CHAPTER 11
States of Matter; Liquids and Solids
■ 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.
11.1. First, calculate the heat required to vaporize 1.00 kg of ammonia:
1000 g 1 mol NH3 23.4 kJ
1.00 kg NH3 = 1374.04 kJ
1 kg 17.03 g NH3 1 mol
The amount of water at 0C that can be frozen to ice at 0C with this heat is
1 mol H 2 O 18.01 g H 2 O
1374.04 kJ = 4117.54g = 4.12 kg H2O
6.01 kJ 1 mol H 2 O
11.2. Use the two-point form of the Clausius-Clapeyron equation to calculate P2:
P2 26.8 103 J / mol 1 1
ln =
760 mmHg 8.31 J / (K mol) 319 K 308 K
1.12 104
= 3225 K = −0.36106
K
Converting to antilogs gives
P2
= antilog(−0.36106) = e−0.36106 = 0.69693
760 mmHg
P2 = 0.69693 760 mmHg = 529.6 = 5.3 102 mmHg
11.3. Use the two-point form of the Clausius-Clapeyron equation to solve for Hvap:
757 mmHg H vap 1 1
ln =
522 mmHg 8.31 J/(K mol) 368 K 378 K
H vap 7.2 105
0.37169 =
8.31 J/(K mol) K
Hvap = 4.3 104 J/mol (4 101 kJ/mol)
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, Chapter 11: States of Matter; Liquids and Solids 377
11.4. a. Liquefy methyl chloride by a sufficient increase in pressure below 144C.
b. Liquefy oxygen by compressing to 50 atm below −119C.
11.5. a. Propanol has a hydrogen atom bonded to an oxygen atom. Therefore, hydrogen bonding is
expected. Because propanol is polar (from the O–H bond), we also expect dipole-dipole
forces. Weak London forces exist, too, because such forces exist between all molecules.
b. Linear carbon dioxide is not polar, so only London forces exist among CO2 molecules.
c. Bent sulfur dioxide is polar, so we expect dipole-dipole forces; we also expect the usual
London forces.
11.6. The order of increasing vapor pressure is butane (C4H10), propane (C3H8), and ethane (C2H6).
Because London forces tend to increase with increasing molecular mass, we would expect the
molecule with the highest molecular mass to have the lowest vapor pressure.
11.7. Because ethanol has an H atom bonded to an O atom, strong hydrogen bonding exists in ethanol
but not in methyl chloride. Hydrogen bonding explains the lower vapor pressure of ethanol
compared to methyl chloride.
11.8. a. Zinc, a metal, is a metallic solid.
b. Sodium iodide, an ionic substance, exists as an ionic solid.
c. Silicon carbide, a compound in which carbon and silicon might be expected to form
covalent bonds to other carbon and silicon atoms, exists as a covalent network solid.
d. Methane, at room temperature a gaseous molecular compound with covalent bonds, freezes
as a molecular solid.
11.9. Only MgSO4 is an ionic solid; C2H5OH, CH4, and CH3Cl form molecular solids; thus MgSO4
should have the highest melting point. Of the molecular solids, CH4 has the lowest molecular
mass (16.0 amu) and would be expected to have the lowest melting point. Both C2H5OH and
CH3Cl have approximately the same molecular masses (46.0 amu vs. 50.5 amu), but C2H5OH
exhibits strong hydrogen bonding and, therefore, would be expected to have the higher melting
point. The order of increasing melting points is CH4, CH3Cl, C2H5OH, and MgSO4.
11.10. Each of the four corners of the cell contains one atom, which is shared by a total of four unit cells.
Therefore, the corners contribute one whole atom.
Atoms 1/4 atom
= 4 corners = 1 atom
Unit cell 1 corner
11.11. Use the edge length to calculate the volume of the unit cell. Then, use the density to determine the
mass of one atom. Divide the molar mass by the mass of one atom.
V = (3.509 10−10 m)3 = 4.321 10−29 m3
3
0.534 g 100 cm
d= 3
= 5.34 10 g/m
5 3
1 cm 1m
Mass of 1 unit cell = d V = (5.34 105 g/m3) (4.321 10−29 m3) = 2.3074 10−23 g
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, 378 Chapter 11: States of Matter; Liquids and Solids
There are two atoms in a body-centered cubic unit cell; thus, the mass of one lithium atom is
