CHAPTER 14
SOLUTIONS
SOLUTIONS TO REVIEW QUESTIONS
1.
O
H
H
H
H
O
H
O
H
H
H
H
Cl–
O
O
Na+
H
H
H
O
O
H
H
H
H
O
These diagrams are intended to illustrate the orientation of the water molecules about the ions, not the
number of water molecules.
2. From Table 14.2, approximately 26.8 g of KBr would dissolve in 50 g water at 0 C.
3. From Figure 14.4, solubilities in water at 25 C:
(a)
NH4Cl
39 g=100 g H2O
(b)
CuSO4
22 g=100 g H2O
(c)
NaNO3
91 g=100 g H2O
4. Going down Group 1A, the solubilities of both the chlorides and bromides decrease.
5. From Fig. 14.4, the solubility, in grams of solute per 100 g of H2O:
(a)
(b)
NH4Cl at 35 C, 43 g
CuSO4 at 60 C, 39 g
(c)
(d)
SO2 gas at 30 C, 8 g at 1 atm
NaNO3 at 15 C, 82 g
6. Li2SO4
7.
8.
6:5 g
100 ¼ 30 mass percent
21:5 g
Cube
Volume
Number=1 cm cube
1 cm
1 cm3
1
0:01 cm
1 106 cm3
106 ð1 cm3 Þ= 1 106 cm3 ¼ 106 cubes
Area of face
Total surface area
1 cm2
6 cm2
1 104 cm2
6 102 cm2
- 158 -
- Chapter 14 1 106 cubes ð6 faces=cubeÞ 1 104 cm2 =face ¼ 6 102 cm2
9.
40: g
100 ¼ 35 mass percent
115 g
10. The rate of dissolving decreases. The rate of dissolving is at its maximum when the solute and solvent are
first mixed.
11. A supersaturated solution of NaC2H3O2 may be prepared in the following sequence:
(a) Determine the mass of NaC2H3O2 necessary to saturate a specific amount of water at room
temperature.
(b) Place a bit more NaC2H3O2 in the water than the amount needed to saturate the solution.
(c) Heat the solution until all the solid dissolves.
(d) Cover the container and allow it to cool undisturbed. The cooled solution, which should contain
no solid NaC2H3O2, is supersaturated.
To test for supersaturation, add one small crystal of NaC2H3O2 to the solution. Immediate crystallization
is an indication that the solution was supersaturated.
12. Because the concentration of water is greater in the thistle tube, the water will flow through the membrane
from the thistle tube to the urea solution in the beaker. The solution level in the thistle tube will fall.
13. A true solution is one in which the size of the particles of solute are between 0.1 – 1 nm. True solutions are
homogeneous and the ratio of solute to solvent can be varied. They can be colored or colorless but
are transparent. The solute remains distributed evenly in the solution, it will not settle out.
14. The two components of a solution are the solute and the solvent. The solute is dissolved into the solvent
or is the least abundant component. The solvent is the dissolving agent or the most abundant component.
15. It is not always apparent which component in a solution is the solute. For example, in a solution
composed of equal volumes of two liquids, the designation of solute and solvent would be simply a
matter of preference on the part of the person making the designation.
16. The ions or molecules of a dissolved solute do not settle out because the individual particles are so
small that the force of molecular collisions is large compared to the force of gravity.
17. Yes. It is possible to have one solid dissolved in another solid. Metal alloys are of this type. Atoms
of one metal are dissolved among atoms of another metal.
18. Orange. The three reference solutions are KC1, KMnO4 and K2Cr207. They all contain Kþ ions in
solution. The different colors must result from the different anions dissolved in the solutions: MnO4
(purple) and Cr2 O72 (orange). Therefore, it is predictable that the Cr2 O72 ion present in an aqueous
solution of Na2Cr2O7 will impart an orange color to the solution.
19. Hexane and benzene are both nonpolar molecules. There are no strong intermolecular forces between
molecules of either substance or with each other, so they are miscible. Sodium chloride consists of
ions strongly attracted to each other by electrical attractions. The hexane molecules, being nonpolar, have
no strong forces to pull the ions apart, so sodium chloride is insoluble in hexane.
- 159 -
- Chapter 14 20. Coca Cola has two main characteristics, taste and fizz (carbonation). The carbonation is due to a
dissolved gas, carbon dioxide. Since dissolved gases become less soluble as temperature increases, warm
Coca Cola would be flat, with little to no carbonation. It is, therefore, unappealing to most people.
21. Air is considered to be a solution because it is a homogeneous mixture of several gaseous substances and
does not have a fixed composition.
22. A teaspoon of sugar would definitely dissolve more rapidly in 200 mL of hot coffee than in 200 mL of
iced tea. The much greater thermal agitation of the hot coffee will help break the sugar molecules
away from the undissolved solid and disperse them throughout the solution. Other solutes in coffee and
tea would have no significant effect. The temperature difference is the critical factor.
23. The solubility of gases in liquids is greatly affected by the pressure of a gas above the liquid. The
greater the pressure, the more soluble the gas. There is very little effect of pressure regarding the
dissolution of solids in liquids.
24. For a given mass of solute, the smaller the particles, the faster the dissolution of the solute. This is due to
the smaller particles having a greater surface area exposed to the dissolving action of the solvent.
25. In a saturated solution, the net rate of dissolution is zero. There is no further increase in the amount
of dissolved solute, even though undissolved solute is continuously dissolving, because dissolved solute
is continuously coming out of solution, crystallizing at a rate equal to the rate of dissolving.
26. When crystals of AgNO3 and NaCl are mixed, the contact between the individual ions is not intimate
enough for the double displacement reaction to occur. When solutions of the two chemicals are
mixed, the ions are free to move and come into intimate contact with each other, allowing the reaction to
occur easily. The AgCl formed is insoluble.
27. A nonvolatile solute (such as salt) lowers the freezing point of water. Adding salt to icy roads in winter
melts the ice because the salt lowers the freezing point of water.
28. A 16 molar solution of nitric acid is a solution that contains 16 moles HNO3 per liter of solution.
29. The two solutions contain the same number of chloride ions. One liter of 1 M NaCl contains 1 mole of
NaCl, therefore 1 mole of chloride ions. 0.5 liter of 1 M MgCl2 contains 0.5 mol of MgCl2 and 1 mole
of chloride ions.
1 mol MgCl2
2 mol Cl
ð0:5 LÞ
¼ 1 mol Cl
L
1 mol MgCl2
30. The champagne would spray out of the bottle all over the place. The rise in temperature and the increase
in kinetic energy of the molecules by shaking both act to decrease the solubility of gas within the liquid.
