Chapter 8 Review, pages 416–421
Knowledge
1. (a)
2. (b)
3. (d)
4. (d)
5. (c)
6. (a)
7. (c)
8. (b)
9. (d)
10. False. Most of the water on Earth is salt water.
11. False. When a solid dissolves in a liquid, the solid is always the solute.
12. False. Liquids that are miscible with each other mix in all proportions.
13. True
14. False. Changes in temperature significantly affect the solubility of a solid in a liquid.
15. False. Amount concentration is the correct IUPAC name for solution concentration.
16. True
17. False. A graduated pipette can deliver a range of volumes.
18. True
19. (a) (v)
(b) (ii)
(c) (i)
(d) (iv)
(e) (iii)
20. (a) (iii)
(b) (vi)
(c) (v)
(d) (i)
(e) (iv)
Understanding
21. Water is a liquid, not a gas, at room temperature because of hydrogen bonding. The hydrogen
bonds hold water molecules together as a liquid up to temperatures of 100 ºC.
22. Potable water is water that is suitable for human consumption.
23. A homogeneous mixture is a solution of two or more substances that contains only one
phase. The components are uniformly mixed, giving a uniform appearance. Apple juice is a
homogeneous mixture. A heterogeneous mixture has substances that are in two or more phases.
Oil and water is a heterogeneous mixture.
24. A solution that has a relatively high quantity of solute compared to the volume of solution is
considered to be a concentrated solution. Since this solution contains twice as much sugar as
water, it is a concentrated solution.
25. (a) An amalgam is an alloy of mercury and another metal.
(b) Silver-coloured dental fillings are amalgams of mercury and silver.
26. During the process of hydration, ions are surrounded by water molecules. Hydration helps
stabilize the ions in solution.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8-2
27. A surfactant reduces the surface tension of water by breaking down the hydrogen bonding
network at the surface. This allows water to mix with a non-polar liquid such as oil.
28. Dispersants work to break up the oil into smaller droplets that gradually sink to the sea floor.
The smaller droplets make it easier for naturally occurring micro-organisms in the ocean to
decompose the oil into simpler, less toxic substances. The combination of the dispersants and the
micro-organisms minimize the damage caused by oil spills.
29. From Figure 2, the solubility of sucrose in water at 5 ºC is 210 g/100 g water. A solution
containing 150 g of sucrose dissolved in 100 g of water would be unsaturated because the point
on the graph would be below the curve.
30. (a) In general, the solubility of a solid in a liquid increases as the temperature increases.
Pressure has no significant effect on the solubility of a solid in a liquid.
(b) In general, the solubility of a gas in a liquid generally decreases as the temperature increases
and the solubility of a gas in a liquid increases as the applied pressure increases.
31. To prepare 1.00 L of solution, a known quantity of solute is dissolved in about one quarter
(250 mL) of the volume of the solvent, such as water. Additional solvent is carefully added until
the bottom of the solution’s meniscus lines up with the 1.00 L calibration mark on the flask.
32. Students do not usually use stock solutions of acids in a high school laboratory because stock
solutions are concentrated solutions.
n
33. (a) c =
V
n
(b) V =
c
(c) n = cV
34. Pipettes are used for analytical work involving small volumes. A volumetric pipette delivers
a fixed volume of solution. A graduated pipette has volume markings or graduations, and can
deliver a range of volumes, from 0.1 mL to 10.0 mL. The graduated pipette is not as precise as a
volumetric pipette.
cV
35. (a) cc = d d
Vc
(b) Vc =
(c) cd =
(d) Vd =
cdVd
cc
ccVc
Vd
ccVc
cd
36. Liquid household cleaners are usually very dilute solutions of a chemical or several
chemicals. The concentration of chemical is very low so that any fumes that may be emitted
from the cleaner will not produce serious or immediate health problems. When diluting
hydrochloric acid in the lab, you use concentrated hydrochloric acid. Concentrated hydrochloric
acid is very noxious and highly corrosive. You must perform this operation under a fume hood
and wear safety eyeware, apron, and gloves.
