CHAPTER
14
SOLUTIONS MANUAL
Mixtures and Solutions
Section 14.1 Types of Mixtures
pages 476 – 479
Section Assessment 14.1
7. Summarize What causes Brownian motion?
Collisions of particles of the dispersion medium
with the dispersed particles results in Brownian
motion.
page 479
1. Explain Use the properties of seawater to
describe the characteristics of mixtures.
Answers will vary but might include that
seawater is a heterogeneous mixture with dirt
and mud particles, and it is a homogeneous
mixture with dissolved substances.
2. Distinguish between suspensions and colloids.
Suspension particles are larger than colloidal
particles. Suspension particles settle out of the
mixture, whereas colloidal particles do not.
8. Compare and Contrast Make a table that
compares the properties of solutions, suspensions, and colloids.
Student tables will vary, but should include
particle size, if the particles settle out, and if the
particles display the Tyndall effect. A sample table
follows.
Suspensions, Colloids, and Solutions
Particle size
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
All solutions are homogeneous mixtures
containing two or more substances. Solutions
may be liquid, solid, or gas. Solution types are
identified in Table 14.2.
4. Explain Use the Tyndall effect to explain why
it is more difficult to drive through fog using
high beams than using low beams.
High beams are aimed farther down the road
than low beams. Because the fog scatters
light, there is less light from the high beams to
illuminate the road than from the low beams.
Also, because the high beams are aimed more
directly into the fog, more of their light is
reflected back toward the driver, making it more
difficult to see.
5. Describe different types of colloids.
See Table 14.1 for descriptions of colloid types.
6. Explain Why do dispersed colloid particles
stay dispersed?
The particles do not settle out because they have
polar or charged layers surrounding them. These
layers repel each other and prevent the particles
from settling or separating.
Tyndall
effect?
Suspensions
Large (wide
variation)
Yes
Yes
Colloids
1 nm–1000 nm
No
Yes
Solutions
Atomic scale
(atoms, ions,
and molecules)
No
No
3. Identify the various types of solutions. Describe
the characteristics of each type of solution.
Particles
settle?
Section 14.2 Solution
Concentration
pages 480–488
Practice Problems
pages 481–488
9. What is the percent by mass of NaHCO3 in a
solution containing 20.0 g NaHCO3 dissolved
in 600.0 mL H2O?
600.0 mL H2O 3 1.0 g/mL 5 600.0 g H2O
20 g NaHCO
___
3 100 5 3%
3
600 g H2O 1 20 g NaHCO3
10. You have 1500.0 g of a bleach solution. The
percent by mass of the solute sodium
hypochlorite, NaOCl, is 3.62%. How many
grams of NaOCl are in the solution?
3.62% 5 100 3
mass NaOCI
__
1500.0 g
mass NaOCl 5 54.3 g
Solutions Manual
Chemistry: Matter and Change • Chapter 14
277
14
SOLUTIONS MANUAL
11. In question 10, how many grams of solvent are
18. What is the molarity of a bleach solution
containing 9.5 g of NaOCl per liter of bleach?
in the solution?
1500.0 g 2 54.3 g 5 1445.7 g solvent
mol NaOCl 5 9.5 g 3
12. Challenge The percent by mass of calcium
chloride in a solution is found to be 2.65%. If
50.0 grams of calcium chloride is used, what is
the mass of the solution?
2.65% 5
molarity 5
1 mol
_
5 0.13 mol
74.44 g
mol NaOCI
0.128 mol
__
5_
1.00 L
1.00 L solution
5 0.13M
19. Challenge How much calcium hydroxide
__
100 3 50 g CaCl2
(Ca(OH)2), in grams, is needed to produce
1.5 L of a 0.25M solution?
mass of solution
mass of solution 5 1886.79 g
13. What is the percent by volume of ethanol
0.25M 5
x mol Ca(OH)
__
2
1.5 L solution
in a solution that contains 35 mL of ethanol
dissolved in 155 mL of water?
x 5 0.38 mol Ca(OH)2
35 mL
__
3 100 5 18%
0.38 mol Ca(OH)2 3
74.08 g
_
5 28 g Ca(OH)
155 mL 1 35 mL
14. What is the percent by volume of isopropyl
alcohol in a solution that contains 24 mL of
isopropyl alcohol in 1.1 L of water?
24 mL
__
3 100 5 2.1%
24 mL 1 1100 mL
mol
20. How many grams of CaCl2 would be dissolved
in 1.0 L of a 0.10M solution of CaCl2?
mol CaCl2 5 (0.10M)(1.0 L) 5 (0.10 mol/L)(1.0 L)
5 0.10 mol CaCl2
mass CaCl2 5 0.10 mol CaCl2 3
110.98 g
_
1 mol
15. Challenge If 18 mL of methanol are used to
make an aqueous solution that is 15% methanol
by volume, how many milliliters of solution are
produced?
mass of CaCl2 5 11 g
21. How many grams of CaCl2 should be dissolved
18 mL
15% 5 __ 3 100
in 500.0 mL of water to make a 0.20M solution
of CaCl2?
x 5 120 mL
mol CaCl2 5 500.0 mL 3
x mL solution
16. What is the molarity of an aqueous solution
containing 40.0 g of glucose (C6H12O6) in 1.5 L
of solution?
mol C6H12O6 5 40.0 g 3
molarity 5
1 mol
_
5 0.222 mol
180.16 g
mol C H O
0.222 mol
__
5 _ 5 0.15M
6 12
6
1.5 L solution
1.5 L
1L
_
3 0.20M
1000 mL
5 0.10 mol
mass CaCl2 5 0.10 mol CaCl2 3
110.98 g
_
22. How many grams of NaOH are in 250 mL of a
3.0M NaOH solution?
mol NaOH 5 250 mL 3
1L
_
3 3.0M
1000 mL
3.0 mol
1L
_
3_
containing 1.55 g of dissolved KBr.
5 250 mL 3
1 mol
mol KBr 5 1.55 g 3 _ 5 0.0130 mol KBr
5 0.75 mol
mol KBr
0.0130 mol
molarity 5 __ 5 __
mass NaOH 5 0.75 mol NaOH 3
1L
1000 mL
119.0 g
1.60 L solution
5 8.13 3 1023M
01.60 L
Chemistry: Matter and Change • Chapter 14
1 mol
mass of CaCl2 5 11 g
17. Calculate the molarity of 1.60 L of a solution
278
2
mass of NaOH 5 3.0 3
101
40.00 g
_
1 mol
g
Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
CHAPTER
14
CHAPTER
SOLUTIONS MANUAL
23. Challenge What volume of ethanol (C2H3OH)
is in 100.0 mL of 0.15M solution? The density
of ethanol is 0.7893 g/mL.
0.15 mol ethanol
1L
_
3 __
1000 mL
1 L solution
46
g
ethanol
1 mL ethanol
3 __ 3 __ 5 0.87 mL
100 mL 3
1 mol ethanol
29. What is the mole fraction of NaOH in an aqueous
solution that contains 22.8% NaOH by mass?
Assume 100.0 g sample.