The pressure inside the bottle would be great. As the cork is popped, much of the gas would escape from
the liquid very rapidly, causing the champagne to spray.
31. The number of grams of NaCl in 750 mL of 5.0 molar solution is
5:0 mol NaCl 58:44 g
¼ 2:2 102 g NaCl
ð0:75 LÞ
L
1 mol
- 160 -
- Chapter 14 Dissolve the 220 g of NaCl in a minimum amount of water, then dilute the resulting solution to a final
volume of 750 mL (0.75 L).
32. A semipermeable membrane will allow water molecules to pass through in both directions. If it has pure
water on one side and 10% sugar solutions on the other side of the membrane, there is a higher
concentration of water molecules on the pure water side. Therefore, there are more water molecule
impacts per second on the pure water side of the membrane. The net result is more water molecules pass
from the pure water to the sugar solution. Osmotic pressure effect.
33. The urea solution will have the greater osmotic pressure because it has 1.67 mol solute/kg H2O, while the
glucose solution has only 0.83 mol solute/kg H2O.
34. A lettuce leaf immersed in salad dressing containing salt and vinegar will become limp and wilted as a
result of osmosis. As the water inside the leaf flows into the dressing where the solute concentration is
higher the leaf becomes limp from fluid loss. In water, osmosis proceeds in the opposite direction flowing
into the lettuce leaf maintaining a high fluid content and crisp leaf.
35. The concentration of solutes (such as salts) is higher in seawater than in body fluids. The survivors who
drank seawater suffered further dehydration from the transfer of water by osmosis from body tissues to
the intestinal tract.
36. Ranking of the specified bases in descending order of the volume of each required to react with 1 liter of
1 M HC1. The volume of each required to yield 1 mole of OH ion is shown.
(a)
(b)
(c)
(d)
1 M NaOH
0.6 M Ba(OH)2
2 M KOH
1.5 M Ca(OH)2
1 liter
0.83 liter
0.50 liter
0.33 liter
37. The boiling point of a liquid or solution is the temperature at which the vapor pressure of the liquid equals
the pressure of the atmosphere. Since a solution containing a nonvolatile solute has a lower vapor
pressure than the pure solvent, the boiling point of the solution must be at a higher temperature than for
the pure solvent. At the higher boiling temperature the vapor pressure of the solution equals the
atmospheric pressure.
38. The freezing point is the temperature at which a liquid changes to a solid. The vapor pressure of a solution
is lower than that of a pure solvent. Therefore, the vapor pressure curve of the solution intersects the
vapor pressure curve of the pure solvent, at a temperature lower than the freezing point of the pure
solvent. (See Figure 14.8b) At this point of intersection, the vapor pressure of the solution equals the
vapor pressure of the pure solvent.
39. Water and ice are different phases of the same substance in equilibrium at the freezing point of water,
0 C. The presence of the methanol lowers the vapor pressure and hence the freezing point of water. If the
ratio of alcohol to water is high, the freezing point can be lowered as much as 10 C or more.
40. Effectiveness in lowering the freezing point of 500. g water:
(a)
(b)
100. g (2.17 mol) of ethyl alcohol is more effective than 100. g (0.292 mol) of sucrose.
20.0 g (0.435 mol) of ethyl alcohol is more effective than 100. g (0.292 mol) of sucrose.
- 161 -
- Chapter 14 (c)
20.0 g (0.625 mol) of methyl alcohol is more effective than 20.0 g (0.435 mol) of ethyl alcohol.
41. Both molarity and molality describe the concentration of a solution. However, molarity is the ratio of
moles of solute per liter of solution, and molality is the ratio of moles of solute per kilogram of solvent.
42. 5 molal NaCl ¼ 5 mol NaCl=kg H2O; 5 molar NaCl ¼ 5 mol NaCl=L of solution. The volume of the 5
molal solution will be larger than 1 liter (1 L H2O þ 5 mol NaCl). The volume of the 5 molar solution is
exactly 1 L (5 mol NaCl þ sufficient H2O to produce 1 L of solution). The molarity of a 5 molal solution
is therefore, less than 5 molar.
- 162 -
- Chapter 14 -
SOLUTIONS TO EXERCISES
1. Reasonably soluble:
Insoluble:
2. Reasonably soluble:
Insoluble:
(b) K2SO4
(c) Na3PO4
(d) NaOH
(a) AgCl
(e) PbI2
(f) SnCO3
(b) Cu(NO3)2
(d) NH4C2H3O2
(f) AgNO3
(a) Ba3(PO4)2
(c) Fe(OH)3
(e) MgO
3. Mass percent calculations
(a) 15.0 g KCl þ 100.0 g H2O ¼ 115.0 g solution
15:0 g
ð100Þ ¼ 13:0% KCl
115:0 g
(b)
(c)
(d)
2.50 g Na3PO4 þ 10.0 g H2O ¼ 12.5 g solution
2:50 g
ð100Þ ¼ 20:0% Na3 PO4
12:5 g
77:09 g
(0.20 mol NH4C2H3O2)
¼ 15 g NH4 C2 H3 O2
1 mol
15 g NH4C2H3O2 þ 125 g H2O ¼ 140. g solution
15 g
ð100Þ ¼ 11% NH4 C2 H3 O2
140 g
40:00 g
(1.50 mol NaOH)
¼ 60:0 g NaOH
1 mol
18:02 g
ð33:0 mol H2 OÞ
¼ 595 g H2 O
1 mol
60.0 g NaOH þ 595 g H2O ¼ 655 g solution
60:0 g
ð100Þ ¼ 9:16% NaOH
655 g
4. Mass percent calculations
(a) 25.0 g NaNO3 in 125.0 g H2O ¼ 150.0
25:0 g
ð100Þ ¼ 16:7% NaNO3
150:0 g
(b)
1.25 g CaCl2 in 35.0 g H2O ¼ 36.3 g solution
1:25 g
ð100Þ ¼ 3:44% CaCl2
36:3 g
- 163 -
- Chapter 14 (c)
(d)
194:2 g
(0.75 mol K2CrO4) þ
¼ 150 g K2 CrO4
1 mol
150 g K2CrO4 þ 225 g H2O ¼ 380 g solution
150 g
ð100Þ ¼ 39% K2 CrO4
380 g
98:09 g
(1.20 mol H2SO4)
¼ 118 g H2 SO4
1 mol
18:02 g
(72.5 mol H2O)
¼ 1:31 103 g H2 O
1 mol
118 g H2SO4 þ 1.31 103 g H2O ¼ 1.43 103 g solution
118 g
ð100Þ ¼ 8:25% H2 SO4
1:43 103 g
5. A 15.5% solution contains 15.5 g AgNO3 per 100. g solution
100: g solution
ð25:2 g AgNO3 Þ
15:5 g AgNO3
¼ 163 g solution
6. A 10.0% NaCl solution contains 10.0 g NaCl per 100. g solution
100: g solution
ð25:0 g NaClÞ
¼ 250: g solution
10:0 g NaCl
7. (a)
(b)
8. (a)
(b)