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Chapter 8: Water and Solutions
8-3
37. (a) Given: mmineral = 2.53 g
cppb = 2.53 ppb
Required: mass of solution, msolution
Analysis: cppb =
msolute
msolution
! 109
Solution:
Step 1. Rearrange the equation to solve for the mass of solution.
m
msolution = solute ! 109
cppb
Step 2. Substitute the correct values and solve.
2.53 g
msolution =
! 109
2.53
msolution = 1.0 ! 109 g
Statement: The mass of solution that contains 2.53 g of the mineral is 1.0 × 109 g.
(b) An aqueous solution has the same density as water, 1.0 g/mL.
1L
Vsolution = 1.0 ! 109 mL !
1000 mL
Vsolution = 1.0 ! 106 L
The volume of solution that contains 2.53 g of the mineral is 1.0 × 106 L.
Understanding
38. The water molecule is highly polar because of the polar H–O bonds and their asymmetrical
arrangement.
39. (a) The two types of intermolecular forces that exist between water molecules are hydrogen
bonding and London dispersion forces. Hydrogen bonds form between the hydrogen atom of one
water molecule and the oxygen atom of a nearby molecule. Hydrogen bonding is a stronger force
than the London dispersion force, holding the water molecules together as a liquid up to high
temperatures (100 ºC).
(b) Most of the physical properties of water are attributed to hydrogen bonding.
40. The high boiling point of water is related to the hydrogen bonding between water molecules
that provide the additional “stickiness” to hold water molecules together as a liquid, up to almost
100 °C.
41. (a) The volume of water needed to produce 1 kg of coffee is 20 000 L, while the volume of
water need to produce 1 kg of rice is 5000 L. Therefore, it takes four times as much water to
produce the same amount of coffee as it does to produce rice.
(b) The volume of water needed to produce 1 kg of sugar plus 1 L of milk is equal to 5000 L
(3000 L + 2000 L). The other product that requires 5000 L of water is 1 kg of cheese or rice.
42. Acetic acid is the solvent and water is the solute because water is used to dilute the acetic
acid.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8-4
43. (a) A tossed salad is a heterogeneous mixture.
(b) A sample of 14-karat gold is an alloy. It is a homogeneous mixture.
(c) Clear, dry air is a homogeneous mixture.
(d) Sweetened iced tea is a homogeneous mixture.
44. (a) Coffee with fully dissolved sugar is a concentrated solution.
(b) Lightly salted chicken soup is a dilute solution.
(c) Steel with small percentages of iron and carbon is a dilute solution.
45. (a) Only one phase is visible in a homogeneous mixture while two or more phases are visible
in a heterogeneous mixture.
(b) A solution is a homogeneous mixture of a solute in a solvent. The solute is the substance that
there is less of and the solvent is the substance that there is a greater quantity of.
(c) A concentrated solution has a relatively high quantity of solute compared to the volume of
solution. A dilute solution has a relatively low quantity of solute compared to the volume of
solution.
(d) A surfactant is a substance that acts on the surface of a liquid, reducing surface tension. This
allows a polar solvent such as water to mix with a non-polar liquid such as oil. A detergent is a
type of surfactant.
(e) Miscible liquids mix together to form a solution while immiscible liquids do not mix
together.
46. When an ionic compound dissolves in water, the water molecules reorient themselves so that
the negative (oxygen) end of each molecule is attracted to a nearby positively charged sodium
ion and the positive (hydrogen) end is attracted to the chloride ion. The water–ion attractions
start pulling the ions away from the crystal and the crystal dissolves. Once in solution, each ion
is surrounded by a layer of water molecules so that the ions cannot attract each other.