½ 2.3074 10−23 g = 1.1537 10−23 g
The known atomic mass of lithium is 6.941 amu, so Avogadro's number is
6.941 g/mol
NA = = 6.016 1023 = 6.02 1023 atoms/mol
1.1537 1023 g/atom
11.12. Use Avogadro's number to convert the molar mass of potassium to the mass per one atom.
39.0983 g K 1 mol K 6.4925 1023 g K
23
=
1 mol K 6.022 10 atoms 1 atom
There are two K atoms per unit cell; therefore, the mass per unit cell is
6.4925 1023 g K 2 atoms 1.2985 1022 g
=
1 atom 1 unit cell 1 unit cell
The density of 0.856 g/cm3 is equal to the mass of one unit cell divided by its unknown volume,
V. After solving for V, determine the edge length from the cube root of the volume.
1.2985 1022 g
0.856 g/cm3 =
V
1.2985 1022 g
V= = 1.517 10−22 cm3 (1.517 10−28 m3)
0.856 g/cm3
Edge length = 3 1.517 1028 m3 = 5.333 10−10 = 5.33 10−10 m (533 pm)
■ ANSWERS TO CONCEPT CHECKS
11.1. a. (1) At t = 0, since the system is not at equilibrium and there are no H2O molecules in the
gaseous state, you would expect the rate of evaporation to exceed the rate of
condensation. At t = 1, since evaporation has proceeded at a greater rate than
condensation, there must now be fewer molecules in the liquid state, resulting in a
lower level of H2O(l).
(2) At t = 1, since some of the H2O has gone into the vapor state, the vapor pressure must
be higher.
(3) At t = 1, since evaporation has occurred, there must be more molecules in the vapor
state.
(4) At t = 1, since the system has not reached equilibrium you would expect the rate of
evaporation to exceed the rate of condensation.
© 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 11: States of Matter; Liquids and Solids 379
(5)
t =1
b. (1) Between t = 1 and t = 2, the system has still not reached equilibrium. Therefore,
because the rate of evaporation continues to exceed the rate of condensation, you
would expect the water level to be lower.
(2) Prior to reaching equilibrium at t = 2, you would continue to observe a rate of
evaporation greater than the rate of condensation, resulting in a higher vapor pressure
than t = 1.
(3) Since evaporation has been occurring at a greater rate than condensation between
points t = 1 and t = 2, you would expect more molecules in the vapor state at t = 2.
(4) When the system has reached equilibrium at t = 2, the rate of evaporation equals the
rate of condensation.
(5)
t=2
11.2. You would have to cook the egg for a longer time. The reason is that, since the boiling
temperature is lower, it would take longer for the egg to become hard, a chemical process (which
slows down at lower temperature).
11.3. a. Because A has stronger intermolecular forces than B, A will boil at a higher temperature.
Statement (a) is true.
b. Because A has stronger intermolecular forces than B, A will be more viscous than B.
Statement (b) is false.
c. Because B has weaker intermolecular forces than A, it will require a lower temperature to
freeze. Statement (c) is true.
© 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.
States of Matter; Liquids and Solids
■ 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.
11.1. First, calculate the heat required to vaporize 1.00 kg of ammonia:
1000 g 1 mol NH3 23.4 kJ
1.00 kg NH3 = 1374.04 kJ
1 kg 17.03 g NH3 1 mol
The amount of water at 0C that can be frozen to ice at 0C with this heat is
1 mol H 2 O 18.01 g H 2 O
1374.04 kJ = 4117.54g = 4.12 kg H2O
6.01 kJ 1 mol H 2 O
11.2. Use the two-point form of the Clausius-Clapeyron equation to calculate P2:
P2 26.8 103 J / mol 1 1
ln =
760 mmHg 8.31 J / (K mol) 319 K 308 K
1.12 104
= 3225 K = −0.36106
K
Converting to antilogs gives
P2
= antilog(−0.36106) = e−0.36106 = 0.69693
760 mmHg
P2 = 0.69693 760 mmHg = 529.6 = 5.3 102 mmHg
11.3. Use the two-point form of the Clausius-Clapeyron equation to solve for Hvap:
757 mmHg H vap 1 1
ln =
522 mmHg 8.31 J/(K mol) 368 K 378 K
H vap 7.2 105
0.37169 =
8.31 J/(K mol) K
Hvap = 4.3 104 J/mol (4 101 kJ/mol)
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, Chapter 11: States of Matter; Liquids and Solids 377
11.4. a. Liquefy methyl chloride by a sufficient increase in pressure below 144C.
b. Liquefy oxygen by compressing to 50 atm below −119C.
11.5. a. Propanol has a hydrogen atom bonded to an oxygen atom. Therefore, hydrogen bonding is
expected. Because propanol is polar (from the O–H bond), we also expect dipole-dipole
forces. Weak London forces exist, too, because such forces exist between all molecules.
b. Linear carbon dioxide is not polar, so only London forces exist among CO2 molecules.
c. Bent sulfur dioxide is polar, so we expect dipole-dipole forces; we also expect the usual
London forces.
11.6. The order of increasing vapor pressure is butane (C4H10), propane (C3H8), and ethane (C2H6).
Because London forces tend to increase with increasing molecular mass, we would expect the
molecule with the highest molecular mass to have the lowest vapor pressure.
11.7. Because ethanol has an H atom bonded to an O atom, strong hydrogen bonding exists in ethanol
but not in methyl chloride. Hydrogen bonding explains the lower vapor pressure of ethanol
compared to methyl chloride.