7:5 g CaSO4
¼ 1:9 g CaSO4
ð25 g solutionÞ
100: g solution
25 g solution 6:9 g solute ¼ 23 g solvent
12:0 g BaCl2
¼ 9:0 g BaCl2
ð75 g solutionÞ
100: g solution
75 g solution 9:0 g solute ¼ 66 g solvent
9. Mass/volume percent.
15:0 g C2 H5 OH
(a)
ð100Þ ¼ 10:0% C2 H5 OH
150:0 mL solution
25:2 g NaCl
(b)
ð100Þ ¼ 20:1% NaCl
125:5 mL solution
- 164 -
- Chapter 14 10. Mass/volume percent.
175:2 g C12 H22 O11
ð100Þ ¼ 63:59% C12 H22 O11
(a)
275:5 mL solution
35:5 g of CH3 OH
(b)
ð100Þ ¼ 47:3% CH3 OH
75:0 mL solution
11. Volume percent.
50:0 mL hexanol
ð100Þ ¼ 40:0% hexanol
(a)
125 mL solution
2:0 mL ethanol
(b)
ð100Þ ¼ 13% ethanol
15:0 mL solution
12. Volume percent.
37:5 mL butanol
ð100Þ ¼ 13:6% butanol
(a)
275 mL solution
4:0 mL methanol
(b)
ð100Þ ¼ 16% methanol
25:0 mL solution
mol
13. Molarity problems M ¼
L
0:25 mol 1000 mL
(a)
¼ 3:3 M
75:0 mL
1L
1:75 mol
(b)
¼ 2:3 M KBr
0:75 L
35:0 g
1 mol
(c)
¼ 0:341 M NaC2 H3 O2
1:25 L 82:03 g
75 g CuSo4 5 H2 O
1 mol
(d)
¼ 0:30 M CuSO4
1:0 L
249:7 g
mol
14. Molarity problems M ¼
L
0:50 mol 1000 mL
(a)
¼ 4:0 M
125 mL
1L
2:25 mol
¼ 1:50 M CaCl2
(b)
1:50 L
275 g
1 mol
1000 mL
(c)
¼ 1:97 M C6 H12 O6
775 mL 180:2 g
1L
125 g MgSO4 7 H2 O
1 mol
(d)
¼ 0:203 M MgSO4
2:50 L
246:5 g
- 165 -
- Chapter 14 mol solute
or mol solute ¼ ðL solutionÞðMolarityÞ
L solution
1:20 mol H2 SO4
¼ 1:8 mol H2 SO4
ð1:5 LÞ
L
1L
0:0015 mol BaCl2
¼ 3:8 105 mol BaCl2
ð25:0 mLÞ
L
1000 mL
1L
0:35 mol K3 PO4
ð125 mLÞ
¼ 0:044 mol K3 PO4
L
1000 mL
15. Molarity ¼
(a)
(b)
(c)
mol solute
or mol solute ¼ ðL solutionÞ ðMolarityÞ
L solution
1:50 mol HNO3
¼ 1:1 mol HNO3
ð0:75 LÞ
L
1L
0:75 mol NaClO3
ð10:0 mLÞ
¼ 7:5 103 mol NaClO3
L
1000 mL
1L
0:50 mol LiBr
ð175 mLÞ
¼ 0:088 mol LiBr
1000 mL
L
16. Molarity ¼
(a)
(b)
(c)
17. (a)
(b)
(c)
18. (a)
(b)
(c)
0:75 mol K2 CrO4
194:2 g
¼ 360 g K2 CrO4
ð2:5 LÞ
L
1 mol
1L
0:050 mol HC2 H3 O2 60:05 g
ð75:2 mLÞ
¼ 0:226 g HC2 H3 O2
L
1000 mL
1 mol
1L
16 mol HNO3 63:02 g
ð250 mLÞ
¼ 250 g HNO3
L
1000 mL
1 mol
18 mol H2 SO4
98:09 g
¼ 2:1 103 g H2 SO4
ð1:20 LÞ
L
1 mol
1L
1:5 mol KMnO4
158:0 g
ð27:5 mLÞ
¼ 6:52 g KMnO4
L
1000 mL
1 mol
1L
0:025 mol Fe2 ðSO4 Þ3
399:9 g
ð120 mLÞ
¼ 1:2 g Fe2 ðSO4 Þ3
L
1000 mL
1 mol
19. (a)
(b)
1L
1000 mL
ð0:15 mol H3 PO4 Þ
¼ 2:0 102 mL
0:750 mol H3 PO4
1L
1 mol
1L
1000 mL
ð35:5 g H3 PO4 Þ
¼ 483 mL
97:99 g 0:750 mol H3 PO4
1L
- 166 -
- Chapter 14 -
20. (a)
(b)
ð0:85 mol NH4 ClÞ
1L
1000 mL
¼ 3:4 103 mL
0:250 mol NH4 Cl
1L
1 mol
1L
1000 mL
ð25:2 gÞNH4 Cl
¼ 1:88 103 mL
53:49 g 0:250 mol NH4 Cl
L
21. Dilution problem V1 M 1 ¼ V2 M 2
(a)
V1 ¼ 125 mL
V2 ¼ ð125 mL þ 775 mLÞ ¼ 900: mL
M 1 ¼ 5:0 M
M2 ¼ M2
ð125 mLÞð5:0 M Þ ¼ ð900: mLÞðM 2 Þ
M2 ¼
(b)
ð125 mLÞð5:0 M Þ
¼ 0:694 M
900: mL
V1 ¼ 250 mL
V2 ¼ ð250 mL þ 750 mLÞ ¼ 1:00 103 mL
M 1 ¼ 0:25 M
M2 ¼ M2
ð250 mLÞð0:25 M Þ ¼ 1:00 103 mL ðM 2 Þ
M2 ¼
(c)
ð250 mLÞð0:25 M Þ
¼ 0:063 M
1:00 103 mL
First calculate the moles of HNO3 in each solution. Then calculate the molarity.
