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C H A P T E R 13
PROPERTIES OF LIQUIDS
SOLUTIONS TO REVIEW QUESTIONS
1. The potential energy is greater in the liquid water than in the ice. The heat necessary to melt the
ice increases the potential energy of the liquid, thus allowing the molecules greater freedom of motion.
The potential energy of steam (gas) is greater than that of liquid water.
2. At 0 C, all three substances, H2S, H2Se, and H2Te, are gases, because they all have boiling points
below 0 C.
3. Liquids are made of particles that are close together, they are not compressible and they have
definite volume. Solids also exhibit similar properties.
4. Liquids take the shape of the container they are in. Gases also exhibit this property.
5. The pressure of the atmosphere must be 1.00 atmosphere, otherwise the water would be boiling at
some other temperature.
–
O
6.
H
H
+
7. Melting point, boiling point, heat of fusion, heat of vaporization, density, and crystal structure in the solid
state are some of the physical properties of water that would be very different, if the molecules were
linear and nonpolar instead of bent and highly polar. For example, the boiling point, melting point, heat of
fusion and heat of vaporization would be lower because linear molecules have no dipole moment and the
attraction among molecules would be much less.
8. Prefixes preceding the word hydrate are used in naming hydrates, indicating the number of molecules of
water present in the formulas. The prefixes used are:
mono ¼ 1
hexa ¼ 6
di ¼ 2
hepta ¼ 7
tri ¼ 3
octa ¼ 8
tetra ¼ 4
nona ¼ 9
penta ¼ 5
deca ¼ 10
9. The distillation setup in Figure 13.13 would be satisfactory for separating salt and water, but not for
separating ethyl alcohol and water. In the first case, the water is easily vaporized, the salt is not, so
the water boils off and condenses to a pure liquid. In the second case, both substances are easily
vaporized, so both would vaporize (though not to an equal degree) and the condensed liquid would
contain both substances.
10. The thermometer would be at about 70 C. The liquid is boiling, which means its vapor pressure equals
the confining pressure. From Table 13.1, we find that ethyl alcohol has a vapor pressure of 543 torr at 70 C.
11. The water in both containers would have the same vapor pressure, for it is a function of the temperature of
the liquid.
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- Chapter 13 12. Vapor pressure is the pressure exerted by a vapor when it is in equilibrium with its liquid. The liquid
molecules on the surface and the gas molecules above the liquid are the important players in vapor
pressure. One liter of water will evaporate at the same rate as 100 mL of water if they are at the same
temperature and have the same surface area. The liquid and vapor in both cases will come to equilibrium
in the same amount of time and they will have the same vapor pressure. Vapor pressure is a characteristic
of the type of liquid, not the number of molecules of a liquid.
13. In Figure 13.1, it would be case (b) in which the atmosphere would reach saturation. The vapor pressure
of water is the same in both (a) and (b), but since (a) is an open container the vapor escapes into the
atmosphere and doesn’t reach saturation.
14. If ethyl ether and ethyl alcohol were both placed in a closed container, (a) both substances would be
present in the vapor, for both are volatile liquids; (b) ethyl ether would have more molecules in the vapor
because it has a higher vapor pressure at a given temperature.
15. At 30 torr, H2O would boil at approximately 29 C, ethyl alcohol at 14 C, and ethyl ether at some
temperature below 0 C.
16. (a)
(b)
(c)
At a pressure of 500 torr, water boils at 88 C.
The normal boiling point of ethyl alcohol is 78 C.
At a pressure of 0.50 atm (380 torr), ethyl ether boils at 16 C.
17. Based on Figure 13.7:
(a) Line BC is horizontal because the temperature remains constant during the entire process of
melting. The energy input is absorbed in changing from the solid to the liquid state.
(b) During BC, both solid and liquid phases are present.
(c) The line DE represents the change from liquid water to steam (vapor) at the boiling temperature of
water.
18. Physical properties of water:
(a) melting point, 0 C
(b) boiling point, 100 C (at 1 atm pressure)
(c) colorless
(d) odorless
(e) tasteless
(f) heat of fusion, 335 J=g (80 cal/g)
(g) heat of vaporization, 2.26 kJ=g (540 cal=g)
(h) density ¼ 1.0 g=mL (at 4 C)
(i) specific heat ¼ 4.184 J=g C
19. For water, to have its maximum density, the temperature must be 4 C, and the pressure sufficient to keep
it liquid, d ¼ 1.0 g=mL
20. Apply heat to an ice-water mixture, the heat energy is absorbed to melt the ice (heat of fusion), rather than
warm the water, so the temperature remains constant until all the ice has melted.