Then, mass NaOH 5 22.8 g
mass H2O 5 100.0 g 2 (mass NaOH) 5 77.2 g
0.7893 g ethanol
mol NaOH 5 22.8 g 3
24. What volume of a 3.00M KI stock solution
mol H2O 5 77.2 g 3
(3.00M)V1 5 (1.25M)(0.300 L)
mol fraction NaOH 5
V1
3.00M
25. How many milliliters of a 5.0M H2SO4 stock
solution would you need to prepare 100.0 mL
of 0.25M H2SO4?
(5.0M)V1 5 (0.25M)(100.0 mL)
V1
5.0 M
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
HCl is diluted to make 2 L of solution, how
much HCl, in grams was in the solution?
mol HCl 5 5M 3 0.5 L 5 2.5 mol HCl
36.45 g HCl
__
1 mol
3 2.5 mol
mass of HCl 5 91.15 g
27. What is the molality of a solution containing
10.0 g Na2SO4 dissolved in 1000.0 g of water?
mol Na2SO4 5 10.0 g Na2SO4
1 mol
3_
142.04 g
5 0.0704 mol Na2SO4
0.0704 mol Na SO
__
5 0.0704 m
2
4
1.0000 Kg H2O
needed to make a 1.00m aqueous solution?
1 mol
__
1 kg solvent
molar mass of Ba(OH)2 5
mass of Ba(OH)2 5 171 g
Solutions Manual
mol NaOH 1 mol H2O
0.570 mol NaOH
0.570
____
5_
4.85
0.570 mol NaOH 1 4.28 mol H2O
5 0.118
The mole fraction of NaOH is 0.118.
30. Challenge If the mole fraction of sulfuric acid
1 2 0.125 5 0.875 mole fraction of water
Assume a sample of the solution totals 100.0 moles.
By definition there would be 87.5 moles of water
and 12.5 moles of sulfuric acid in the sample.
87.5 mol of H2O 3
18.02 g
_
5 1580 g H O
12.5 mol H2SO4 3
98.08 g
_
5 1230 g H SO
2
1 mol
1 mol
2
4
percent by mass H2SO4
1230 g H2SO4
3 100
5
(1580 1 1230) g solution
5 43.8% H2SO4 by mass
___
Section Assessment 14.2
page 488
28. Challenge How much (Ba(OH)2), in grams, is
mol Ba(OH)2 5
mol NaOH
___
0.125 5 mole fraction of H2SO4
26. Challenge If 0.5 L of 5M stock solution of
molality 5
2
18.02 g
(H2SO4) in an aqueous solution is 0.125, what
is the percent by mass of H2SO4?
(0.25M)(100.0 mL)
5 __ 5 5.0 mL
mass of HCl 5
5
40.00 g
1 mol
_
5 4.28 mol H O
would you use to make 0.300 L of a 1.25M KI
solution?
(1.25M)(0.300 L)
5 __ 5 0.125 L 5 125 mL
1 mol
_
5 0.570 mol NaOH
171 g Ba(OH)
__
2
1 mol
31. Compare and contrast five quantitative
ways to describe the composition of solutions.
molarity, molality, and mole fraction are based
on moles of solute per some other quantity;
percent by volume and molarity are defined on a
per volume of solution basis; molality and mole
fraction are based on a per quantity of solvent
basis; percent by mass and percent by volume are
the only ratios involving percentages
Chemistry: Matter and Change • Chapter 14
279
14
SOLUTIONS MANUAL
S1 5
32. Explain the similarities and differences
between a 1M solution of NaOH and a 1m
solution of NaOH.
33. Calculate A can of chicken broth contains
450 mg of sodium chloride in 240.0 g of broth.
What is the percent by mass of sodium chloride
in the broth?
1g
450 mg NaCl 3 _ 5 0.45 g NaCl
240.0 g
34. Solve How much ammonium chloride
(NH4Cl), in grams, is needed to produce 2.5 L
of a 0.5M aqueous solution?
0.5M
_
3 2.5 L
20.0 kPa
37. A gas has a solubility of 0.66 g/L at 10.0 atm
of pressure. What is the pressure on a 1.0-L
sample that contains 1.5 g of gas?
_
S2 5 1.5 g 5 1.5 g/L
1.0 L
P2 5 P1 3
S
1.5 g/L
_
5 10.0 atm 3 _ 5 23 atm
2
S1
0.66 g/L
P2 5 P1 1 (P1)(0.400)
5 (7.0 atm) 1 (7.0 atm)(0.400)
5 9.8 atm
S2 5 S1 3
1L
5 1.25 mol of NH4Cl
53.49 g NH Cl
mass of NH Cl 5 1.25 mol NH Cl 3 __
4
4
2
P1
of pressure is 0.52 g/L. How many grams of the
gas would be dissolved per 1 L if the pressure
increased 40.0 percent?
0.45 g
_
3 100 5 0.19%
4
P
110.0 kPa
_
5 0.55 g/L 3 _ 5 3.0 g/L
38. Challenge The solubility of a gas at 7.0 atm
1000 mg
mol of NH4Cl 5
1.0 L
S2 5 S1 3
Both solutions contain NaOH (solute) dissolved in
water (solvent). The 1 m solution contains 1 mole
of NaOH per kilogram of water; the 1M solution
contains 1 mole of NaOH per liter of solution.
percent by mass 5
0.55 g
_
5 0.55 g/L
1 mol
mass of NH4Cl 5 66.86 g
35. Outline the laboratory procedure for preparing
a specific volume of a dilute solution from a
concentrated stock solution.
Calculate the volume of stock solution needed
and add it to a volumetric flask. Add water up to
the flask’s calibration line.
P
_
2
P1
S2 5 (0.52 g/L) 3
9.8 atm
_
7.0 atm
S2 5 0.73 g/L
Section Assessment 14.3
page 497
39. Describe factors that affect the formation of
solutions.
Surface area, temperature, and pressure affect the
formation of solutions.
40. Define solubility.
Section 14.3 Factors Affecting
Solvation
pages 489–497
Practice Problems
page 497
36. If 0.55 g of a gas dissolves in 1.0 L of water at
20.0 kPa of pressure, how much will dissolve at
110.0 kPa of pressure?
280
Chemistry: Matter and Change • Chapter 14
Solubility refers to the maximum amount of
solute that can dissolve in a given amount of
solvent at a particular temperature and pressure.
41. Describe how intermolecular forces affect
solvation?
The attractive forces between solute and solvent
particles overcome the forces holding the solute
particles together, thus, pulling the solute
particles apart.
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CHAPTER
14
CHAPTER
SOLUTIONS MANUAL
42. Explain on a particle basis why the vapor pres-
sure of a solution is lower than a pure solvent.
Tb 5 100°C 1 0.320°C 5 100.320°C
When a solvent contains a solute, fewer solvent
particles occupy the surface. Fewer particles
escape into the gaseous state.
DTf 5 1.86°C/m 3 0.625m 5 1.16°C
43. Sumarize If a seed crystal was added to a
Tf 5 0.0°C 2 1.16°C 5 21.16°C
46. What are the boiling point and freezing point of
supersaturated solution, how would you characterize the resulting solution?
a 0.40m solution of sucrose in ethanol?
After the excess solute particles crystallize out of
solution, the solution is saturated.