1L
0:50 mol
ð75 mLÞ
¼ 0:038 mol HNO3
1000 mL
1L
1L
1:5 mol
ð75 mLÞ
¼ 0:11 mol HNO3
1000 mL
1L
Total mol ¼ 0.15 mol
Total volume ¼ 75 mL þ 75 mL ¼ 150. mL ¼ 0.150 L
0:150 mol
¼ 1:00 M
0:150 L
22. Dilution problem V1 M 1 ¼ V2 M 2
(a)
V1 ¼ 175 mL
V2 ¼ ð175 mL þ 275 mLÞ ¼ 450: mL
M 1 ¼ 3:0 M
M2 ¼ M2
ð175 mLÞð3:0 M Þ ¼ ð450: mLÞðM 2 Þ
M2 ¼
(b)
ð175 mLÞð3:0 M Þ
¼ 1:2 M
450: mL
V1 ¼ 350 mL
V2 ¼ ð350 mL þ 150 mLÞ ¼ 5:0 102 mL
M 1 ¼ 0:10 M
M2 ¼ M2
ð350 mLÞð0:10 M Þ ¼ 5:0 102 mL ðM 2 Þ
M2 ¼
ð350 mLÞð0:10 M Þ
¼ 0:070 M
5:0 102 mL
- 167 -
- Chapter 14 (c)
First calculate the moles of HCl in each solution. Then calculate the molarity.
1L
0:250 mol
ð50:0 mLÞ
¼ 0:0125 mol HCl
1000 mL
1L
1L
0:500 mol
ð25:0 mLÞ
¼ 0:0125 mol HCl
1000 mL
1L
Total mol ¼ 0.0250 mol
Total volume ¼ 50.0 mL þ 25.0 mL ¼ 75.0 mL ¼ 0.0750 L
0:250 mol
¼ 0:333 M
0:0750 L
23. V1 M 1 ¼ V2 M 2
(a)
ðV1 Þð15 M Þ ¼ ð750 mLÞð3:0 M Þ
V1 ¼
(b)
ð750 mLÞð3:0 M Þ
¼ 150 mL 15 M H3 PO4
15 M
ðV1 Þð16 M Þ ¼ ð250 mLÞð0:50 M Þ
V1 ¼
ð250 mLÞð0:50 M Þ
¼ 7:8 mL 16 M HNO3
16 M
24. V1 M 1 ¼ V2 M 2
(a)
ðV1 Þð18 M Þ ¼ ð225 mLÞð2:0 M Þ
V1 ¼
(b)
ð225 mLÞð2:0 M Þ
¼ 25 mL 18 M H2 SO4
18 M
ðV1 Þð15 M Þ ¼ ð75 mLÞð1:0 M Þ
V1 ¼
ð75 mLÞð1:0 M Þ
¼ 5:0 mL 15 M NH3
15 M
6:0 mol HC2 H3 O2
25. ð0:125 LÞ
¼ 0:75 mol HC2 H3 O2
L
(a)
(b)
Final volume after mixing
125 mL þ 525 mL ¼ 650: mL ¼ 0:650 L
0:75 mol HC2 H3 O2
¼ 1:2 M HC2 H3 O2
0:650 L
1L
1:5 mol HC2 H3 O2
ð175 mLÞ
¼ 0:26 mol HC2 H3 O2
L
1000 mL
Total moles ¼ 0:75 mol þ 0:26 mol ¼ 1:01 mol HC2 H3 O2
Finfal volume ¼ 125 mL þ 175 mL ¼ 300: mL ¼ 0:300 L
1:01 mol HC2 H3 O2
¼ 3:37 M HC2 H3 O2
0:300 L
- 168 -
- Chapter 14 3:0 mol HCl
26. ð0:175 LÞ
¼ 0:53 mol HCl
L
(a)
(b)
Final volume after mixing
175 mL þ 250 mL ¼ 425 mL ¼ 0:425 L
0:53 mol HCl
¼ 1:2 M HCl
0:425 L
1L
6:0 mol HCl
ð115 mLÞ
¼ 0:69 mol HCl
1000 mL
L
Total moles ¼ 0:53 mol þ 0:69 mol ¼ 1:22 mol HCl
Final volume ¼ 175 mL þ 115 mL ¼ 290: mL ¼ 0:290 L
1:22 mol HCl
¼ 4:21 M HCl
0:290 L
27. 3 CaðNO3 Þ2 ðaqÞ þ 2 Na3 PO4 ðaqÞ ! Ca3 ðPO4 Þ2 ðsÞ þ 6 NaNO3 ðaqÞ,
1 mol Ca3 ðPO4 Þ2
¼ 1:4 mol Ca3 ðPO4 Þ2
(a) ð2:7 mol Na3 PO4 Þ
2 mol Na3 PO3
6 mol NaNO3
¼ 1:5 mol NaNO3
(b)
0:75 mol CaðNO3 Þ2
3 mol CaðNO3 Þ2
(c)
L CaðNO3 Þ2 ! mol CaðNO3 Þ2 ! mol Na3 PO4
0:225 mol
2 mol Na3 PO4
¼ 0:218 mol Na3 PO4
1:45 L CaðNO3 Þ2
3 mol CaðNO3 Þ2
L
(d)
mL CaðNO3 Þ2 ! mol CaðNO3 Þ2 ! mol Ca3 ðPO4 Þ2 ! g Ca3 ðPO4 Þ2
0:500 mol 1 mol Ca3 ðPO4 Þ2
310:18 g
125 mL CaðNO3 Þ2
¼ 6:46 g Ca3 ðPO4 Þ2
3 mol CaðNO3 Þ2
1000 mL
mol
(e)
mL CaðNO3 Þ2 ! mol CaðNO3 Þ2 ! mol Na3 PO4 ! mL Na3 PO4
0:50 mol
2 mol Na3 PO4
1000 mL
15:0 mL CaðNO3 Þ2
¼ 20: mL Na3 PO4
1000 mL 3 mol CaðNO3 Þ2
0:25 mol
(f)
Find mol CalðNO3 Þ2 mL Na3 PO4 ! mol Na3PO4 ! mol CaðNO3 Þ2
2:0 mol
3 mol CaðNO3 Þ2
¼ 0:15 mol CaðNO3 Þ2
ð50:0 mL Na3 PO4 Þ