21. Ice at 0 C contains less heat energy than water at 0 C. Heat must be added to convert ice to water, so the
water will contain that much additional heat energy.
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- Chapter 13 22. Ice floats in water because it is less dense than water. The density of ice at 0 C is 0.915 g=mL.
Liquid water, however, has a density of 1.00 g=mL. Ice will sink in ethyl alcohol, which has a density
of 0.789 g=mL.
23. The heat of vaporization of water would be lower if water molecules were linear instead of bent. If
linear, the molecules of water would be nonpolar. The relatively high heat of vaporization of water is a
result of the molecule being highly polar and having strong dipole-dipole and hydrogen bonding
attraction for other water molecules.
24. Ethyl alcohol exhibits hydrogen bonding; ethyl ether does not. This is indicated by the high heat of
vaporization of ethyl alcohol, even though its molar mass is much less than the molar mass of ethyl ether.
25. A linear water molecule, being nonpolar, would exhibit less hydrogen bonding than the highly polar,
bent, water molecule. The polar molecule has a greater intermolecular attractive force than a nonpolar
molecule.
26. Water, at 80 C, will have fewer hydrogen bonds than water at 40 C. At the higher temperature, the
molecules of water are moving faster than at the lower temperature. This results in less hydrogen bonding
at the higher temperature.
27. H2NCH2CH2NH2 has two polar NH2 groups. It should, therefore, show more hydrogen bonding and a
higher boiling point (117 C) versus 49 C for CH3CH2CH2NH2.
28. Rubbing alcohol feels cold when applied to the skin, because the evaporation of the alcohol absorbs heat
from the skin. The alcohol has a fairly high vapor pressure (low boiling point) and evaporates quite
rapidly. This produces the cooling effect.
29. (a)
(b)
Order of increasing rate of evaporation: mercury, acetic acid, water, toluene, benzene, carbon
tetrachloride, methyl alcohol, bromine.
Highest boiling point is mercury. Lowest boiling point is bromine.
30. Water boils when its vapor pressure equals the prevailing atmospheric pressure over the water. In order
for water to boil at 50 C, the pressure over the water would need to be reduced to a point equal to the
vapor pressure of the water (92.5 torr).
31. In a pressure cooker, the temperature at which the water boils increases above its normal boiling point,
because the water vapor (steam) formed by boiling cannot escape. This results in an increased pressure
over water and, consequently, an increased boiling temperature.
32. Vapor pressure varies with temperature. The temperature at which the vapor pressure of a liquid equals
the prevailing pressure is the boiling point of the liquid.
33. As temperature increases, molecular velocities increase. At higher molecular velocities, it becomes
easier for molecules to break away from the attractive forces in the liquid.
34. Water has a relatively high boiling point because there is a high attraction between molecules due to
hydrogen bonding.
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- Chapter 13 35. Ammonia would have a higher vapor pressure than SO2 at 40 C because it has a lower boiling point
(NH3 is more volatile than SO2).
36. As the temperature of a liquid increases, the kinetic energy of the molecules as well as the vapor
pressure of the liquid increases. When the vapor pressure of the liquid equals the external pressure,
boiling begins with many of the molecules having enough energy to escape from the liquid. Bubbles of
vapor are formed throughout the liquid and these bubbles rise to the surface, escaping as boiling
continues.
37. HF has a higher boiling point than HCl because of the strong hydrogen bonding in HF (F is the
most electronegative element). Neither F2 nor Cl2 will have hydrogen bonding, so the compound,
F2, with the lower molar mass, has the lower boiling point.
38. The boiling liquid remains at constant temperature because the added heat energy is being used to convert
the liquid to a gas, i.e., to supply the heat of vaporization for the liquid at its boiling point.
39. 34.6 C, the boiling point of ethyl ether. (See Table 13.2)
40. If the lake is in an area where the temperature is below freezing for part of the year, the expected
temperature would be 4 C at the bottom of the lake. This is because the surface water would cool
to 4 C (maximum density) and sink.