47. (a) KCl(s) → K+(aq) + Cl–(aq)
(b) MgBr2(s) → Mg2+(aq) + 2 Br–(aq)
(c) Li2SO4(s) → 2 Li+(aq) + SO42–(aq)
48. Water is not written as part of a chemical equation that shows the dissociation of an ionic
compound in an aqueous solution because it does not undergo a chemical change during the
reaction.
49. Ethanol, unlike most molecular compounds, mixes readily with water because both
compounds are V-shaped molecules with a central oxygen atom. Water has two O–H bonds
while ethanol has one O–H bond. Both compounds can form hydrogen bonds with neighbouring
molecules as well as with each other when they are mixed together to form an aqueous solution.
50. The statement “like dissolves like” means that solutes dissolve in solvents of similar polarity.
Therefore, polar solutes dissolve in polar solvents and non-polar solutes dissolve in non-polar
solvents.
51. (a) When the crystal dissolves quickly, the solution is unsaturated.
(b) When the crystal does not grow, the solution is saturated.
(c) When the crystal grows rapidly, the solution is supersaturated.
52. (a) Thermal pollution is excess thermal energy released into water. It can result when warm
water that has been used in an industrial process is discharged into the environment.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8-5
53. Since the solubility of a gas increases as the temperature decreases, cooling the drink while
adding carbon dioxide gas would help obtain a high concentration of gas in the pop. Since the
solubility of a gas in a liquid increases as the pressure of the gas is increased, pumping the
carbon dioxide under pressure into the drink would also obtain a high concentration of carbon
dioxide gas in the pop.
54. Given: VCaCl = 1500 L
2
nCaCl = 245 mol
2
Required: amount concentration of calcium chloride solution, cCaCl
Analysis: cCaCl =
2
nCaCl
2
VCaCl
2
2
Solution: Substitute the correct values into the equation and solve.
245 mol
cCaCl =
2
1500 L
mol
cCaCl = 0.16
2
L
Statement: The amount concentration of calcium chloride solution is 0.16 mol/L.
55. Given: VNaOH = 125 mL
mNaOH = 8.2 g
Required: amount concentration of sodium hydroxide solution, cNaOH
Analysis: cNaOH =
nNaOH
VNaOH
Solution:
Step 1. Convert the volume from millilitres to litres.
1L
VNaOH = 125 mL !
1000 mL
VNaOH = 0.125 L
Step 2. Determine the molar mass of sodium hydroxide.
M NaOH = M Na + M O + M H
g
g
g
+ 16.00
+ 1.01
mol
mol
mol
g
M NaOH = 40.00
mol
Step 3. Determine the amount of sodium hydroxide.
1 mol
nNaOH = 8.2 g !
40.00 g
= 22.99
nNaOH = 0.2050 mol [2 extra digits carried]
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Chapter 8: Water and Solutions
8-6
Step 4. Determine the concentration of the sodium hydroxide.
0.2050 mol
cNaOH =
0.125 L
mol
cNaOH = 1.6
L
Statement: The amount concentration of sodium hydroxide solution is 1.6 mol/L.
56. (a) Yes, it does make a difference whether you use the volume of the final solution or the
volume of the solvent added. The volume of the final solution is needed because the solute has a
mass, which means that it will add to the volume of the final solution.
(b) You cannot just add 500 mL of water to the solute because the solute itself adds volume to
the solution. Therefore, you must add less than 500 mL of water to obtain a final volume of
500 mL after the solute is added.
57. Given: VKNO = 250 mL
3
mKNO = 25 g
3
Required: amount concentration of potassium nitrate solution, cKNO
Analysis: cKNO =
3
nKNO
3
VKNO
3
3
Solution:
Step 1. Convert volume from millilitres to litres.
1L
VKNO = 250 mL !
3
1000 mL
VKNO = 0.25 L
3
Step 2. Determine the molar mass of potassium nitrate.
