the mole fractions are
21.706 mol C2 H5OH
Mole fraction C2H5OH = = 0.994501 = 0.995
21.826 mol
0.120 mol CH3OH
Mole fraction CH3OH = = 0.005498 = 0.00550
21.826 mol
The sum of the mole fractions is 1.000.
12.9. One mole of solution contains 0.250 mol methanol and 0.750 mol ethanol. The mass of this
amount of ethanol, the solvent, is
46.07 g C2 H5OH
0.750 mol C2H5OH = 34.55 g C2H5OH (0.03455 kg)
1 mol C2 H5OH
The molality of methanol in the ethanol solvent is
0.250 mol CH3OH
= 7.2358 = 7.24 m CH3OH
0.03455 kg C2 H5OH
12.10. Assume an amount of solution contains 1 kg of water. The mass of urea in this mass is
60.05 g urea
3.42 mol urea = 205.4 g urea
1 mol urea
The total mass of solution is 205.4 + 1000.0 g = 1205.4 g. The volume and molarity are
1 mL
Volume of solution = 1205.4 g = 1153.49 mL = 1.15349 L
1.045 g
3.42 mol urea
Molarity = = 2.9649 mol/L = 2.96 M
1.15349 L solution
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 12: Solutions 415
12.11. Assume a volume equal to 1.000 L of solution. Then
Mass of solution = 1.029 g/mL (1.000 103 mL) = 1029 g
60.05 g urea
Mass of urea = 2.00 mol urea = 120.1 g urea
1 mol urea
Mass of water = (1029 − 120.1) g = 908.9 g water (0.9089 kg)
2.00 mol urea
Molality = = 2.2004 = 2.20 m urea
0.9089 kg solvent
12.12. Calculate the moles of naphthalene and the moles of chloroform:
1 mol C10 H8
0.515 g C10H8 = 0.004018 mol C10H8
128.17 g C10 H8
1 mol CHCl3
60.8 g CHCl3 = 0.50929 mol CHCl3
119.38 g CHCl3
The total number of moles is 0.004018 + 0.50929 = 0.5133 mol, and the mole fraction of
chloroform is
0.50929 mol CHCl3
Mole fraction CHCl3 = = 0.9921
0.5133 mol
0.004018 mol C10 H8
Mole fraction C10H8 = = 0.007828
0.5133 mol
The vapor-pressure lowering is
P = P X C10 H8 = (156 mmHg)(0.007828) = 1.221 = 1.22 mmHg
Use Raoult's law to calculate the vapor pressure of chloroform:
P = P X CHCl3 = (156 mmHg)(0.9921) = 154.7 = 155 mmHg
12.13. Solve for cm in the freezing-point-depression equation (T = Kfcm; Kf in Table 12.3):
T 0.150C
cm = = = 0.8073 m
Kf 1.858C/m
Use the molal concentration to solve for the mass of ethylene glycol:
0.08073 mol glycol
0.0378 kg solvent = 0.003051 mol glycol
1 kg solvent
62.1 g glycol
0.003051 mol glycol = 1.894 10−1 = 1.89 10−1 g glycol
1 mol glycol
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 416 Chapter 12: Solutions
12.14. Calculate the moles of ascorbic acid (vitamin C) from the molality, and then divide the mass of
0.930 g by the number of moles to obtain the molar mass:
0.0555 mol vit. C
0.0950 kg H2O = 0.005272 mol vit C
1 kg H 2 O
0.930 g vit. C
= 176.4 = 176 g/mol
0.005272 mol vit. C
The molecular weight of ascorbic acid, or vitamin C, is 176 amu.
12.15. The molal concentration of white phosphorus is
T 0.159°C
cm = = = 0.06625 m
Kb 2.40C/m
The number of moles of white phosphorus (Px) present in this solution is
0.06625 mol Px
0.0250 kg CS2 = 0.001656 mol Px
1 kg CS2
The molar mass of white phosphorus equals the mass divided by moles:
0.205 g 0.001656 mol = 123.77 = 124 g/mol
Thus, the molecular weight of Px is 124 amu. The number of P atoms in the molecule of white
phosphorus is obtained by dividing the molecular weight by the atomic weight of P:
123.77 amu Px
= 3.9965 = 4.00
30.97 amu P
Hence, the molecular formula is P4 (x = 4).
12.16. The number of moles of sucrose is
1 mol sucrose
5.0 g sucrose = 0.0146 mol sucrose
342.3 g sucrose
The molarity of the solution is
0.0146 mol sucrose
= 0.146 M sucrose
0.100 L
The osmotic pressure, , is equal to MRT and is calculated as follows:
0.146 mol sucrose 0.0821 L atm
293 K = 3.51 = 3.5 atm
1L K mol
12.17. The number of ions from each formula unit is i. Here,
i=1+2=3
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