41. The formation of hydrogen and oxygen from water is an endothermic reaction, due to the following
evidence:
(a) Energy must continually be provided to the system for the reaction to proceed. The reaction will
cease when the energy source is removed.
(b) The reverse reaction, burning hydrogen in oxygen, releases energy as heat.
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- Chapter 13 -
SOLUTIONS TO EXERCISES
1. Acid anhydride: [HClO4, Cl2O7]
2. Acid anhydride: [H2SO3, SO2]
[H2CO3, CO2] [H3PO4, P2O5]
[H2SO4, SO3] [HNO3, N2O5]
3. Basic anhydrides: [LiOH, Li2O] [NaOH, Na2O] [Mg(OH)2, MgO]
4. Basic anhydrides: [KOH, K2O] [Ba(OH)2, BaO] [Ca(OH)2, CaO]
5. (a)
(b)
(c)
D
C2 H5 OH þ 3 O2 ! 2 CO2 þ 3 H2 O
Ca þ 2 H2 O ! H2 þ CaðOHÞ2
D
2 LiOH ! Li2 O þ H2 O
(d) Na2 CO3 10 H2 O ! Na2 CO3 þ 10 H2 O
!HClO þ HCl
(e) Cl2 þ H2 O !2 NaOH
(f) Na2 O þ H2 O D
6. (a)
CO2 þ H2 O !H2 CO3
(b)
MgSO4 7 H2 O ! MgSO4 þ 7 H2 O
(c)
(d)
(e)
2 C3 H7 OH þ 9 O2 ! 6 CO2 þ 8 H2 O
!BaðNO3 Þ2 þ 2 H2 O
2 HNO3 þ BaðOHÞ2 BaO þ H2 O !BaðOHÞ2
(f)
CaðOHÞ2 ! CaO þ H2 O
D
D
D
7. (a)
(b)
(c)
barium bromide dihydrate
aluminum chloride hexahydrate
iron(III) phosphate tetrahydrate
8. (a)
(b)
(c)
magnesium ammonium phosphate hexahydrate
iron(II) sulfate heptahydrate
tin(IV) chloride pentahydrate
9. Deionized water is water from which the ions have been removed.
(a) Hard water contains dissolved calcium and magnesium salts.
(b) Soft water is free of ions that cause hardness (Ca2+ and Mg2+) but it may contain other ions such
as Naþ and Kþ.
10. Deionized water is water from which the ions have been removed.
(a) Distilled water has been vaporized by boiling and recondensed. It is free of nonvolatile impurities,
but may still contain any volatile impurities that were initially present in the water.
(b) Natural waters are generally not pure, but contain dissolved minerals and suspended matter, and
can even contain harmful bacteria.
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- Chapter 13 11. In which of the following substances would you expect to find hydrogen bonding?
(a) C3H7OH will hydrogen bond; one of the hydrogens is bonded to oxygen.
(b) H2O2 will hydrogen bond; hydrogen is bonded to oxygen.
(c) CHCl3 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen.
(d) PH3 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen.
(e) HF will hydrogen bond; hydrogen is bonded to fluorine.
12. (a)
(b)
(c)
(d)
(e)
HI will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen.
NH3 will hydrogen bond; hydrogen is bonded to nitrogen.
CH2F2 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen.
C2H5OH will hydrogen bond; one of the hydrogens is bonded to oxygen.
H2O will hydrogen bond; hydrogen is bonded to oxygen.
H
13. (a)
C3 H7
O H
O
C3H7OH
C3 H7
H
(b)
H O O H
(c)
H F
H
14. (a)
H
H
H
H
C
C
H
H
N
HF
H F
H
(b)
H2O2
O O H
O
H
O
H
H
C
C
H
H
H
C2H5OH
H
H
N
H
NH3
H
(c)
H
O
H
H
H2O
O
H
15. Water spreading out on a glass plate is an example of adhesive forces. The water molecules have stronger
attractive forces toward the glass surface than they do to other water molecules.
16. Water forming beaded droplets is an example of cohesive forces. The water molecules have stronger
attractive forces for other water molecules that they do for the surface.