Tb 5 78.5°C 1 0.49°C 5 79.0°C
44. Make and Use Graphs Use the informa-
tion in Table 14.4 to graph the solubilities of
aluminum sulfate, lithium sulfate, and potassium chloride at 0°C, 20°C, 60°C, and 100°C.
Which substance’s solubility is most affected by
increasing temperature?
Solubility v. Temperature
90
Solubility (g/100 g H2O)
DTb 5 1.22°C/m 3 0.40m 5 0.49°C
DTf 5 1.99°C/m 3 0.40m 5 0.80°C
Tf 5 2114.1°C 2 0.80°C 5 2114.9°C
47. Challenge If a 0.045m solution (consisting of a
nonvolatile, nonelectrolyte solute) is experimentally found to have a freezing point depression of
0.080ºC. What is the freezing point depression
constant (Kf)? Which is most likely to be the
solvent—water, ethanol, or chloroform?
DT
_
m
0.080ºC
_
5
Kf 5
80
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
DTb 5 0.512°C/m 3 0.625m 5 0.320°C
Al2(SO4 )3
70
60
KCl
f
0.045m
5 1.8ºC/m
It is most likely water because the calculated
value is closest to 1.86°C/m
50
40
Li2SO4
30
Section Assessment 14.4
page 504
20
48. Explain the nature of colligative properties.
10
0
20
60
100
Temperature (°C)
Aluminum sulfate shows the greatest change in
solubility over the temperature range.
Section 14.4 Colligative Properties
of Solutions
pages 498–504
Practice Problems
page 503
Colligative properties depend on the number of
solute particles in a solution.
49. Describe four colligative properties of solutions.
vapor pressure lowering: the decrease in vapor
pressure with increasing solute particles in
solution; boiling point elevation: the increase in
boiling point with increasing solute particles in
solution; freezing point depression: the decrease
in freezing point with increasing solute particles
in solution; osmotic pressure: the change in
osmotic pressure with increasing solute particles
in solution
45. What are the boiling point and freezing point of
a 0.625m aqueous solution of any nonvolatile,
nonelectrolyte solute?
Solutions Manual
Chemistry: Matter and Change • Chapter 14
281
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SOLUTIONS MANUAL
50. Explain why a solution has a higher boiling
point than the pure solvent.
pages 508–511
Solute particles in solution decrease the vapor
pressure above the solution. Because a solution
boils when its vapor pressure equals the external
pressure, this decrease in vapor results in the
need for a higher temperature in order for the
solution to boil.
51. Solve An aqueous solution of calcium chloride
(CaCl2) boils at 101.3ºC. How many kilograms
of calcium chloride were dissolved in 1000 grams
of the solvent?
m5
DT
1.3ºC
_
5_
0.512ºC/m
Kb
5 2.53m 5 2.53 moles solute particles/1 kg
solvent
1 mol CaCl
110.98 g
__
3_
2
3 mol particles
1 kg
3 _ 5 0.0936 kg
1 mol
1000 g
52. Calculate the boiling point elevation of a solu-
tion containing 50.0 g of glucose (C6H12O6)
dissolved in 500.0 g of water. Calculate the
freezing point depression for the same solution.
50.0 g glucose 3
molality 5
Section 14.1
Mastering Concepts
54. Explain what is meant by the statement “not
all mixtures are solutions.”
Solutions are homogeneous mixtures that are
uniform in composition with a single phase.
Mixtures can also be heterogeneous, where the
substances that make them up remain distinct.
55. What is the difference between solute and
b
2.53 mol particles 3
Chapter 14 Assessment
1 mol
_
5 0.278 mol glucose
180.15 g
0.278 mol glucose
__
5 0.556m
0.5000 kg H2O
DTb 5 (0.512°C/m)(0.556m) 5 0.285°C
Tb 5 100.000°C 1 0.285°C 5 100.285°C
DTf 5 (1.86°C/m)(0.556m) 5 1.03°C
Tf 5 0.00°C 2 1.03°C 5 21.03°C
53. Investigate A lab technician determines the
solvent?
A solute is the substance being dissolved.
The solvent is the substance in which the
solute dissolves.
56. What is a suspension and how does it differ
from a colloid?
A suspension is a heterogeneous mixture that
settles out if left undisturbed. The particles
dispersed in a colloid are much smaller than those
in a suspension and do not settle out.
57. How can the Tyndall effect be used to distin-
guish between a colloid and a solution? Why?
A beam of light is visible in a colloid but not in
a solution. Dispersed colloid particles are large
enough to scatter light (Tyndall effect).
58. Name a colloid formed from a gas dispersed in
a liquid.
Student answers may include whipped cream or
beaten egg whites.
59. Salad dressing What type of heterogeneous
boiling point elevation of an aqueous solution
of a nonvolatile, nonelectrolyte to be 1.12°C.
What is the solution’s molality?
mixture is shown in Figure 14.24 on page 508?
What characteristic is most useful in classifying
the mixture?
1.12°C 5 0.512°C/m 3 m
The mixture is a suspension. Left undisturbed, the
mixture components settle out.
m 5 2.19m
60. What causes Brownian motion observed in
liquid colloids?
The random particle movements in liquid colloids
result from collisions between particles in the
mixture.
282
Chemistry: Matter and Change • Chapter 14
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CHAPTER
CHAPTER
14
61. Aerosol sprays are categorized as colloids.
Identify the phases of an aerosol spray.
The most abundant mixture component is in
the gas phase. The dispersed particles are in the
liquid phase.
Section 14.2
SOLUTIONS MANUAL
Mastering Problems
67. According to lab procedure, you stir 25.0 g
of MgCl2 into 550 mL of water. What is the
percent by mass of MgCl2 in the solution?
percent by mass of MgCl2
25.0 g MgCl
5
3 100 5 4.3%
25.0 g MgCl 1 550 g H2O
___
68. How many grams of LiCl are in 275 g of a 15%
Mastering Concepts
62. What is the difference between percent by mass
and percent by volume?
Percent by mass is a comparison between the
mass of solute and the total mass of the solution.
Percent by volume is a comparison between the
volume of the solute and the total volume of the
solution.
63. What is the difference between molarity and
aqueous solution of LiCl?
mass of LiCl 5
275 g 3 15
__
5 41 g
100
69. You need to make a large quantity of a 5%
solution of HCl but have only 25 mL HCl.
What volume of 5% solution can be made from
this volume of HCl?
volume of solution 5
25 mL HCl
_
3 100 5 500 mL
5
molality?
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Molarity is solution concentration expressed
as the moles of solute per volume of solution.
Molality expresses concentration as moles of
solute per kilogram of solvent. Molality does not
depend upon the temperature of the solution.
64. What factors must be considered when creating
a dilute solution from a stock solution?
The molarity and volume of both stock solution
and dilute solution are required in the formula
M1V1 5 M2V2.
65. How do 0.5M and 2.0M aqueous solutions of
NaCl differ?
The 2M solution contains more moles of NaCl per
volume of water than the 0.5M solution.