2 mol Na3 PO4
1000 mL
mol
0:15 mol CaðNO3 Þ2
M¼
¼ 3:0 M CaðNO3 Þ2
M¼
0:0500 L
L
28. 2 NaOHðaqÞ þ H2 SO4 ðaqÞ ! Na2 SO4 ðaqÞ þ 2 H2 Oðl Þ,
1 mol Na2 SO4
¼ 3:6 mol Na2 SO4
(a) ð3:6 mol H2 SO4 Þ
1 mol H2 SO4
2 mol H2 O
(b) ð0:025 mol NaOHÞ
¼ 0:025 mol H2 O
2 mol NaOH
- 169 -
- Chapter 14 (c)
L H2 SO4 ! mol H2 SO4 ! mol NaOH
0:125 mol
2 mol NaOH
¼ 0:625 mol NaOH
ð2:50 L H2 SO4 Þ
1L
1 mol H2 SO4
(d)
mL NaOH ! mol NaOH ! mol Na2 SO4 ! g Na2 SO4
0:050 mol 1 mol Na2 SO4
142:05 g
ð25 mL NaOHÞ
¼ 0:089 g Na2 SO4
2 mol NaOH
1000 mL
mol
(e)
mL NaOH ! mol NaOH ! mol H2 SO4 ! mL H2 SO4
0:750 mol 1 mol H2 SO4
1000 mL
ð25:5 mL NaOHÞ
¼ 38:25 mL H2 SO4
2 mol NaOH
1000 mL
0:250 mol
(f)
Find mol NaOH
mL H2 SO4 ! mol H2 SO4 ! mol NaOH
0:125 mol
2 mol NaOH
¼ 8:93 103 mol NaOH
ð35:72 mL H2 SO4 Þ
1000 mL
1 mol H2 SO4
mol
8:93 103 mol NaOH
M¼
M¼
¼ 0:185 M NaOH
L
0:04820 L
29. 2 KMnO4 ðaqÞ þ 16 HClðaqÞ ! 2 MnCl2 ðaqÞ þ 5 Cl2 ðgÞ þ 8 H2 Oðl Þ þ 2 KClðaqÞ
0:250 mol
8 mol H2 O
(a) ð15:0 mL HClÞ
¼ 1:88 103 mol H2 O
1000 mL
16 mol HCl
2 mol KMnO4
1L
¼ 12:3 L KMnO4
(b) ð1:85 mol MnCl2 Þ
2 mol MnCl2
0:150 mol KMnO4
0:525 mol 16 mol HCl 1000 mL
(c) ð125 mL KClÞ
¼ 210: mL HCl
1000 mL
2 mol KCl
2:50 mol
0:250 mol
16 mol HCl
¼ 0:0312 mol HCl
(d) ð15:60 mL KMnO4 Þ
1000 mL
2 mol KMnO4
0:0312 mol HCl
M¼
¼ 1:41 M HCl
0:02220 L
(e)
(f)
! L Cl2 ðgas at STPÞ
mL HCl ! mol HCl ! mol Cl2 2:5 mol
5 mol Cl2
22:4 L
ð125 mL HClÞ
¼ 2:2 L Cl2
1000 mL 16 mol HCl
mol
Limiting reactant problem. Convert volume of both reactants to liters of Cl2 gas.
0:750 mol
5 mol Cl2
22:4 L
ð15:0 mL HClÞ
¼ 0:0788 L Cl2
16 mol HCl
1000 mL
mol
0:550 mol
5 mol Cl2
22:4 L
ð12:0 mL KMnO4 Þ
¼ 0:373 L Cl2
2 mol KMnO4
1000 mL
mol
HCL is limiting reactant. 0.0788 L of Cl2 are produced.
! 2 KC2 H3 O2 ðaqÞ þ H2 Oðl Þ þ CO2 ðgÞ
30. K2 CO3 ðaqÞ þ 2 HC2 H3 O2 ðaqÞ 0:150 mol
1 mol H2 O
¼ 1:88 103 mol H2 O
(a) ð25:0 mL HC2 H3 O2 Þ
1000 mL
2 mol HC2 H3 O2
- 170 -
- Chapter 14 -
(b)
(c)
(d)
1 mol K2 CO3
1L
¼ 41:7 L K2 CO3
ð17:5 mol KC2 H3 O2 Þ
2 mol KC2 H3 O2
0:210 mol K2 CO3
0:750 mol 2 mol HC2 H3 O2
1000 mL
¼ 90:2 mL HC2 H3 O2
ð75:2 mL K2 CO3 Þ
1 mol K2 CO3
1000 mL
1:25 mol HC2 H3 O2
0:250 mol 2 mol HC2 H3 O2
¼ 9:25 103 mol HC2 H3 O2
ð18:50 mL K2 CO3 Þ
1 mol K2 CO3
1000 mL
9:25 103 mol HC2 H3 O2
M¼
¼ 0:911 M HC2 H3 O2
0:01015 L
(e)
mL HCL ! mol HCl ! mol Cl2 ! L Cl2 ðgas at STPÞ
1:5 mol
1 mol CO2
22:4 L
ð105 mL of HC2 H3 O2 Þ
¼ 1:8 L CO2
1000 mL 2 mol HC2 H3 O2
mol
(f)
Limiting reactant problem. Convert volume of both reactant to liters of CO2 gas.
0:350 mol
1 mol CO2
22:4 L
ð25:0 mL K2 CO3 Þ
¼ 0:196 L CO2
1 mol K2 CO3
1000 mL
mol
0:250 mol
1 mol CO2
22:4 L
ð25:0 mL HC2 H3 O2 Þ
¼ 0:0700 L CO2
2 mol HC2 H3 O2
1000 mL
mol
HC2 H3 O2 is the limiting reactant. 0.0700 L of CO2 are produced.