1 mol
¼ 0:0874 mol Na2 CO3 10 H2 O
17. ð25:0 g Na2 CO3 10 H2 OÞ
286:2 g
1 mol
18. ð25:0 g Na2 B4 O7 10 H2 OÞ
¼ 0:0655 mol Na2 B4 O7 10 H2 O
381:4 g
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- Chapter 13 19. (a)
(b)
20. (a)
(b)
1 mol
7 mol H2 O
18:02 g
ð125 g MgSO4 7 H2 OÞ
¼ 64:0 g H2 O
246:5 g 1 mol MgSO4 7 H2 O mol 1 mol
1 mol MgSO4
120:4 g
ð125 g MgSO4 7 H2 OÞ
¼ 61:1 g MgSO4
246:5 g 1 mol MgSO4 7 H2 O
mol
1 mol
6 mol H2 O
18:02 g
ð125 g AlCl3 6 H2 OÞ
¼ 56:0 g H2 O
241:4 g 1 mol AlCl3 6 H2 O mol 1 mol
1 mol AlCl3
133:3 g
ð125 g AlCl3 6 H2 OÞ
¼ 69:0 g AlCl3
241:4 g 1 mol AlCl3 6 H2 O
mol
21. Assume 1 mol of the compound which contains 10 mol of water.
g H2 O
ð10Þð18:02 gÞ
% H2 O ¼
ð100Þ ¼ 47:25% H2 O
ð100Þ ¼
g Na2 B4 O7 10 H2 O
ð381:4 gÞ
22. Assume 1 mol of the compound which contains 2 mol of water.
g H2 O
ð2Þð18:02 gÞ
% H2 O ¼
ð100Þ ¼ 20:93% H2 O
ð100Þ ¼
g CaSO4 2 H2 O
ð172:2 gÞ
23. Assume 100. g of the compound.
ð0:2066Þð100: gÞ ¼ 20:66 g Fe ð0:3935Þð100: gÞ ¼ 39:35 g Cl
ð0:3999Þð100: gÞ ¼ 39:99 g H2 O
1 mol
ð20:66 g FeÞ
¼ 0:3699 mol Fe
55:85 g
1 mol
ð39:35 g ClÞ
¼ 1:110 mol Cl
35:45 g
1 mol
ð39:99 g H2 OÞ
¼ 2:219 mol H2 O
18:02 g
0:3699
¼ 1:000
0:3699
1:110
¼ 3:001
0:3699
2:219
¼ 5:999
0:3699
The formula is FeCl3 6 H2 O
24. Assume 100. g of the compound.
ð0:2469Þð100: gÞ ¼ 24:69 g Ni ð0:2983Þð100: gÞ ¼ 29:83 g Cl
ð0:4548Þð100: gÞ ¼ 45:48 g H2 O
1 mol
ð24:69 g NiÞ
¼ 0:4207 mol Ni
58:69 g
1 mol
ð29:83 g ClÞ
¼ 0:8415 mol Cl
35:45 g
1 mol
ð45:48 g H2 OÞ
¼ 2:524 mol H2 O
18:02 g
0:4207
¼ 1:000
0:4207
0:8415
¼ 2:000
0:4207
2:524
¼ 6:000
0:4207
The formula is NiCl2 6 H2 O
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- Chapter 13 25. Energy (Ea) to heat the water to steam from 15 C ! 100. C
4:184 J
Ea ¼ ðmÞðspecific heatÞðDtÞ ¼ ð275 gÞ
ð85 CÞ ¼ 9:8 104 J
g C
Energy (Eb) to convert water at 100 C to steam: heat of vaporization ¼ 2.26 103 J=g
Eb ¼ ðmÞðheat of vaporizationÞ ¼ ð275 gÞ 2:26 103 J=g ¼ 6:22 105 J
Etotal ¼ Ea þ Eb ¼ 9:8 104 J þ 6:22 105 J ¼ 7:20 105 J
26. Energy (Ea) to cool the water from 35 C ! 0 C
4:184 J
Ea ¼ ðmÞðspecific heatÞðDtÞ ¼ ð325 gÞ
ð35 CÞ ¼ 4:8 104 J
gC
Energy (Eb) to convert water to ice: heat of fusion ¼ 335 J=g
Eb ¼ ðmÞðheat of fusionÞ ¼ ð325 gÞð335 J=gÞ ¼ 1:09 105 J
Etotal ¼ Ea þ Eb ¼ 4:8 104 J þ 1:09 105 J ¼ 1:57 105 J
27. Energy released in cooling the water: 25 C to 0 C
4:184 J
ð25 CÞ ¼ 3:1 104 J
E ¼ ðmÞðspecific heatÞðDtÞ ¼ ð300: gÞ
g C
Energy required to melt the ice
E ¼ ðmÞðheat of fusionÞ ¼ ð100: gÞð335 J=gÞ ¼ 3:35 104 J
Less energy is released in cooling the water than is required to melt the ice. Ice will remain and the water
will be at 0 C.