70. Calculate the percentage by volume of a solu-
tion created by adding 75 mL of acetic acid to
725 mL of water.
percent volume
5
75 mL CH COOH
____
3 100
3
75 mL CH3OOH 1 725 mL solution
5 9.4%
71. Calculate the molarity of a solution that
contains 15.7 g of CaCO3 dissolved in 275 mL
of water.
mol of CaCO3 5 15.7 g CaCO3
1 mol CaCO3
3
5 0.157 mol CaCO3
100.01 g CaCO3
__
275 mL 3
66. Under what conditions might a chemist describe
a solution in terms of molality? Why?
Under conditions of changing temperature.
Because molality is based on mass, it does not
change with temperature.
1L
_
5 0.275 L
1000 mL
molarity 5
0.157 mol CaCO
__
5 0.571M
3
0.275 L solution
72. What is the volume of a 3.00M solution made
with 122 g of LiF?
mol of LiF 5 122 g LiF 3
volume of solution 5
Solutions Manual
1 mol LiF
_
5 4.71 mol LiF
25.9 g LiF
4.71 mol
_
5 1.57 L
3.00M
Chemistry: Matter and Change • Chapter 14
283
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SOLUTIONS MANUAL
73. How many moles of BaS would be used to
make 1.5 3 103 mL of a 10.0M solution?
1.5 3 103 mL 3
mol BaS 5
1L
_
5 1.5 L
0.17 L HCl 3
1000 mL
10.0 mol
_
3 1.5 L
0.42 L HCl 3
2.0 L of a 3.5M solution?
mol of CaCl2 5
3.5 mol
_
3 2.0 L 5 7.0 mol CaCl
2
1L
mass of CaCl2 5 7.0 mol CaCl2 3
__
110.1 g CaCl2
1 mol CaCl2
mass of CaCl2 5 770 g
are frequently prepared. Complete Table 14.7
by calculating the volume of concentrated, or
12M, hydrochloric acid that should be used to
make 1.0 L of HCl solution with each molarity
listed.
Molarity of HCl
desired
Volume of 12M HCl stock
solution needed (mL)
0.50
42 mL
1.0
83 mL
1.5
130 mL
2.0
170 mL
5.0
420 mL
0.50 mol/L 3 1 L
__
5 0.042 L HCl
12 mol/L
1000 mL
__
5 42 mL HCl
V1 5
1L
__
1.0 mol/L 3 1 L
5 0.083 L HCl
12 mol/L
0.083 L HCl 3
1000 mL
_
5 83 mL HCl
1L
__
1.5 mol/L 3 1 L
5 0.13 L HCl
V1 5
12 mol/L
0.13 L HCl 3
284
1000 mL
_
5 420 mL HCl
1L
76. How much 5.0M nitric acid (HNO3), in
milliliters, is needed to make 225 mL of
1.0M HNO3?
__
1.0 M 3 225 mL
5.0 M
V1 5
volume of HNO3 5 45 mL
77. Experiment In the lab, you dilute 55 mL of
75. Stock solutions of HCl with various molarities
0.042 L HCl 3
1L
__
1L
74. How many grams of CaCl2 are needed to make
1000 mL
_
5 170 mL HCl
5.0 mol/L 3 1 L
5 0.42 L HCl
12 mol/L
V1 5
mol BaS 5 15 mol
V1 5
__
2.0 mol/L 3 1 L
5 0.17 L HCl
12 mol/L
V1 5
1000 mL
_
5 130 mL HCl
a 4.0M solution to make 250 mL of solution.
Calculate the molarity of the new solution.
M2 5
4.0M 3 55 mL
__
5 0.88M
250 mL
78. How many milliliters of 3.0M phosphoric acid
(H3PO4) can be made from 95 mL of a 5.0M
H3PO4 solution?
V2 5
__
5.0 M 3 95 mL
5 160 mL
3.0 M
79. If you dilute 20.0 mL of a 3.5M solution
to make 100.0 mL of solution, what is the
molarity of the dilute solution?
M2 5
3.5M 3 20 mL
__
5 0.70M
100 mL
80. What is the molality of a solution containing
75.3 grams of KCl dissolved in 95.0 grams of
water?
_
mol KCl 5 75.3 g KCl 3 1 mol KCl
74.6 g KCl
5 1.01 mol KCl
95.0 g H2O 3
m5
1 kg
_
5 0.0950 kg H O
2
1000 g
1.01 mol KCl
__
5 10.6 mol/kg
0.0950 kg H2O
1L
Chemistry: Matter and Change • Chapter 14
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SOLUTIONS MANUAL
84. What is the mole fraction of H2SO4 in a
81. How many grams of Na2CO3 must be dissolved
into 155 grams of water to create a solution
with a molality of 8.20 mol/kg?
155 g H2O 3
solution containing the percentage of sulfuric
acid and water shown in Figure 14.25?
1 kg
_
5 0.155 kg H O
2
1000 g
H2SO4
27.3%
mol NaCO3 5 8.20 mol/kg 3 0.155 kg 5 1.27 mol
mass of NaCO3 5 1.27 mol NaCO3 3
83.00 g NaCO
__
H2O
72.7%
3
mol NaCO3
5 105 g
82. What is the molality of a solution containing
30.0 g of naphthalene (C10H8) dissolved in
500.0 g of toluene?
30.0 g C10H8 3
1 mol C H
__
5 0.234 mol C
10 8
128 g C10H8
500.0 g toluene 3
m5
10H8
1 kg
_
5 0.5000 kg
0.234 mol C H
__
5 0.468m
0.5000 kg toluene
2
4
2
97.1 g H 2SO 4
4
1 mol H O
__
5 4.034 mol H O
2
2
18.02 g H 2O
0.281 mol H SO
____
4
2
0.281 mol H 2SO 4 1 4.034 mol H 2O
85. Calculate the mole fraction of MgCl2 in a
83. What are the molality and mole fraction of
solute in a 35.5 percent by mass aqueous
solution of formic acid (HCOOH)?
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
1 mol H SO
__
5 0.281 mol H SO
5 0.0650
10 8
__
35.5 g HCOOH
35.5% means
100.0 g solution
solution created by dissolving 132.1 g MgCl2
into 175 mL of water.
132.1 g MgCl 2 3
175 mL H 2O 3
1 mol HCOOH
35.5 g HCOOH 3 __
1 mol MgCl
__
5 1.387 mol MgCl
2
mass of water 5 100.0 g 2 35.5 g 5 64.5 g
5 6.45 3 1022 kg
1 mol H O
__
2
XMgCl2 5
2
95.21 g
1.0 g H O
1 mol H O
_
3 __
2
1 mL H 2O
5 9.72 mol H2O
46.03 g HCOOH
5 0.771 mol HCOOH
2
18.0 g H 2O
1.387 mol MgCl
____
5 0.125
2
1.387 mol MgCl2 1 9.72 mol H2O
Section 14.3
18.02 g H2O
5 3.58 mol H2O
molality 5
72.7 g H 2O 3
X H 2S O 4 5
1000 g
moles of water 5 64.5 g H2O 3
27.3 g H 2SO 4 3
0.771 mol HCOOH
__
5 12.0m
6.45 3 1022 kg H2O
0.771 mol
mole fraction 5 ___ 5 0.177
0.771 mol 1 3.58 mol
Mastering Concepts
86. Describe the process of solvation.
A solute introduced into a solvent is surrounded
by solvent particles. Due to the attraction
between solute and solvent particles, solute
particles are pulled apart and surrounded by
solvent particles. Once separated, solute particles
disperse into solution.