mol solute
kg solvent
2:0 mol HCl 1000 g
¼ 11 m HCl
175 g H2 O
kg
14:5 g C12 H22 O11
1000 g
1 mol
¼ 0:0770 m C12 H22 O11
550:0 g H2 O
kg
342:3 g
25:2 mL CH3 OH
0:791 g 1000 g
1 mol
¼ 0:985 m CH3 OH
595 g CH3 CH2 OH
mL
kg
32:04 g
31. Molality ¼ m ¼
(a)
(b)
(c)
mol solute
kg solvent
125 mol CaCl2
1000 g
¼ 1:67 m CaCl2
750:0 g H2 O
kg
2:5 g C6 H12 O6
1000 g
1 mol
¼ 0:026 m C6 H12 O6
525 g H2 O
kg
180:2 g
17:5 mL ðCH3 Þ2 CHOH 0:785 g
1 mL
1000 g
1 mol
¼ 6:44 m ðCH3 Þ2 CHOH
35:5 mL H2 O
mL
1:00 g
kg
60:09 g
32. Molality ¼ m ¼
(a)
(b)
(c)
- 171 -
- Chapter 14 33. (a)
(b)
(c)
34. (a)
(b)
(c)
2:68 g C10 H8
38:4 g C6 H6
1 mol
128:2 g C10 H8
1000 g C6 H6
¼ 0:544 m
kg
5:1 C
Kf (for benzene) ¼
Freezing point of benzene ¼ 5.5 C
m
5:1 C
Dtf ¼ ð0:544 mÞ
¼ 2:8 C
m
Freezing point of solution ¼ 5:5 C 2:8 C ¼ 2:7 C
2:53 C
Kb (for benzene) ¼
Boiling point of benzene ¼ 80.1 C
m
2:53 C
Dtb ¼ ð0:544 mÞ
¼ 1:38 C
m
Boiling point of solution ¼ 80:1 C þ 1:38 C ¼ 81:5 C
100:0 g C2 H6 O2
1 mol
1000 g
¼ 10:74 m
150:0 g H2 O
62:07 g
kg
0:512 C
Dtb ¼ mKb ¼ ð10:74 mÞ
¼ 5:50 C ðIncrease in boiling pointÞ
m
Boiling point ¼ 100:00 C þ 5:50 C ¼ 105:50 C
1:86 C
Dtf ¼ mKf ¼ ð10:74 mÞ
¼ 20:0 C ðDecrease in freezing pointÞ
m
Freezing point ¼ 0:00 C 20:0 C ¼ 20:0 C
35. Freezing point of acetic acid is 16.6 C
Kf acetic acid ¼
3:90 C
m
Dtf ¼ 16:6 C 13:2 C ¼ 3:4 C
Dtf ¼ mKf
3:4 C
m¼
¼ 0:87 m
3:90 C=m
Convert 8.00 g unknown/60.0 g HC2 H3 O2 to g/mol (molar mass)
g unknown
g unknown
g
!
!
g HC2 H3 O2
kg HC2 H3 O2
mol
8:00 g unknown
1000 g
1 kg CH2 H3 O2
¼ 153 g=mol
0:87 mol unknown
60:0 g HC2 H3 O2
kg
Conversion:
36. Dtf ¼ 2:50 C
Kf ðfor H2 OÞ ¼
1:86 C
m
Dtf ¼ mKf
m¼
2:50 C
¼ 1:34 m
1:86 C=m
- 172 -
- Chapter 14 Convert 4.80 g unknown=22.0 g H2 O to g=mol (molar mass)
4:80 g unknown 1000 g
1 kg H2 O
¼ 163 g=mol
22:0 g H2 O
kg
1:34 mol unknown
37. Salt (NaCl) is an ionic compound. When it is dissolved in water the sodium and chloride ions separate or
dissociate. The polar water molecules are attracted to the polar sodium and chloride ions and are
hydrated (surrounded by water molecules). The sodium and chloride ions are separated from one another
and distributed throughout the water in this way. Naþ and Cl are hydrated in aqueous solution: See
Question 1 in the Review Questions.
38. Sugar molecules are not ionic; therefore they do not dissociate when dissolved in water. However,
sugar molecules are polar, so water molecules are attracted to them and the sugar becomes
hydrated. The water molecules help separate the sugar molecules from each other and distribute
them throughout the water.
39. Sugar and salt behave differently when dissolved in water because salt is an ionic compound and sugar is
a molecular compound.
40. An isotonic sodium chloride solution has the same osmotic pressure as human blood plasma. When blood
cells are placed in an isotonic solution the osmotic pressure inside the cells is equal to the osmotic
pressure outside the cells so there is no change in the appearance of the blood cells.
41. The KMnO4 crystals give the solution its purple color. The purple streaks are formed because the solute
has not been evenly distributed throughout that solvent yet. The MnO4 has a purple color in solution.
42. The line for KNO3 slopes upward, because the solubility increases as the temperature increases. KNO3
has the steepest slope of all the compounds given in the diagram. It exhibits the greatest increase in
the number of grams of solute that is able to dissolve in 100 g of water than any other compound in the
diagram as the temperature increases.
mol solute 13:5 g C6 H12 O6
1 mol C6 H12 O6
43. Molarity ¼ M ¼
¼ 0:0749 M C6 H12 O6
;
L
180:2 g
L solution
44. 1:0 m HCl ¼
1 mol HCl 36:46 g HCl
¼
1 kg H2 O
1000 g H2 O
Total mass of solution ¼ 1000 g þ 36:46 g ¼ 1036:46 g
Therefore, 1:0 m HCl ¼
1 mol HCl
1036:46 g HCl solution
NaOH þ HCl ! NaCl þ H2 O
Calculate the grams NaOH to neutralize HCl
ð250:0 g solutionÞ
1 mol HCl
1036:46 g solution
1 mol NaOH 40:00 g
¼ 9:648 g NaOH
1 mol HCl
mol
- 173 -
- Chapter 14 Calculate the grams of 10.0% NaOH solution that contains 9.648 g NaOH.
9:648 g NaOH
10:0 g NaOH
¼
x
100:0 g 10:0% NaOH solution
x ¼ 96:5 g 10% NaOH solution
45. (a)
(b)
(c)
1000 mL 1:06 g 15:0 g sugar
¼ 1:6 102 g sugar
ð1:0 L syrupÞ
L
mL
100: g syrup
1:6 102 g C12 H22 O11
1 mol
¼ 0:47 M
L
342:3 g
m¼
46. Kf ¼
mol sugar
kg H2 O
15% sugar by mass ¼ 15:0 g C12 H22 O11 þ 85:0 g H2 O
15:0 g C12 H22 O11
85:0 g H2 O
5:1 C
m
3:84 g C4 H2 N
250:0 g C6 H6
1000 g H2 O
1 kg H2 O
1 mol
342:3 g C12 H22 O11
¼ 0:516 m
Dtf ¼ 0:614 C
1000 g
15:4 g C4 H2 N
¼
kg
kg C6 H6
Dtf ¼ mKf
0:614 C
0:12 mol C4 H2 N
¼ 0:12 m ¼
5:1 C=m
kg C6 H6
15:4 g C4 H2 N
1 kg C6 H6
¼ 128 g=mol ¼ 1:3 102 g=mol
0:12 mol C4 H2 N
kg C6 H6
m¼
Empirical mass ðC4 H2 NÞ ¼ 64:07 g
130 g
¼ 2:0 (number of empirical formulas per molecular formula)
64:07 g
Therefore, the molecular formula is twice the empirical formula, or C8 H4 N2.
36:46 g
47. ð12:0 mol HClÞ
¼ 438 g HCl in 1:00 L solution
mol
1:18 g solution 1000 mL
ð1:00 LÞ
¼ 1180 g solution
mL
L
1180 g solution 438 g HCl ¼ 742 g H2 O ð0:742 kg H2 OÞ
Since molality ¼
mol HCl 12:0 mol HCl
¼
¼ 16:2 m HCl
kg H2 O
0:742 kg H2 O
- 174 -
- Chapter 14 48. First calculate the g KNO3 in the solution.
mg Kþ
g Kþ
g KNO3
!