11.8. a. Zinc, a metal, is a metallic solid.
b. Sodium iodide, an ionic substance, exists as an ionic solid.
c. Silicon carbide, a compound in which carbon and silicon might be expected to form
covalent bonds to other carbon and silicon atoms, exists as a covalent network solid.
d. Methane, at room temperature a gaseous molecular compound with covalent bonds, freezes
as a molecular solid.
11.9. Only MgSO4 is an ionic solid; C2H5OH, CH4, and CH3Cl form molecular solids; thus MgSO4
should have the highest melting point. Of the molecular solids, CH4 has the lowest molecular
mass (16.0 amu) and would be expected to have the lowest melting point. Both C2H5OH and
CH3Cl have approximately the same molecular masses (46.0 amu vs. 50.5 amu), but C2H5OH
exhibits strong hydrogen bonding and, therefore, would be expected to have the higher melting
point. The order of increasing melting points is CH4, CH3Cl, C2H5OH, and MgSO4.
11.10. Each of the four corners of the cell contains one atom, which is shared by a total of four unit cells.
Therefore, the corners contribute one whole atom.
Atoms 1/4 atom
= 4 corners = 1 atom
Unit cell 1 corner
11.11. Use the edge length to calculate the volume of the unit cell. Then, use the density to determine the
mass of one atom. Divide the molar mass by the mass of one atom.
V = (3.509 10−10 m)3 = 4.321 10−29 m3
3
0.534 g 100 cm
d= 3
= 5.34 10 g/m
5 3
1 cm 1m
Mass of 1 unit cell = d V = (5.34 105 g/m3) (4.321 10−29 m3) = 2.3074 10−23 g
© 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.
, 378 Chapter 11: States of Matter; Liquids and Solids
There are two atoms in a body-centered cubic unit cell; thus, the mass of one lithium atom is
½ 2.3074 10−23 g = 1.1537 10−23 g
The known atomic mass of lithium is 6.941 amu, so Avogadro's number is
6.941 g/mol
NA = = 6.016 1023 = 6.02 1023 atoms/mol
1.1537 1023 g/atom
11.12. Use Avogadro's number to convert the molar mass of potassium to the mass per one atom.
39.0983 g K 1 mol K 6.4925 1023 g K
23
=
1 mol K 6.022 10 atoms 1 atom
There are two K atoms per unit cell; therefore, the mass per unit cell is
6.4925 1023 g K 2 atoms 1.2985 1022 g
=
1 atom 1 unit cell 1 unit cell
The density of 0.856 g/cm3 is equal to the mass of one unit cell divided by its unknown volume,
V. After solving for V, determine the edge length from the cube root of the volume.
1.2985 1022 g
0.856 g/cm3 =
V
1.2985 1022 g
V= = 1.517 10−22 cm3 (1.517 10−28 m3)
0.856 g/cm3
Edge length = 3 1.517 1028 m3 = 5.333 10−10 = 5.33 10−10 m (533 pm)
■ ANSWERS TO CONCEPT CHECKS
11.1. a. (1) At t = 0, since the system is not at equilibrium and there are no H2O molecules in the
gaseous state, you would expect the rate of evaporation to exceed the rate of
condensation. At t = 1, since evaporation has proceeded at a greater rate than
condensation, there must now be fewer molecules in the liquid state, resulting in a
lower level of H2O(l).
(2) At t = 1, since some of the H2O has gone into the vapor state, the vapor pressure must
be higher.
(3) At t = 1, since evaporation has occurred, there must be more molecules in the vapor
state.
(4) At t = 1, since the system has not reached equilibrium you would expect the rate of
evaporation to exceed the rate of condensation.
© 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 11: States of Matter; Liquids and Solids 379
(5)
t =1
b. (1) Between t = 1 and t = 2, the system has still not reached equilibrium. Therefore,
because the rate of evaporation continues to exceed the rate of condensation, you
would expect the water level to be lower.
(2) Prior to reaching equilibrium at t = 2, you would continue to observe a rate of
evaporation greater than the rate of condensation, resulting in a higher vapor pressure
than t = 1.
(3) Since evaporation has been occurring at a greater rate than condensation between
points t = 1 and t = 2, you would expect more molecules in the vapor state at t = 2.
(4) When the system has reached equilibrium at t = 2, the rate of evaporation equals the
rate of condensation.
(5)
t=2
11.2. You would have to cook the egg for a longer time. The reason is that, since the boiling
temperature is lower, it would take longer for the egg to become hard, a chemical process (which
slows down at lower temperature).
11.3. a. Because A has stronger intermolecular forces than B, A will boil at a higher temperature.
Statement (a) is true.
b. Because A has stronger intermolecular forces than B, A will be more viscous than B.
Statement (b) is false.
c. Because B has weaker intermolecular forces than A, it will require a lower temperature to
freeze. Statement (c) is true.
© 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.