87. What are three ways to increase the rate of
solvation?
increase the temperature of the solvent, increase
the surface area of the solute, agitation
Solutions Manual
Chemistry: Matter and Change • Chapter 14
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14
SOLUTIONS MANUAL
88. Explain the difference between saturated and
unsaturated solutions.
A saturated solution contains the maximum
amount of solute under a given set of conditions.
An unsaturated solution contains less than the
maximum amount.
Mastering Problems
89. At a pressure of 1.5 atm, the solubility of a gas
is 0.54 g/L. Calculate the solubility when the
pressure is doubled.
92. The solubility of a gas at 37.0 kPa is 1.80 g/L.
At what pressure will the solubility reach
9.00 g/L?
P2 5
P 2 5 185 kPa
93. Use Henry’s Law to complete the Table 14.8.
Solubility and Pressure
Solubility (g/L)
__
0.54 g/L 3 3.0 atm
S2 5
1.5 atm
S 2 5 1.08 g/L
90. At 4.5 atm of pressure, the solubility of a gas is
9.5 g/L. How many grams of gas will dissolve
in 1L if the pressure is reduced by 3.5 atm?
S2 5
__
9.5 g/L 3 1.0 atm
4.5 atm
P2 5
Pressure (kPa)
2.9
25
3.7
32
4.5
39
__
32 kPa 3 2.9 g/L
3.7 g/L
P 2 5 25 kPa
S2 5
__
3.7 g/L 3 39 kPa
32 kPa
S 2 5 4.5 g/L
S 2 5 2.1 g/L
91. Using Figure 14.26, compare the solubility of
potassium bromide (KBr) and potassium nitrate
(KNO3) at 80°Celsius.
Solubility v. Temperature
Solubility (g/100 g of water)
__
37.0 kPa 3 9.00 g/L
1.8 g/L
240
220
200
180
160
140
120
100
80
60
40
20
0
S5
KNO3
KBr
60
80
___
44.01 g CO
__ 3 __
9.0 3 10–6 mol CO 2
NaCl
40
inside a bottle of soft drink is 4.0 atm at 25°C.
The solubility of CO2 is 0.12 mol/L. When the
bottle is opened, the partial pressure drops
to 3.0 3 1024 atm. What is the solubility of
CO2 in the open drink? Express your answer in
grams per liter.
(0.12 mol/L)(3.0 3 10–4 atm)
4.0 atm
5 9.0 3 10–6 mol/L CO2
NaClO2
20
94. Soft Drinks The partial pressure of CO2
100
120
1L
5 4.0 3 10–4 g/L CO2
2
1 mol CO2
Section 14.4
Temperature (ºC)
The solubility of KBr is 95 g/100 g H 2O. The
solubility of KNO 3 is nearly twice as high at the
same temperature, at nearly 170 g/100 g H 2O.
286
Chemistry: Matter and Change • Chapter 14
Mastering Concepts
95. Define the term colligative property.
A physical property of a solution that is affected
by the number of solute particles but not their
nature.
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SOLUTIONS MANUAL
96. Use the terms dilute and concentrated to
compare the solution on both sides of a
membrane.
If there is a concentration gradient, the solution
is more dilute on one side of the membrane
and more concentrated on the other side of the
membrane.
97. Identify each variable in the following formula.
DT b 5 K bm
DT b represents the difference between the boiling
points of a solution and the pure solvent; K b is
the molal boiling point elevation constant; m
represents the solution molality.
98. Define the term osmotic pressure, and explain
why it is considered a colligative property.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Osmotic pressure is the pressure exerted by
water molecules that move into solution through
osmosis. Osmotic pressure is a colligative property
because it depends on the number of solute
particles dissolved in solution.
Mastering Problems
99. Calculate the freezing point of a solution of
12.1 grams of naphthalene (C 10H 8) dissolved
into 0.175 kg of benzene (C 6H 6). Refer to
Table 14.6 for the necessary constant.
mol C 10H 8 5 12.1 g C10H8 3
1 mol C H
__
10
5 0.0945 mol C 10H 8
m5
into 1.00 liter of water. Use Table 14.6 to find
the freezing point of the solution.
mol MgCl 2 5
179 g MgCl
__
5 1.88 mol MgCl
2
kg H 2O 5 1.00 L H 2O 3
3
1gH O
1000 mL
_
3_
2
1L
1 kg
_
5 1.00 kg H O
1 mL H 2O
2
1000 g
m5
2
95.3 g/mol
1.88 mol MgCl
__
5 1.88m
2
1 kg H 2O
particle m 5 1.88m 3 3 5 5.64m
DTf 5 1.86°C/m 3 5.64m 5 10.5°C
Tf 5 0.0°C 2 10.5°C 5 210.5°C
101. Cooking A cook prepares a solution for
boiling by adding 12.5 grams of NaCl to a pot
holding 0.750 liters of water. At what temperature should the solution in the pot boil? Use
Table 14.5 for the necessary constant.
mol NaCl 5
12.5 g NaCl
__
5 0.214 mol NaCl
58.44 g/mol
kg H2O 5 0.750 L H2O 3
3
1gH O
1000 mL
_
3_
1 kg
_
5 0.750 kg H O
1000 g
solution molality 5
2
1L
1 mL H 2O
2
0.214 mol NaCl
__
5 0.285m
0.750 kg H 2O
particle molality 5 0.285m 3 2 5 0.570m
0.0945 mol C H
__
5 0.540m
10
8
128.08 g C 10H 8
100. In the lab, you dissolve 179 grams of MgCl 2
8
0.175 kg C 6H 6
DT f 5 5.12°C/m 3 0.540m 5 2.76°C
DTb 5 0.512°C/ m 3 0.570 m 5 0.292°C
Tb 5 100.00°C 1 0.292°C 5 100.29°C
T f 5 5.5°C 2 2.76°C 5 2.74°C
Solutions Manual
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14
SOLUTIONS MANUAL
102. The boiling point of ethanol (C 2H 5OH)
105. Household Paint Some types of paint
changes from 78.5°C to 85.2°C when an
amount of naphthalene (C 10H 8) is added to
1.00 kg of ethanol. How much naphthalene, in
grams, is required to cause this change? Refer
to Table 14.5 for needed data.
are colloids composed of pigment particles
dispersed in oil. Based on what you know
about colloids, recommend an appropriate
location for storing cans of leftover household
paint. Justify your recommendation.
DT b 5 85.2°C 2 78.5°C 5 6.70°C
When a colloid is exposed to heat, suspended
particles can settle out. Paint should be stored
in a cool location where it cannot freeze, and
away from direct sunlight and objects like water
heaters or furnaces that generate heat.
solution molality 5
moles of solute 5
6.70°C
_
5 5.49m
1.22°C/m
5.49 mol C H
__
10
8
1 kg C 2H 5OH
3 1.00 kg C 2H 5OH 5 5.49 mol C10H8
grams of solute 5 5.49 mol C 10H 8 3
__
128 g C 10H 8
1 mol C 10H 8
5 703 g C10H8
103. Ice Cream A rock salt (NaCl), ice, and water
mixture is used to cool milk and cream to
make homemade ice cream. How many grams
of rock salt must be added to water to lower
the freezing point by 10.0°C?