M KNO = M K + M N + 3M O
3
= 39.10
!
g
g
g $
+ 14.01
+ 3# 16.00
mol
mol
mol &%
"
g
mol
Step 3. Determine the amount of potassium nitrate.
1 mol
nKNO = 25 g !
3
101.11 g
M KNO = 101.11
3
nKNO = 0.2473 mol [2 extra digits carried]
3
Step 4. Determine the concentration of potassium nitrate.
0.2473 mol
cKNO =
3
0.25 L
mol
cKNO = 0.99
3
L
Statement: The amount concentration of potassium nitrate solution is 0.99 mol/L.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8-7
58. Given: VKMnO = 14.5 L
4
cKMnO = 0.585 mol/L
4
Required: mass of potassium permanganate, mKMnO
Analysis: cKMnO =
4
nKMnO
4
VKMnO
4
4
Solution:
Step 1. Rearrange the formula to solve for amount.
nKMnO = cKMnO VKMnO
4
4
4
Step 2. Substitute the given values and solve.
0.585 mol
nKMnO =
! 14.5 L
4
1L
nKMnO = 8.4825 mol [2 extra digits carried]
4
Step 3. Determine the molar mass of potassium permanganate.
M KMnO = M K + M Mn + 4 M O
4
= 39.10
!
g
g
g $
+ 54.94
+ 4 # 16.00
mol
mol
mol &%
"
g
mol
Step 4. Determine the mass of potassium permanganate.
158.04 g
mKMnO = 8.4825 mol !
4
1 mol
mKMnO = 1340 g
M KNO = 158.04
3
4
Statement: The mass of potassium permanganate in 14.5 L of a 0.585 mol/L solution is 1340 g.
59. Table 1 Unknown and Known Quantities in Equations
Unknown
quantity
Given (known) quantity
Equation
n
amount
volume of solution, amount of
(a) c =
concentration solute
V
n
volume of
amount concentration, amount of
(b) V =
solution
solute
c
concentration of concentrated
concentration
solution, volume of concentrated (c) c = ccVc
of dilute
d
solution, volume of dilute
Vd
solution
solution
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Chapter 8: Water and Solutions
8-8
60. (a) ccVc = cdVd
(b) Yes, the equation takes into account the amount of solute in moles because the units on both
sides of the equation are mol/L × L. Since the volume units cancel, we are left with the final unit
of moles for both sides of the equation.
61. The definition given for dilution is incorrect because it assumes that water is involved in all
dilutions. Water is not involved in all dilutions. Dilution is the process of reducing the
concentration of a solution by adding more solvent to a solution.
62. (a) Given: cc = 18.0 mol/L
cd = 4.0 mol/L
Vd = 1.00 L
Required: volume of the concentrated solution, Vc
Analysis: ccVc = cdVd
Solution:
Step 1. Rearrange the equation to solve for initial volume.
cV
Vc = d d
cc
Step 2. Substitute the correct values and solve.
!
mol $
# 4.0 L & 1.00 L
"
%
Vc =
mol
18.0
L
Vc = 0.22 L
Statement: To prepare this solution, 0.22 L (or 220 mL) of 18.0 mol/L sulfuric acid solution
must be diluted to 1.00 L.
(b) Given: cc = 4.0 mol/L
(
)
cd = 3.3 mol/L
Vd = 100 mL
Required: volume of the concentrated solution, Vc
Analysis: ccVc = cdVd
Solution:
Step 1. Rearrange the equation to solve for initial volume.
cV
Vc = d d
cc
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Chapter 8: Water and Solutions
8-9
Step 2. Substitute the correct values and solve.
!
mol $
3.3
100 mL
#
L &%
"
Vc =
mol
4.0
L
Vc = 82.5 mL
Statement: The volume of fresh battery acid solution containing the same amount of sulfuric
acid is 82.5 mL.