28. Energy to heat the water ¼ energy to condense the steam
4:184 J
ð100: C 25 CÞ ¼ ðmÞð2259 J=gÞ
ð300: gÞ
g C
m ¼ 42 g (grams of steam required to heat the water to 100. C) 42 g of steam are required to heat 300. g of
water to 100. C. Since only 35 g of steam are added to the system, the final temperature will be less than
100. C. Not sufficient steam.
29. Energy lost by warm water ¼ energy gained by the ice
x ¼ final temperature
1000 mL 1:0 g
massðH2 OÞ ¼ ð1:5 L H2 OÞ
¼ 1500 g
L
mL
4:184 J
J
4:184 J
ð1500 gÞ
ð75 C xÞ ¼ ð75 gÞ 335
ð x 0 CÞ
þ ð75 gÞ
g C
g
g C
4:707 105 J ð6276x J= CÞ ¼ 2:51 104 J þ 313:8x J= C
4:46 105 J ¼ 6:5898 103 x J= C
x ¼ 68 C
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- Chapter 13 30. E ¼ (m)(heat of fusion)
(500. g)(335 J/g) ¼ 167,000 J needed to melt the ice
9560 J < 167,500 J
Since 167,500 J are required to melt all the ice, and only 9560 J are available, the system will be at 0 C. It
will be a mixture of ice and water.
31. (a)
2 K þ 2
H2 O ! 2KOH
þ H 2 1 mol
2 mol H2 O ð18:02 gÞ
ð25 g KÞ
¼ 12 g H2 O
39:10 g
2 mol K
mol
(b)
Na2 O þ H2 O ! 2 NaOH
1 mol
1 mol H2 O
ð18:02 gÞ
ð25 g Na2 OÞ
¼ 7:3 g H2 O
61:98 g 1 mol Na2 O
mol
(c)
SO3 þ H2 O ! H2 SO
4
1 mol
1 mol H2 O ð18:02 gÞ
ð25 g SO3 Þ
¼ 5:6 g H2 O
80:07 g
1 mol SO3
mol
32. (a)
CaO þ H2 O ! CaðOH
Þ2
1 mol
1 mol H2 O ð18:02 gÞ
ð15 g CaOÞ
¼ 4:8 g H2 O
56:08 g 1 mol CaO
mol
(b)
PO4
P2 O5 þ 3 H
2O ! 2 H
3
ð18:02 gÞ
1 mol
3 mol H2 O
ð15 g P2 O5 Þ
¼ 5:7 g H2 O
141:9 g 1 mol P2 O5
mol
(c)
þ H 2 2 Li þ 2 H2 O ! 2LiOH
1 mol
2 mol H2 O ð18:02 gÞ
ð15 g LiÞ
¼ 39 g H2 O
6:941 g
2 mol Li
mol
33. Water forms droplets because of surface tension, or the desire for a droplet of water to minimize its
ratio of surface area to volume. The molecules of water inside the drop are attracted to other water
molecules all around them, but on the surface of the droplet the water molecules feel an inward attraction
only. This inward attraction of surface water molecules for internal water molecules is what holds the
droplets together and minimizes their surface area.
34. Steam molecules will cause a more severe burn. Steam molecules contain more energy at 100 C
than water molecules at 100 C due to the energy absorbed during the vaporization stage (heat of
vaporization).
35. The alcohol has a higher vapor pressure than water and thus evaporates faster than water. When the
alcohol evaporates it absorbs energy from the water, cooling the water. Eventually the water will lose
enough energy to change from a liquid to a solid (freeze).
36. When one leaves the swimming pool, water starts to evaporate from the skin of the body. Part of
the energy needed for evaporation is absorbed from the skin, resulting in the cool feeling.
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- Chapter 13 37.