D T f 5 K fm
_ _
__ __
DT f
10.0°C
5 5.38m ions of Na1 and Cl2
5
m5
1.86°C/m
Kf
2.69 mol NaCl
moles of solute
5
molality 5
1 kg solvent
kilograms of solvent
58.44 g NaCl
2.69 mol NaCl
__
3 __
1 kg H2O
1 mol NaCl
5 157 g NaCl per 1 kg H 2O
106. Which solute has the greatest effect on the
boiling point of 1.00 kg of water: 50.0 grams
of strontium chloride (SrCl 2) or 150.0 grams
of carbon tetrachloride (CCl 4)? Justify your
answer.
50.0 grams SrCl 2 has the greatest effect.
mol SrCl 2 5
50.0 g mol SrCl
__
5 0.315 mol SrCl
2
158.6 g mol SrCl2
solution molality 5
MgCl2(s) in H2O(l): Yes. NH3(l) in C6H6(l): No.
H2(g) in H2O(l): No. I2(l) in Br2(l): Yes. Predictions
are based on the general rule “like dissolves
like.” A polar solvent like water will dissolve
a polar solute like magnesium chloride, and
a nonpolar solvent like liquid bromine will
dissolve a nonpolar solute like liquid iodine.
Ammonia is a polar molecule, while benzene
is nonpolar. Water is a polar molecule while
diatomic hydrogen is nonpolar.
288
Chemistry: Matter and Change • Chapter 14
0.315 mol
_
5 0.315m
1.00 kg
particle molality 5 0.315m 3 3 5 0.945m
DT b 5 0.512°C/m 3 0.945m 5 0.484°C
T b SrCl 2 solution 5 100.0°C 1 0.484°C
5 100.484°C
mol CCl 4 5
150.0 g mol CCl
__
5 0.974 mol CCl
4
154 g mol
solution molality 5
Mixed Review
104. Apply your knowledge of polarity and solubility to predict whether solvation is possible
in each situation shown in Table 14.9. Explain
your answers.
2
4
0.974 mol
_
5 0.974m
1.00 kg
particle molality 5 0.974m 3 1 5 0.974m
DT b 5 0.512°C/m 3 0.974m 5 0.310°C
T b CCl 4 solution 5 100.0°C 1 0.310°C 5 100.31°C
107. Study Table 14.4. Analyze solubility and
temperature data to determine the general trend
followed by the gases (NH3, CO2, O2) in the
chart. Compare this trend to the trend followed
by most of the solids in the chart. Identify the
solids listed that do not follow the general trend
followed by most of the solids in the chart.
For the gases, solubility decreases as
temperature increases. For most solids, solubility
increases as temperature increases. Ca(OH)2 and
Li2SO4 do not follow the general trend for solids.
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SOLUTIONS MANUAL
108. An air sample yields the percent composition
shown in Figure 14.27. Calculate the mole
fraction of each gas present in the sample.
Argon
1.00%
111. What would be the molality of the solu-
tion described in the previous problem? The
density of the Ca(NO 3) 2 solution is 1.08 kg/L.
mass of solution 5 3.00 L 3
Oxygen
21.0%
5 3.24 kg solution
mass of solute 5 246 g 3
Nitrogen
78.0%
78.0 g N 2
1 kg
_
5 0.246 kg
1000 g
mass of solvent 5 3.24 kg solution 2 0.246 kg
solute
1 mol N
3 _ 5 2.79 mol N
moles of Ca(NO3)2 5 3.00 L 3
2
28.0 g N2
2
21.0 g O 2 3
_ 5 0.656 mol O
1.00 g Ar 3
1 mol Ar
_
5 0.0251 mol Ar
32.0 g O 2
0.500 mol Ca(NO )
__
3 2
1L
5 1.50 mol Ca(NO3)2
1 mol O 2
2
molality, m 5
1.50 mol Ca(NO )
____
3 2
3.24 kg solution 2 0.246 kg solute
5 0.501m
39.9 g Ar
XN2 5
_____
2.79 mol N2
2.79 mol N 2 1 0.656 mol O 2 1 0.0251 mol Ar
5 0.804
XO2 5
0.656 mol O
_____
Think Critically
112. Develop a plan for making 1000 mL of a 5%
by volume solution of hydrochloric acid in
water. Your plan should describe the amounts
of solute and solvent necessary, as well as the
steps involved in making the solution.
2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
__
1.08 kg solution
1 L solution
2.79 mol N 2 1 0.656 mol O 2 1 0.0251 mol Ar
5 0.189
% by volume 5
X Ar 5
5% 5
_____
0.0251 mol N 2
2.79 mol N 2 1 0.656 mol O 2 1 0.0251 mol Ar
5 0.00723
109. If you prepared a saturated aqueous solution
of potassium chloride at 25°C and then heated
it to 50°C, would you describe the solution
as unsaturated, saturated, or supersaturated?
Explain.
volume solute
__
3 100
volume solution
volume solute
__
3 100
1000 mL
vol solute 5 50 mL
50 mL HCl needed. Subtract the volume of HCl
from the total solution volume to determine
a volume of 950 mL H2O needed. Dissolve
50 mL HCl in somewhat less than 950 mL H2O. Add
water until the volume of the solution is
1000 mL.
unsaturated; the solubility of KCl in water
increases with temperature. A solution at 50°C
holds more solute than one at 25°C.
110. How many grams of calcium nitrate
(Ca(NO3)2) would you need to prepare 3.00 L
of a 0.500M solution?
3.00 L 3
__ __
0.500 mol Ca(NO 3) 2
1L
3
164.09 g Ca(NO3)2
1 mol Ca(NO 3) 2
5 246 g Ca(NO3)2
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SOLUTIONS MANUAL
114. Extrapolate The solubility of argon in water
113. Compare and infer Study the phase
diagram in Figure 14.21. Compare the dotted
lines surrounding DTf and DTb and describe
the differences you observe. How might these
lines be positioned differently for solutions of
electrolytes and nonelectrolytes? Why?
at various pressures is shown in
Figure 14.28. Extrapolate the data to 15 atm.
Use Henry’s law to verify the solubility determined by your extrapolation.
Solubility (mg gas/100 g water)
Phase Diagram
Pure solvent
1 atm
Solution
Increasing Pressure
SOLID
LIQUID
∆P
∆Tf
Boiling
point of
solution
NO
60
50
Ar
40
30
O2
CH4
20
10
N2
H2
0
2.0
4.0
6.0
8.0
10.0
S
S
_
5_
GAS
Normal
freezing point
of water
70
Gas pressure (atm)
Normal
boiling point
of water
Freezing
point of
solution
Solubility v. Gas Pressure
∆Tb
Increasing Temperature
The freezing point of the solution is below the
normal freezing point of water, while the boiling
point of the solution is above the normal boiling
point of water. DTf and DTb would be larger for
electrolytes than nonelectrolytes. Electrolytes
dissociate in water, resulting in a larger number
of particles in solution.