63. Given: cc = 1.50 mol/L
(
)
Vc = 245 mL
Vd = 568 mL
Required: concentration of diluted solution, cd
Analysis: ccVc = cdVd
Solution:
Step 1. Rearrange the equation to solve for initial volume.
cV
Vc = d d
cc
Step 2. Substitute the correct values and solve.
1L
Vc = 245 mL !
1000 mL
Vc = 0.245 L
Step 3. Convert volume from millilitres to litres.
1L
Vd = 568 mL !
1000 mL
Vd = 0.568 L
Step 4. Rearrange the equation to solve for final concentration.
cV
cd = c c
Vd
Step 5. Substitute the correct values and solve.
!
mol $
1.50
0.245 L
#"
L &%
cd =
0.568 L
mol
cd = 0.647
L
Statement: The final amount concentration of the solution is 0.647 mol/L.
(
)
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8-10
64. (a) The nurse needs to know the molar mass of the glucose to calculate the percentage
weight/volume of the solution.
(b) Given: nglucose = 1.2 mol
Vglucose = 1200 mL
Required: concentration of final solution as percentage weight/volume, cW/V
Analysis: cW/V =
mglucose
Vsolution
! 100 %
Solution:
Step 1. Determine the molar mass of glucose.
M C H O = 6 M C + 12 M H + 6 M O
6
12
6
!
!
!
g $
g $
g $
= 6 # 12.01
+
12
1.01
+
6
16.00
#"
#"
mol &%
mol &%
mol &%
"
g
mol
Step 2. Determine the mass of glucose.
180.18 g
mglucose = 1.2 mol !
1 mol
mglucose = 216.2 g
MC H
6
12 O6
= 180.18
Step 3. Determine the concentration of glucose.
216.2 g
cW/V =
! 100 %
1200 mL
cW/V = 18.0 %
Statement: The percentage weight/volume of glucose is 18.0 %.
65. Percentage concentration is always less than 100 % because the volume of the final solution
in the denominator includes the volume of the solute in the numerator.
66. (a) Convert the concentration in ppm to ppb.
1000 ppb
cppb = cppm !
1 ppm
1000 ppb
= 290 ppm !
1 ppm
cppb = 2.9 ! 105 ppb
The concentration in ppb is 2.9 × 105 ppb.
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Chapter 8: Water and Solutions
8-11
(b) Convert the concentration in ppm to ppt.
1 ! 106 ppt
cppt = cppm !
1 ppm
6
= 290 ppm !
1 ! 10 ppt
1 ppm
cppt = 2.9 ! 108 ppt
The concentration in ppt is 2.9 × 108 ppt.
Analysis and Application
67. (a) Water is used in almost all manufacturing and power generation processes. As the
demand for manufactured goods increases, the demand for industrial use of water also increases.
Water is used in fabricating, processing, washing, diluting, cooling, and transporting a product.
(b) Answers may vary. Sample answers: The oil and gas industry could reduce its consumption
of fresh water by using saline groundwater, which is not suitable for drinking, or recycled water.
Other industries could improve methods of recycling and reusing water in the manufacturing
process.
68. (a) Glucose is soluble in water because it contains five polar O–H bonds. The oxygen atoms,
which have partial negative charges, form hydrogen bonds with the hydrogen atoms of the water
molecules.
(b) It is essential that glucose is water soluble because glucose is the most common energy-rich
fuel used by cells in the body. Brain cells use glucose exclusively as fuel.
69. I would expect bromine, Br, to more soluble in a liquid hydrocarbon, like carbon tetrachloride,
CCl4, than in water, H2O.
Examine the electronegativity values for an atom of bromine, carbon, chloride, hydrogen, and
oxygen.
Br = 3.0
C = 2.6
H = 2.2
Cl = 3.2
O = 3.4
Difference
0.6
1.2
The electronegativity difference for a C-Cl bond indicates the bond is mostly covalent.
The electronegativity difference for a H-O bond indicates the bond is mostly polar covalent.