!
! g KNO3
mL
mL
mL
5:5 mg Kþ
1g
101:1 g KNO3
ð450 mLÞ ¼ 6:4 g KNO3
mL
39:10 g Kþ
1000 mg
The conversion is:
Now calculate the mol KNO3 and the molarity.
1 mol
ð6:4 g KNO3 Þ
¼ 0:063 mol KNO3
101:1 g
0:063 mol KNO3
¼ 0:14 M
0:450 L
1 qt
946:1 mL 14 mL ethyl alcohol
49. ð16 fl: oz witch hazelÞ
¼ 66 mL ethyl alcohol
32 oz
qt
100 mL witch hazel
50. Verification of Kb for water
Dtb ¼ mKb
Dtb ¼ 101:62 C 100 C ¼ 1:62 C
Kb ¼
Dtb
m
First calculate the molality of the solution.
m¼
16:10 g C2 H6 O2
3:16 mol C2 H6 O2
¼
ð62:07 g=molÞð0:0820 kg H2 OÞ
kg H2 O
Kb ¼
51. (a)
(b)
Dtb
1:62 C
0:513 C kg H2 O
¼
¼
m
3:16 mol=kg H2 O
mol
ð500:0 mL solutionÞ
0:90 g NaCl
¼ 4:5 g NaCl
100: mL solution
4:5 g NaCl
ð100Þ ¼ 9:0%
x mL
x¼
x ¼ volume of 9:0% solution
4:5 g NaCl
¼ 50: mLð4:5 g NaCl in solutionÞ
9:0%
500: mL 50: mL ¼ 450: mL H2 O must evaporate
52. From Figure 14.4, the solubility of KNO3 in H2 O at 20 C is 32 g per 100. g H2 O.
ð50:0 g KNO3 Þ
100: g H2 O
32:0 g KNO3
¼ 156 g H2 O to produce a saturated solution:
175 g H2 O 156 g H2 O ¼ 19 g H2 O must be evaporated:
100: mL solution
53. ð150 mL alcoholÞ
¼ 210 mL solution
70:0 mL alcohol
- 175 -
- Chapter 14 -
54. (a)
(b)
1000 mL solution 1:21 g
35:0 g HNO3
¼ 424 g HNO3
ð1:00 L solutionÞ
100: g solution
L solution
mL
1000 mL solution
1:00 L
ð500: g HNO3 Þ
¼ 1:18 L solution
424 g HNO3
1000 mL
mol solute
L solution
85 g H3 PO4
1:7 g solution 1000 mL 1 mol H3 PO4
¼ 15 M H3 PO4
100 g solution
97:99 g
mL solution
L
55. Molarity ¼ M ¼
56. First calculate the molarity of the solution
80:0 g H2 SO4
1000 mL
1 mol
¼ 1:63 M H2 SO4
500: mL
L
98:09 g
M 1 V1 ¼ M 2 V2
ð1:63 M Þð500: mLÞ ¼ ð0:10 M ÞðV2 Þ
V2 ¼
ð1:63 M Þð500: mLÞ
¼ 8:2 103 mL ¼ 8:2 L
0:10 M
4 qt
1L
5:25 mol HOCH2 CH2 OH
62:07 g
ð4:0 galÞ
1 gal 1:057 qt
L
1 mol HOCH2 CH2 OH
57.
¼ 4:9 103 g HOCH2 CH2 OH
58. Mg þ 2 HCl ! MgCl2 þ H2 ðgÞ
(a) mL HCl ! mol HCl ! mol H2
3:00 mol
1 mol H2
ð200:0 mL HClÞ
¼ 0:300 mol H2
1000 mL 2 mol HCl
(b)
PV ¼ nRT
1 atm
P ¼ ð720 torrÞ
¼ 0:95 atm
760 torr
T ¼ 27 C ¼ 300: K
n ¼ 0:300 mol
V¼
nRT ð0:300 molÞð0:0821 L atm=mol KÞð300: KÞ
¼
¼ 7:8 L H2
P
0:95 atm
59. MgðOHÞ2 þ 2 HCl ! MgCl2 þ 2 H2 O
! AlCl3 þ 3 H2 O
AlðOHÞ3 þ 3 HCl Calculate the moles of HCl neutralized by each base.
- 176 -
- Chapter 14 1 mol
2 mol HCl
¼ 0:0411 mol HCl
58:33 g 1 mol MgðOHÞ2
1 mol
3 mol HCl
¼ 0:0385 mol HCl
1:00 g AlðOHÞ3
78:00 g 1 mol AlðOHÞ3
1:20 g MgðOHÞ2
1.20 g Mg(OH)2 reacts with more HCl than 1.00 g Al(OH)3. Therefore, Mg(OH)2 is more effective in
neutralizing stomach acid.
60. (a)
(b)
With equal masses of CH3OH and C2H5OH, the substance with the lower molar mass will represent more moles of solute in solution. Therefore, the CH3OH will be more effective than C2H5OH
as an antifreeze.
Equal molal solutions will lower the freezing point of the solution by the same amount.
61. Calculate molarity and molality. Assume 1000 mL of solution to calculate the amounts of H2SO4 and
H2O in the solution.