1
2
P1
P2
S2 5
___
(55 mg Ar/100 g H 2O)(15 atm)
(10.0 atm)
5 82 mg Ar/100 g H2O
115. Infer Dehydration occurs when more fluid
is lost from the body than is taken in. Scuba
divers are advised to hydrate their bodies
before diving. Use your knowledge of the
relationship between pressure and gas solubility
to explain the importance of hydration prior
to a dive.
As pressure increases with water depth during
a dive, gas concentration in the blood increases.
If blood (solvent) volume is low, the gas (solute)
concentration will be higher than normal levels
at specific depths. A well-hydrated diver has a
greater amount of solvent in which gases can be
dissolved.
290
Chemistry: Matter and Change • Chapter 14
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SOLUTIONS MANUAL
116. Graph Table 14.10 shows solubility data was
collected in an experiment. Plot a graph of the
molarity of KI versus temperature. What is the
solubility of KI at 55°C?
Solubility vs. Temperature
13
Solubility (M)
12
has the highest concentration? Rank the solutions from the greatest to the smallest boiling
point depression. Explain your answer.
a. 0.10 mol NaBr in 100.0 mL solution
b. 2.1 mol KOH in 1.00 L solution
c. 1.2 mol KMnO4 in 3.00 L solution
The molarities are 1.0M NaBr, 2.1M KOH, and
0.40M KMnO4. Because the KOH solution
contributes the greatest concentration of
particles to solution, it has the greatest
boiling point elevation; KMnO4 has the lowest
concentration of particles and the smallest
boiling point depression. Boiling point elevation
depends only upon concentration.
11
10
9
8
0.10 mol NaBr
__
5 1.0M NaBr
0
0.1000 L
0
20
40
60
80
Temperature (°C)
100
Molarity equals 8.67M, 9.76M, 10.6M, 11.6M,
and 12.4M at 20°C, 40°C, 60°C, 80°C, and 100°C,
respectively. The solubility of KI at 55°C is about
10.4M.
117. Design an Experiment You are given a
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
118. Compare Which of the following solutions
sample of a solid solute and three aqueous
solutions containing that solute. How would
you determine which solution is saturated,
unsaturated, and supersaturated?
Add a pinch of solute to each container. If
the solution is supersaturated, crystallization
will occur; saturated, no solute will dissolve;
unsaturated, solute will dissolve.
2.1 mol KOH
__
5 2.1M KOH
1.00 L
1.2 mol KMnO
__
5 0.40M KMnO
4
3.00 L
4
Challenge
Measurements of Solubility of a Gas
Measurement
Solubility
1
0.225
2
0.45
3
0.9
4
1.8
5
3.6
119. Interpret the solubility data in Table 14.11
using the concept of Henry’s Law.
In Henry’s law, solubility is directly proportional
to pressure. In this example, each measurement
indicates a doubling of the solubility
value. This indicates that the pressure is
also doubling between measurements. An
additional observation might include that from
Measurement 1 to Measurement 5, the solubility
has increased by a factor of 16. Therefore, the
pressure would do the same.
Solutions Manual
Chemistry: Matter and Change • Chapter 14
291
14
SOLUTIONS MANUAL
120. You have a solution containing 135.2 grams
122. Identify which of the following molecules is
of dissolved KBr in 2.3 liters of water. What
volume of this solution, in mL, would you use
to make 1.5 liters of a 0.1M KBr? What is the
boiling point of this new solution?
polar. (Chapter 8)
a. SiH4
Step1: Calculate molarity of original solution
b. NO2
135.2 g KBr 3
M5
1 mol KBr
_
5 1.14 mol KBr
polar
119 g KBr
c. H2S
1.14 mol KBr
__
5 0.496 M
2.3 L H2O
polar
Step 2: Dilute the solution – Calculate required
volume
123. Name the following compounds. (Chapter 7)
a. NaBr
1L
Step 3: Calculate boiling point of new solution
ΔTb 5 Kbm
0.10 mol KBr
__
3 __ 3
1LH O
1000 mL H O
1000
g
H
O
1
mL
H
_ 1 __O 5 0.10m
1 L H2O
m5
2
d. NCl3
polar
0.10M 3 1.5 L
__
0.496M
1000 mL 5 300 mL
5 0.30 L 3 _
V1 5
sodium bromide
b. Pb(CH3COO)2
lead(II) acetate
2
2
1 g H2O
nonpolar
2
1 kg H2O
particle molality 5 0.10m 3 2 5 0.20m
ΔTb 5 0.512°C/m 3 0.20m 5 0.10°C
Tb 5 100.0°C 1 0.10°C 5 100.1°C
Cumulative Review
121. The radius of an argon atom is 94 pm.
Assuming the atom is spherical, what is the
volume of an argon atom in nm3? V 5 4/3πr3.
(Chapter 3)
1 nm
94 pm 3 _ 5 0.094 nm
c. (NH4)2CO3
ammonium carbonate
124. A 12.0-g sample of an element contains
5.94 3 1022 atoms. What is the unknown
element? (Chapter 10)
5.94 3 1022 atoms 3
1 mol
__
6.02 3 1023 atoms
5 0.0987 mol
12.0 g
__
5 122 g/mol
0.0987 mol
The atomic mass is 122 amu. The element is
antimony.
1000 pm
V 5 (4/3)(3.14)(0.094 nm)3 5 3.5 3 1023 nm3
125. Pure bismuth can be produced by the reaction
of bismuth oxide with carbon at high
temperatures.
2Bi2O3 1 3C 0 4Bi 1 3CO2
How many moles of Bi2O3 reacted to produce
12.6 mol of CO2? (Chapter 11)
12.6 mol CO2 3
292
Chemistry: Matter and Change • Chapter 14
2 mol Bi O
__
5 8.40 mol Bi O
2
3
3 mol CO2
2
3
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14
Additional Assessment
SOLUTIONS MANUAL
128. At what latitude are average dissolved oxygen
values the lowest?
Writing in Chemistry
126. Homogenized Milk The first homogenized
milk was sold in the United States around 1919.
Today, almost all milk sold in this country is
homogenized, in the form of a colloidal
emulsion. Research the homogenization
process. Write a brief article describing the
process. The article may include a flowchart or
diagram of the process, as well as a discussion
of the reputed benefits and/or drawbacks
associated with drinking homogenized milk.
Student answers will vary. Students should note
that raw milk contains fat dispersed throughout.
If left to stand, the fat separates out, leaving a
cream layer and a skim milk layer. The process
of homogenization breaks the fat globules into
smaller sizes and reduces their tendency to form
a cream layer.
Values are lowest near the equator.
129. Describe the general trend defined by the data.
Relate the trend to the relationship between
gas solubility and temperature.
In general, dissolved oxygen in surface
ocean waters increases as latitude increases
towards both north and south. Surface water
temperatures are greatest near the equator.
Surface water temperature decreases toward the
poles. As temperature decreases, gas solubility
generally increases.