The electronegativity difference for a Br-C-Cl bond is 2.4.
The electronegativity difference for a Br-H-O bond is 1.8.
The difference is greater with a Br-C-Cl bond so it is more likely that bromine will dissolve more
readily in carbon tetrachloride.
70. (a) A surfactant is a substance that acts on the surface of a liquid, reducing surface tension.
Soaps and detergents are examples of surfactants.
(b) Answers may vary. Sample answer: Two different uses of surfactants are to remove greasy
stains from clothes and to clean up oil spills.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8-12
(c) Adding a surfactant such as detergent to laundry removes a grease stain from clothes because
the detergent contains a long hydrophobic hydrocarbon tail end and a hydrophilic head. The nonpolar hydrophobic tails are attracted to the grease while the hydrophilic heads remain surrounded
by the water molecules to which they are attracted. As the water molecules move due to the
agitation of the washing machine, they pull the detergent ions along. This helps dislodge the
grease from the surface. In the water, the grease becomes coated with more detergent ions.
71. The solubility of ammonium chloride, NH4Cl(s), at 50 ºC is 50 g/100 g H2O. Originally,
60 g of ammonium chloride was dissolved in 100 g of water. Therefore, the amount of
ammonium chloride that crystallizes out of solution is:
mNH Cl = 60 g ! 50 g
4
mNH Cl = 10 g
4
Ten grams of ammonium chloride crystallizes out of solution, leaving a saturated solution of
50 g/100 g H2O.
72. The aquarium that is in direct sunlight would probably need a pump to bubble air through the
water. The aquarium that is in the shade would contain more oxygen in the water than the
aquarium in direct sunlight because it would have cooler water. Since the solubility of gases
decreases as the temperature increases, the cool water contains more dissolved oxygen than
warm water.
73. Given: cc = 1.5 g/L
cd = 1.3 g/L
Vd = 20.0 L
Required: volume required to bring the volume back to 20.0 L, Vr
Analysis: Use the dilution equation ccVc = cdVd , then solve for Vr = Vd ! Vc .
Solution:
Step 1. Rearrange the dilution equation to solve for the initial volume.
cV
ccVc = d d
cc
Step 2. Substitute the correct values and solve.
!
g$
# 1.3 L & 20.0 L
"
%
Vc =
g
1.5
L
Vc = 17.3 L
Step 3. Determine the required volume.
Vr = 20.0 L ! 17.3 L
(
)
Vr = 2.7 L
Statement: The volume of water that must be added to the tank is 2.7 L.
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Chapter 8: Water and Solutions
8-13
74. (a) Given: msugar = 88 g
Vsolution = 439 mL
Required: concentration of final solution as percentage weight/volume, cW/V
Analysis: cW/V =
msugar
! 100 %
Vsolution
Solution: Substitute the correct values into the equation and solve.
msugar
cW/V =
! 100 %
Vsolution
88 g
! 100 %
439 mL
cW/V = 20 %
Statement: The percentage weight/volume of the solution is 20 %.
(b) The concentration is measured in percentage weight/volume because the mass of the solute is
given.
75. The units to represent ppm of 1 g/106 mL and 1 g/106 g are equivalent if the solvent is water.
The density of water is 1 g/mL.
76. (a) The benefits associated with using dispersants include protecting the shoreline
ecosystems from oil, breaking up the oil into smaller particles that sink to the floor, and making
it easier for micro-organisms to decompose the oil into less toxic substances. Using dispersants
also reduces the damage caused by floating oil. The risks associated with using dispersants are
that it does not eliminate the oil and that it causes the smaller oil particles to contaminate the
ocean floor, destroying that ecosystem. It is also easier for oil particles to get into the food chain
once the particles are consumed by plankton.
(b) Answers may vary. Students may agree with the decision to use dispersants since the
shoreline ecosystem was saved.