(a)
100
80
60
T(ºC)
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From 0 C to 40. C solid X warms until at 40. C it
begins to melt. The temperature remains at 40.0 C
until all of X is melted. After that, liquid X will
warm steadily to 65 C where it will boil and remain
at 65 C until all of the liquid becomes vapor. Beyond
65 C, the vapor will warm steadily until 100 C.
40
20
0
Joules
(b)
¼ (60. g)(3.5 J=g C)(40. C)
Joules needed (0 C to 40 C)
¼ (60. g)(80. J=g)
Joules needed at 40 C
¼ (60. g)(3.5 J=g C)(25 C)
Joules needed (40 C to 65 C)
¼ (60. g)(190 J=g)
Joules needed at 65 C
Joules needed (65 C to 100 C) ¼ (60. g)(3.5 J=g C)(35 C)
Total Joules needed
(each step rounded to two significant figures)
¼
¼
¼
¼
¼
8400 J
4800 J
5300 J
11,000 J
7400 L
37,000 J
38. During phase changes (ice melting to liquid water or liquid water evaporating to steam), all the
heat energy is used to cause the phase change. Once the phase change is complete the heat energy
is once again used to increase the temperature of the substance.
39. As the temperature of a liquid increases, the molecules gain kinetic energy thereby increasing their
escaping tendency (vapor pressure).
40. Since boiling occurs when vapor pressure equals atmospheric pressure, the graph in Figure 13.4
indicates that water will boil at about 75 C at 270 torr pressure.
41. CuSO4 (anhydrous) is gray white. When exposed to moisture, it turns bright blue forming
CuSO4 5 H2 O. The color change is an indicator of moisture in the environment.
42.
MgSO4 7 H2 O Na2 HPO4 12 H2 O
43. Soap can soften hard water by forming a precipitate with, and thus removing, the calcium and
magnesium ions. This precipitate is a greasy scum and is very objectionable, so it is a poor way to
soften water.
44. Chlorine is commonly used to destroy bacteria in water. Ozone and ultraviolet radiation are also used in
some places.
45. Ozone, O3
46. When organic pollutants in water are oxidized by dissolved oxygen, there may not be sufficient dissolved
oxygen to sustain marine life, such as fish. Most marine life forms depend on dissolved oxygen for
cellular respiration.
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- Chapter 13 47. Liquids that are stored in ceramic containers should never be drunk, for they are likely to have dissolved some
of the lead from the ceramic. If the ceramic is glazed, the liquid is less apt to dissolve lead from the ceramic.
48. Na2 zeoliteðsÞ þ Mg2þ ðaqÞ ! Mg zeoliteðsÞ þ 2 Naþ ðaqÞ
49. Softening of hard water using sodium carbonate:
CaCl2 ðaqÞ þ Na2 CO3 ðaqÞ ! CaCO3 ðsÞ þ 2 NaClðaqÞ
50.
–50
Liquid
–60
Temperature (ºC)
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–70
–80
–90
gas (phase change)
gas
Liquid
Solid
liquid (phase change)
solid
–100
Heat added
51. (a)
(b)
(c)
Melt ice: Ea ¼ (m)(heat of fusion) ¼ (225 g)(335 J=g) ¼ 7.54 104 J
Warm the water: Eb¼ ðmÞðspecific heatÞðDtÞ
4:184 J
ð100: CÞ ¼ 9:41 104 J
¼ ð225 gÞ
g C
Vaporize the water:
Ec ¼ ðmÞðheat of vaporizationÞ ¼ ð225 gÞð2259 J=gÞ ¼ 5:08 105 J
Etotal ¼ Ea þ Eb þ Ec ¼ 6:78 105 J
52. The heat of vaporization of water is 2.26 kJ=g.
18:02 g
ð2:26 kJ=gÞ
¼ 40:7 kJ=mol
mol
53.