1:29 g
ð1000 mL solutionÞ
¼ 1:29 103 g solution
mL
38 g H2 SO4
3
¼ 4:9 102 g H2 SO4
1:29 10 g solution
100 g solution
1:29 103 g solution 4:9 102 g H2 SO4 ¼ 8:0 102 g H2 O in the solution
490 g H2 SO4
1000 g
1 mol
m¼
¼ 6:2 m H2 SO4
kg
98:09 g
8:0 102 g H2 O
4:9 102 g H2 SO4
1 mol
M¼
¼ 5:0 M H2 SO4
L
98:09 g
62. Freezing point depression is 5.4 C
(a) Dtf ¼ mKf
m¼
(b)
Dtf
5:4 C
¼
¼ 2:9 m
Kf
1:86 C kg solvent=mol solute
0:512 C kg solvent 0:512 C
¼
mol solute
m
0:512 C
Dtb ¼ mKb ¼ ð2:9 mÞ
¼ 1:5 C
m
Kb ðfor H2 OÞ ¼
Boiling point ¼ 100 C þ 1:5 C ¼ 101:5 C
63. Freezing point depression ¼ 0:372 C Kf ¼
1:86 C
m
Dtf ¼ mKf
0:372 C
¼ 0:200 m
m ¼
1:86 C=m
- 177 -
- Chapter 14 1 mol
ð6:20 g C2 H6 O2 Þ
¼ 0:100 mol C2 H6 O2
62:07 g
1 kg H2 O
1000 g H2 O
ð0:100 mol C2 H6 O2 Þ
¼ 500: g H2 O
0:200 mol C2 H6 O2
kg H2 O
64. (a)
(b)
(c)
Freezing point depression ¼ 20.0 C
1000 mL 1:00 g
12:0 L H2 O
¼ 1:20 104 g H2 O
L
mL
Dtf ¼ mKf
20:0 C
m ¼
¼ 10:8 m
1:86 C=m 10:8 mol C2 H6 O2
62:07 g
4
1:20 10 g H2 O
¼ 8:04 103 g C2 H6 O2
1000 g H2 O
mol
1:00 mL
3
8:04 10 g C2 H6 O2
¼ 7:24 103 mL C2 H6 O2
1:11 g
F ¼ 1:8ð CÞ þ 32 ¼ 1:8ð20:0Þ þ 32 ¼ 4:0 F
65. HNO3 þ NaHCO3 ! NaNO3 þ H2 O þ CO2
First calculate the grams of NaHCO3 in the sample.
! L HNO3 ! mol HNO3 ! mol NaHCO3 ! g NaHCO3
mL HNO3 1L
0:055 mol 1 mol NaHCO3
84:01 g
ð150 mL HNO3 Þ
1 mol HNO3
1000 mL
L
mol
¼ 0:69 g NaHCO3 in the sample
0:69 g NaHCO3
ð100Þ ¼ 47% NaHCO3
1:48 g sample
66. (a)
Dilution problem: M 1 V1 ¼ M 2 V2
ð1:5 M Þð8:4 LÞ ¼ ð17:8 M ÞðV2 Þ
V2 ¼
(b)
(c)
ð1:5 M Þð8:4 LÞ
¼ 0:71 L
17:8 M
0.71 L of 17.8 M H2 SO4 must be added (assume volumes are additive)
17:8 mol
ð1:00 mLÞ ¼ 0:0178 mol
1000: mL
1:5 mol
ð1:00 mLÞ ¼ 0:0015 mol H2 SO4 in each mL
1000: mL
67. moles HNO3 total ¼ moles HNO3 from 3.00 M þ moles HNO3 from 12.0 M
M T VT ¼ M 3:00 M V3:00 M þ M 12:0 M V12:0 M
- 178 -
- Chapter 14 Assume preparation of 1000. mL of 6 M solution
Let y ¼ volume of 3.00 M solution; volume of 12.0 M ¼ 1000. mL y
(6.00 M )(1000. mL) ¼ (3.00 M)(y) þ (12.0 M )(1000. mL y)
6000. mL ¼ 3.00 y mL þ 12,000 mL 12.0 y
6000: mL
¼ 667 mL 3 M
9:00
1000: mL 667 mL ¼ 333 mL 12 M
6000: mL ¼ 9:00 y
y¼
Mix together 667 mL 3.00 M HNO3 and 333 mL of 12.0 M HNO3 to get 1000. mL of 6.00 M HNO3.
68. HBr þ NaOH ! NaBr þ H2 O
First calculate the molarity of the diluted HBr solution.
The reaction is 1 mol HBr to 1 mol NaOH, so
M A VA ¼ M B VB
ðM A Þð100:0 mLÞ ¼ ð0:37 M Þð88:4 mLÞ
ð0:37 M Þð88:4 mLÞ
¼ 0:33 M HBr ðdiluted solutionÞ
100:00 mL
Now calculate the molarity of the HBr before dilution.
MA ¼
M 1 V1 ¼ M 2 V2
ðM 1 Þð20:0 mLÞ ¼ ð0:33 M Þð240: mLÞ
M1 ¼
ð0:33 M Þð240: mLÞ
¼ 4:0 M HBr ðoriginal solutionÞ
20:0 mL
69. BaðNO3 Þ2 þ 2 KOH ! BaðOHÞ2 þ 2 KNO3
This is a limiting reactant problem. First calculate the moles of each reactant and determine the limiting
reactant.
moles
ML¼
ðLÞ ¼ moles
L
0:642 mol
ð0:0805 LÞ ¼ 0:0517 mol BaðNO3 Þ2
L
0:743 mol
ð0:0445 LÞ ¼ 0:0331 mol KOH
L
According to the equation, twice as many moles of KOH as Ba(NO3)2 are needed, so KOH is the limiting
reactant.
0:25 mol
70. (a)
ð0:0458 LÞ ¼ 0:011 mol Li2 CO3
L
0:25 mol
73:89 g
(b)
ð0:75 LÞ
¼ 14 g Li2 CO3
L
mol
- 179 -
- Chapter 14 -
(c)
(d)
1 mol
1000: mL
ð6:0 g Li2 CO3 Þ
¼ 3:2 102 mL solution
73:89 g
0:25 mol
Assume 1000. mL solution
1:22 g
ð1000: mLÞ ¼ 1220 g solution
mL
0:25 mol 73:89 g Li2 CO3
¼ 18 g Li2 CO3 per L solution
mol
L
g solute
18 g
ð100Þ ¼
ð100Þ ¼ 1:5% (mass percent)
%¼
g solution
1220 g
71. The balanced equation is
! 2 NaClðaqÞ þ H2 Oðl Þ þ SO2 ðgÞ
2 HClðaqÞ þ Na2 SO3 ðaqÞ 2:50 mol 1 mol SO2
ð125 mL HClÞ
¼ 0:156 mol SO2
2
mol HCl
1000
mL 1:75 mol
1 mol SO2
¼ 0:131 mol SO2
ð75 mL Na2 SO3 Þ
1000 mL 1 mol Na2 SO3
Na2 SO3 is the limiting reactant 0.131 mol of SO2 gas will be produced
The gas is at non-standard conditions, so use PV ¼ nRT to find the liters of SO2.
V¼
nRT
P
V¼
ð0:131 molÞð0:0821 L atm=mol KÞð295 KÞ
¼ 3:11 L SO2
ð775 torrÞð1 atm=760 torrÞ
72. mass of solute ¼ mass of container & solute mass of container mass of water
mass of water ¼ (5.549 moles) (18.02 g=mol) ¼ 100.0 g
mass of solute ¼ 563 g 375 g 100.0 g ¼ 88 g
solubility in water ¼ g solute=100 g H2O at 20 C
Using this data, solubility is 88 g solute=100.0 g water ¼ NaNO3 (see Table 14.3).
- 180 -