77. (a)
=
(b) The line can be extended by fitting a straight line to the data as indicated by the dashed line.
(c) Based on the fitted straight line, the solubility at 80 ºC is 220 g/100 g H2O. Answers may
vary depending on the method used to extend the curve.
(d) Answers may vary. Using the straight line fit, the predicted value is 12 g/100 g H2O above
the actual value (220 g/100 g H2O – 208 g/100 g H2O).
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Chapter 8: Water and Solutions
8-14
(e) Answers may vary. The percent error can be calculated as
(220 g/100 g H 2O –208 g/100 g H 2O)
percent error =
! 100 %
208 g/100 g H 2O
percent error = 6 %
The percent difference is 6 %.
78. Yes, both of these statements are true because the solubility curves shown in Figure 4 of
Section 8.5 show that the solubility of potassium nitrate lies above the curve of potassium
chlorate and rises much more steeply with temperature.
79. To prepare 250 mL of a standard solution, we use a volumetric flask, distilled water, a
dropper, and a required mass of solute. The solid is dissolved in about 100 mL of distilled water
in a volumetric flask. Then distilled water is added using the dropper until the bottom of the
solution’s meniscus lines up with the calibration mark on the flask. A volumetric flask is used
rather than a beaker because the volumetric flask is designed to hold a certain volume of
solution, +/– 0.1 mL at a specific temperature (often 20 °C). This allows for greater accuracy in
measuring.
80. (a) The label “70 % alcohol” may not be accurate depending on the method in which
concentration was determined. The concentration could be by percentage concentration,
percentage volume/volume, percentage weight/volume, or percentage weight/weight.
(b) Since the isopropyl alcohol is a dilute solution of concentrated isopropyl alcohol and water, it
should be labelled percentage volume/volume 70 %/30 % isopropyl alcohol.
Reflect on Your Learning
81. (a) Three important ways that water is naturally purified are through evaporation and
condensation, filtration, and bacterial action. Evaporation leaves solutes behind. Water is filtered
as it passes through sand and gravel in the ground. Soil bacteria decompose organic contaminates
into less toxic substances.
(b) Answers may vary. Students may indicate that the original source of water for their home is
surface water. In that case, evaporation and condensation is most important for purifying the
water source. Students who obtain their water from groundwater, may indicate that filtration is
the most important way to purify the water.
82. (a) Population growth is a challenge in providing potable water because it means that more
people are sharing a finite resource. As well, an increase in population means an increase in
demand for manufactured goods, which also require water. Aging or inadequate water
distribution and sanitation systems will also be affected by population growth.
(b) Increasing industrialization is a challenge in providing potable water because water is used in
almost all manufacturing and power generation processes. As industrialization increases, the
demand for water also increases.
(c) Pollution of surface water contaminates the water and may, ultimately, affect the health of
various organisms in an ecosystem. For example, if excess fertilizer is washed into a body of
water, it stimulates the growth of algae and water plants. When the organisms die, the oxygen in
the water is depleted, which produces an aquatic dead zone. As well, contaminated water reduces
the amount of potable water.
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Chapter 8: Water and Solutions
8-15
83. (a) In bronze, solid tin is the solute, solid copper is the solvent, and the bronze is solid.
(b) In mercury amalgams, liquid mercury is the solute, solid silver is the solvent, and the
amalgam is solid
(c) In a mixture of molecules in a scuba tank, oxygen gas is the solute, nitrogen gas is the
solvent, and the solution is a gas.
(d) In pop sweetened with sucrose, carbon dioxide gas is the solute, solid sucrose is the solute,
liquid water is the solvent, and the solution is liquid.
84. (a) Answers may vary. Sample answer: Five aqueous solutions that have been used or
observed are shampoo, vinegar, fruit juice, rainwater, and tap water.
(b) These solutions are aqueous solutions because the solvent in each case is water.
85. Recaptured heat could be used to heat buildings. The heated water could also be used to
power turbines to generate electricity.