0:096 cal
ð150: 20:0 CÞ
E ¼ ðmÞðspecific heatÞðDtÞ ¼ ð250: gÞ
g C
¼ 3:1 103 calð3:1 kcalÞ
54. Energy liberated when steam at 100.0 C condenses to water at 100.0 C
18:02 g 2:26 kJ 1000 J
ð50:0 mol steamÞ
¼ 2:04 106 J
mol
g
kJ
Energy liberated in cooling water from 100.0 C to 30.0 C
18:02 g 4:184 J
ð50:0 mol H2 OÞ
ð100:0 C 30:0 CÞ ¼ 2:64 105 J
mol
g C
Total energy liberated ¼ 2.04 106 J þ 2.64 105 J ¼ 2.30 106 J
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- Chapter 13 55. Energy to warm the ice from 10.0 C to 0 C
2:01 J
ð100: gÞ
ð10:0 CÞ ¼ 2010 J
g C
Energy to melt the ice at 0 C
ð100: gÞð335 J=gÞ ¼ 33,500 J
Energy to heat the water from 0 C to 20.0 C
4:184 J
ð100: gÞ
ð20:0 CÞ ¼ 8370 J
g C
Etotal ¼ 2010 J þ 33,500 J þ 8370 J ¼ 4:39 104 J ¼ 43:9 kJ
56. 2 H2 Oðl Þ ! 2 H2 ðgÞ þ O2 ðgÞ
The conversion is L O2 ! mol O2 ! mol H2O ! g H2O
1 mol
2 mol H2 O 18:02 g
ð25:0 L O2 Þ
¼ 40:2 g H2 O
22:4 L
1 mol O2
mol
57. The conversion is
mol
molecules
molecules
molecules
molecules
!
!
!
!
day
day
hr
min
s
1:00 mol H2 O
day
6:022 1023 molecules 1:00 day
1 hr
1 min
¼ 6:97 1018 molecules=s
mol
24 hr
60 min
60 s
58. Liquid water has a density of 1.00 g/mL.
d¼
m
m
18:02 g
V¼ ¼
¼ 18:0 mL
V
d 1:00 g=mL
ðvolume of 1 moleÞ
1.00 mole of water vapor at STP has a volume of 22.4 L (gas)
59. 2 H2 ðgÞ þ O2 ðgÞ ! 2 H2 OðgÞ
1 mL O2
(a) ð80:0 mL H2 Þ
¼ 40:0 mL O2 react with 80.0 mL of H2
2 mL H2
Since 60.0 mL of O2 are available, some oxygen remains unreacted.
(b)
60.0 mL 40.0 mL ¼ 20.0 mL O2 unreacted.
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E1C13
05/17/2010
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Page 157
- Chapter 13 60. Energy absorbed by the student when steam at 100. C changes to water at 100. C
2:26 kJ
ð1:5 g steamÞ
¼ 3:4 kJ 3:4 103 J
g
Energy absorbed when water cools from 100. C to 20.0 C
E ¼ ðmÞðspecific heatÞðDtÞ
4:184 J
E ¼ ð1:5 gÞ
ð100: C 20:0 CÞ ¼ 5:0 102 J
g C
Etotal ¼ 3:4 103 J þ 5:0 102 J ¼ 3:9 103 J
61. (a)
(b)
won’t hydrogen bond – no hydrogen in the molecule
will hydrogen bond
CH3
O
H
O
CH3
H
(c)
(d)
won’t hydrogen bond; no hydrogen covalently bonded to a strongly electronegative element
will hydrogen bond
H
(e)
O
H
H
H
won’t hydrogen bond; hydrogen must be attached to N, O, or F
62. During the fusion (melting) of a substance the temperature remains constant so a temperature factor is not
needed.
63. Energy needed to heat Cu to its melting point:
E ¼ (50.0 g)(0.385 J=g C)(1083 C 25.0 C) ¼ 2.04 104 J
Energy needed to melt the Cu.
E ¼ ð50:0 gÞð134 J=gÞ ¼ 6:70 103 J
Etotal ¼ 2:04 104 J þ 6:70 103 J ¼ 2:71 104 J
64. A lake freezes from the top down because the density of water as a solid is lower than that of the liquid,
causing the ice to float on the top of the liquid. Since the liquid remains below the solid, marine, and
aquatic life continues.
65. Heat lost by warm water ¼ heat gained by ice
m ¼ grams of ice to lower temperature of water to 0.0 C.
4:184 J
ð120: gÞ
ð45 C 0:0 CÞ ¼ ðmÞð335 J=gÞ
g C
68 g ¼ m (grams of ice melted)
68 g of ice melted. Therefore, 150. g 68 g ¼ 82 g ice remain.
66. Mass solution mass H2O ¼ mass H2SO4
(122 mL)(1.26 g=mL) (100. mL)(1.00 g=mL) ¼ 54 g H2SO4 added
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