Standardized Test Practice
pages 512–513
Bromine (Br2) Concentration of
Four Aqueous Solutions
0.9000
Document-Based Questions
Percent by mass
Percent by volume
0.6000
0.4779
0.5000
0.0000
1
2
3
0.1545
0.1000
0.1596
0.2000
0.1030
0.2575
0.3000
0.0515
0.4000
0.3189
Percent
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
0.7000
0.7947
0.8000
4
Solution number
127. Are dissolved oxygen values most closely
related to latitude or longitude? Why do you
think this is true?
Dissolved oxygen values are most closely related
to latitude. Surface land and water temperatures
are more closely correlated to latitude than
longitude.
1. What is the volume of bromine (Br2) in 7.000 L
of Solution 1?
a. 55.63 mL
b. 8.808 mL
c. 18.03 mL
d. 27.18 mL
c
Volume of Br2 5 (7.000 L) 3 (0.002575) 5
0.01803 L 5 18.03 mL
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SOLUTIONS MANUAL
2. How many grams of Br2 are in 55.00 g of
H
11.7%
Solution 4?
a. 3.560 g
b. 0.084 98 g
c. 1.151 g
d. 0.2628 g
O
10.4%
C
77.9%
d
Mass of Br2 5 (55.00 g)
3 (0.004779) 5 0.2628 g
3. Which one is an intensive physical property?
a.
b.
c.
d.
volume
length
hardness
mass
c
6. What is the empirical formula for this
substance?
a. CH2O
b. C8HO
c. C10H18O
d. C7H12O
c
a.
b.
c.
d.
NCl2
2NOCl
N2O2
2ClO
Assume a 100.0 g sample.
Determine the number of moles.
1 mol C
77.9 g C 3
5 6.486 mol C
12.01 g C
_
10.4 g O 3
1 mol O
_
5 0.65 mol O
11.7 g H 3
1 mol H
_
5 11.607 mol H
b
5. If 1 mole of each of the solutes listed below
is dissolved in 1 L of water, which solute will
have the greatest effect on the vapor pressure of
its respective solution?
a. KBr
b. C6H12O6
c. MgCl2
d. CaSO4
16.00 g O
1.008 g H
Calculate the simplest ratio of moles.
6.486 mol C
0.65 mol O
__
5 9.98; __ 5 1.00;
0.65
0.65
11.607 mol H
__
5 17.86
0.65
The empirical formula is C10H18O.
7. What is the correct chemical formula for the
c
MgCl2 will produce the most number of particles
in solution: 1 mol Mg21, 2 mol Cl2
ionic compound formed by the calcium ion
(Ca21) and the acetate ion (C2H3O22)?
a. CaC2H3O2
b. CaC4H6O3
c. (Ca)2C2H3O2
d. Ca(C2H3O2)2
d
294
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Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
4. What is the product of this synthesis reaction?
Cl2(g) 1 2NO(g) 0?
14
CHAPTER
SOLUTIONS MANUAL
8. If 16 moles of H2 are used, how many moles of
Fe will be produced?
a. 6
b. 3
c. 12
d. 9
11. How many moles of KClO3 can be dissolved in
100 g of water at 60°C?
21 grams
12. Which can hold more solute at 208C: NaCl
or KCl? How does this compare to their
solubilities at 808C?
c
Fe3O4(s) 1 4 H2(g) → 3 Fe(s) 1 4 H2O(l)
16 mol H2 3
At 20°C, the NaCl solution can hold more solute.
At 80°C, the solubilities are reversed and KCl is
more soluble than NaCl.
3 mol Fe
_
5 12 mol Fe
4 mol H2
9. If 7 moles of Fe3O4 are mixed with 30 moles of
H2, what will be true?
a. There will be no reactants left.
b. 2 moles of hydrogen gas will be left over
c. 30 moles of water will be produced
d. 7 moles of Fe will be produced
b
Fe3O4(s) 1 4 H2(g) → 3 Fe(s) 1 4 H2O(l)
7 mol Fe3O4 3
4 mol H
__
5 28 mol H
2
2
1 mol Fe3O4
used
13. How many moles of KClO3 would be required
to make 1 L of a saturated solution of KClO3 at
75°C?
(30 g/L)(1 mol/122.55 g KClO3) 5 0.245 mol KClO3
in 1 liter.
Use the information below to answer
Questions 14 and 15.
The electron configuration for silicon is 1s2 2s2 2p6
3s2 3p2.
14. Explain how this configuration demonstrates
the aufbau principle.
The aufbau principle dictates that electrons must
fill the lowest available energy levels before
filling any higher energy levels.
10. What is the molar mass of Fe3O4?
a.
b.
c.
d.
231.56 g/mol
71.85 g/mol
287.40 g/mol
215.56 g/mol
15. Draw the orbital diagram for silicon. Explain
how Hund’s rule and the Pauli exclusion
principle are used in constructing the orbital
diagram.
a
3 mol Fe 3 55.847 g/mol 5 167.541 g Fe
Hund’s rule mandates that the last two electrons
will be placed in separate p-orbitals. The Pauli
exclusion principal determines that shared electrons in any given orbital must have opposite
spins, as shown by up and down arrows.
4 mol O 3 15.999 g/mol 5 63.996 g O
molar mass = 167.541 g 1 63.996 g 5 231.537 g/mol
Solubility (g of solute/100 g H2O)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
30 mol H2 2 28 mol used 5 2 mol H2 remaining
Solubilities as a Function
of Temperature
100
90
80
70
CaCl2
60
50
40
30
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
1s
2s
2p
3s
↑
↑
3p
KCl
NaCl
20
10
0
KClO3
Ce2(SO4)3
0 10 20 30 40 50 60 70 80 90 100
Temperature (°C)
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14
SOLUTIONS MANUAL
18. What is the electronegativity difference in the
16. What volume of a 0.125M NiCl2 solution
contains 3.25 g NiCl2?
a. 406 mL
b. 32.5 mL
c. 38.5 mL
d. 26.0 mL
e. 201 mL
compound Li2O?
a. 1.48
b. 2.46
c. 3.4
d. 4.42
e. 5.19
e
b
3.25 g 3
1 mol
1000 mL
1L
_
3_ 3_
129.6 g
5 201 mL
3.44 – 0.98 5 2.46
1L
0.125 mol
19. Which bond has the greatest polarity?
a.
b.
c.
d.
e.
17. Which is NOT a colligative property?
a.
b.
c.
d.
e.
boiling point elevation
freezing point depression
vapor pressure increase
osmotic pressure
heat of solution
C2H
Si2O
Mg2Cl
Al2N
H2Cl
c
C-H: 2.55 2 2.20 5 0.35
c
Si-O: 3.44 2 1.90 5 1.54
Use the data table below to answer
Questions 18 and 19.
Mg-Cl: 3.16 2 1.31 5 1.85
Al-N: 3.04 2 1.61 5 1.43
Electronegativities of Selected Elements
Mg-Cl has the greatest polarity.
2.20
296
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
H-Cl: 3.16 2 2.20 5 0.96
H
Li
Be
B
C
N
O
F
0.98
1.57
2.04
2.55
3.04
3.44
3.93
Na
Mg
Al
Si
P
S
Cl
0.93
1.31
1.61
1.90
2.19
2.58
3.16
Chemistry: Matter and Change • Chapter 14
Solutions Manual