86. Kitchen measuring devices can be used to prepare food because accuracy in measuring is not
as important. When preparing a standard solution, accuracy in measuring is important. That is
why volumetric flasks, which deliver a certain volume of solution +/– 0.01 mL at a specific
temperature, are used in the lab.
87. To dilute the amount of sodium that you ingest, you can drink a large amount of water over a
short time or drink water frequently throughout the day.
88. (a) Rubbing alcohol and vinegar are two solutions where the concentrations do not have to be
exact.
(b) If the percentage concentration of chlorine bleach were too high, it could cause health and
safety problems for the consumer. Concentrated chlorine bleach emits noxious fumes and is
corrosive. It can burn eyes, skin, and cause respiratory problems.
89. (a) Large volumes of water are used in oil and gas production to separate the bitumen from
sand, to force oil out of cracks so that it can be pumped to the surface from underground, and in
the drilling of oil and gas wells.
(b) Saline groundwater is sometimes used as an alternative to fresh water.
(c) Various measures are taken to prevent releasing polluted water into the environment. These
measures include placing a cement and steel casing around an oil and gas well. This casing
prevents pollutants from moving into the fresh water. In the oil sands, cut-off walls, ditches, and
interception wells are built around tailings ponds to capture seepage and runoff.
(d) One water-free process involves using carbon dioxide instead of water to recover oil from
wells. Another process involves using combustion instead of steam to liquefy the bitumen found
underground.
90. (a) An isotonic glucose solution has the same glucose concentration as the cells in the body.
(b) It is important for an IV solution to be isotonic so that fluids are replaced in a person without
diluting the glucose concentration in the body.
(c) The solute in a saline solution is sodium chloride.
(d) An isotonic saline solution is 0.9 % sodium chloride, which reflects the natural salinity of
the body.
91. (a) Dispersants were not very effective in treating the Exxon Valdez oil spill because
dispersants require a combination of wind and wave action to mix the dispersant with the oil. In
Prince William Sound, two days after the Exxon Valdez oil spill occurred, a large storm changed
the oil into mousse and distributed it over a large area. Dispersants do not work on oil mousse.
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Chapter 8: Water and Solutions
8-16
(b) During the cleanup of the Exxon Valdez oil spill, booms were deployed to protect fish
hatcheries and salmon streams, which were deemed to be the highest priority. Containment
booms allowed for the recovery of the oil. Oil in the open water was recovered using skimmers.
Sorbent booms were also used to soak up the oil, but produced extra waste. Hoses that sprayed
seawater or heated seawater were used to clean up the shoreline areas.
92. (a) In a Canadian nuclear reactor, heat energy is produced when uranium atoms split (fission
reaction). This heat energy is carried by heavy water to boilers that contain water, which is
heated to produce steam. The steam is used to drive turbines to produce electricity. Water from a
water source is used to cool the steam and condense it.
(b) One method of cooling is using water from a source, such as a lake or river, to cool the steam
and condense it. Another method is trapping the steam in a cooling chamber first, letting the
steam condense to water, then storing the water in another chamber, or expelling the water from
the plant.
93. (a) A tincture is an alcohol-based substance containing a solute in alcohol.
(b) Given: cW/V = 5 %
Vsolution = 1.5 L
Required: mass of iodine, miodine
Analysis: cW/V =
miodine
Vsolution
! 100 %
Solution:
Step 1. Convert litres to millilitres.
1000 mL
Vsolution = 1.5 L !
1 L
Vsolution = 1500 mL
Step 2. Rearrange the equation to solve for mass of iodine.
c
miodine = W/V ! Vsolution
100 %
Step 3. Substitute the correct values and solve.
5 %
miodine =
! 1500 mL
100 %
miodine = 75 g
Statement: The mass of iodine in 1.5 L of solution is 75 g.
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Chapter 8: Water and Solutions
8-17