C H A P T E R 1 4

SOLUTIONS

SOLUTIONS TO REVIEW QUESTIONS

1.

Na+ Cl–

O

HH

O

HH

O

HH

O

HH

O

HH

O

HH

O

H H

O

H H

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) NH4Cl at 35�C, 43 g (c) SO2 gas at 30

�C, 8 g at 1 atm

(b) CuSO4 at 60�C, 39 g (d) NaNO3 at 15

�C, 82 g

6. Li2SO4

7.6:5 g

21:5 g� 100 ¼ 30 mass percent

8. Cube 1 cm 0:01 cm

Volume 1 cm3 1� 10�6 cm3

Number=1 cm cube 1 106 ð1 cm3Þ= 1� 10�6cm3� � ¼ 106cubes

Area of face 1 cm2 1� 10�4cm2

Total surface area 6 cm2 6� 102 cm2

- 158 -

1� 106 cubes� �

6 faces=cubeð Þ 1� 10�4cm2=face� � ¼ 6� 102 cm2

9.40: g

115 g� 100 ¼ 35 mass percent

10. The rate of dissolving decreases. The rate of dissolving is at its maximumwhen 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: MnO �4

(purple) and Cr2O2�7 (orange). Therefore, it is predictable that the Cr2O

2�7 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 hexanemolecules, being nonpolar, have

no strong forces to pull the ions apart, so sodium chloride is insoluble in hexane.

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- 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.

0:5 Lð Þ 1 mol MgCl2

L

� �2 mol Cl�

1 mol MgCl2

� �¼ 1 mol Cl�

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

0:75 Lð Þ 5:0 mol NaCl

L

� �58:44 g

1 mol

� �¼ 2:2� 102 g NaCl

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- 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) 1 M NaOH 1 liter

(b) 0.6 M Ba(OH)2 0.83 liter

(c) 2 M KOH 0.50 liter

(d) 1.5 M Ca(OH)2 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) 100. g (2.17 mol) of ethyl alcohol is more effective than 100. g (0.292 mol) of sucrose.

(b) 20.0 g (0.435 mol) of ethyl alcohol is more effective than 100. g (0.292 mol) of sucrose.

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- 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 NaCl¼ 5 mol NaCl=kg H2O; 5 molar NaCl¼ 5 mol NaCl=L of solution. The volume of the 5

molal solution will be larger than 1 liter (1 L H2Oþ 5 mol NaCl). The volume of the 5 molar solution is

exactly 1 L (5 mol NaClþ sufficient H2O to produce 1 L of solution). The molarity of a 5 molal solution

is therefore, less than 5 molar.

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- Chapter 14 -

SOLUTIONS TO EXERCISES

1. Reasonably soluble: (b) K2SO4 (c) Na3PO4 (d) NaOH

Insoluble: (a) AgCl (e) PbI2 (f) SnCO3

2. Reasonably soluble: (b) Cu(NO3)2 (d) NH4C2H3O2 (f) AgNO3

Insoluble: (a) Ba3(PO4)2 (c) Fe(OH)3 (e) MgO

3. Mass percent calculations

(a) 15.0 g KClþ 100.0 g H2O¼ 115.0 g solution

15:0 g

115:0 g

� �100ð Þ ¼ 13:0%KCl

(b) 2.50 g Na3PO4þ 10.0 g H2O¼ 12.5 g solution

2:50 g

12:5 g

� �100ð Þ ¼ 20:0%Na3PO4

(c) (0.20 mol NH4C2H3O2)77:09 g

1 mol

� �¼ 15 g NH4C2H3O2

15 g NH4C2H3O2þ 125 g H2O¼ 140. g solution

15 g

140 g

� �100ð Þ ¼ 11%NH4C2H3O2

(d) (1.50 mol NaOH)40:00 g

1 mol

� �¼ 60:0 g NaOH

33:0 mol H2Oð Þ 18:02 g

1 mol

� �¼ 595 g H2O

60.0 g NaOHþ 595 g H2O¼ 655 g solution

60:0 g

655 g

� �100ð Þ ¼ 9:16%NaOH

4. Mass percent calculations

(a) 25.0 g NaNO3 in 125.0 g H2O¼ 150.0

25:0 g

150:0 g

� �100ð Þ ¼ 16:7%NaNO3

(b) 1.25 g CaCl2 in 35.0 g H2O¼ 36.3 g solution

1:25 g

36:3 g

� �100ð Þ ¼ 3:44%CaCl2

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- Chapter 14 -

(c) (0.75 mol K2CrO4)þ 194:2 g

1 mol

� �¼ 150 g K2CrO4

150 g K2CrO4þ 225 g H2O¼ 380 g solution

150 g

380 g

� �100ð Þ ¼ 39%K2CrO4

(d) (1.20 mol H2SO4)98:09 g

1 mol

� �¼ 118 g H2SO4

(72.5 mol H2O)18:02 g

1 mol

� �¼ 1:31� 103 g H2O

118 g H2SO4þ 1.31� 103 g H2O¼ 1.43� 103 g solution

118 g

1:43� 103 g

� �100ð Þ ¼ 8:25%H2SO4

5. A 15.5% solution contains 15.5 g AgNO3 per 100. g solution

25:2 g AgNO3ð Þ 100: g solution

15:5 g AgNO3

� �¼ 163 g solution

6. A 10.0% NaCl solution contains 10.0 g NaCl per 100. g solution

25:0 g NaClð Þ 100: g solution

10:0 g NaCl

� �¼ 250: g solution

7. (a) 25 g solutionð Þ 7:5 g CaSO4

100: g solution

� �¼ 1:9 g CaSO4

(b) 25 g solution� 6:9 g solute ¼ 23 g solvent

8. (a) 75 g solutionð Þ 12:0 g BaCl2100: g solution

� �¼ 9:0 g BaCl2

(b) 75 g solution� 9:0 g solute ¼ 66 g solvent

9. Mass/volume percent.

(a)15:0 g C2H5OH

150:0 mL solution

� �100ð Þ ¼ 10:0%C2H5OH

(b)25:2 g NaCl

125:5 mL solution

� �100ð Þ ¼ 20:1%NaCl

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- Chapter 14 -

10. Mass/volume percent.

(a)175:2 g C12H22O11

275:5 mL solution

� �100ð Þ ¼ 63:59% C12H22O11

(b)35:5 g of CH3OH

75:0 mL solution

� �100ð Þ ¼ 47:3% CH3OH

11. Volume percent.

(a)50:0 mL hexanol

125 mL solution

� �100ð Þ ¼ 40:0% hexanol

(b)2:0 mL ethanol

15:0 mL solution

� �100ð Þ ¼ 13% ethanol

12. Volume percent.

(a)37:5 mL butanol

275 mL solution

� �100ð Þ ¼ 13:6% butanol

(b)4:0 mLmethanol

25:0 mL solution

� �100ð Þ ¼ 16%methanol

13. Molarity problems M ¼ mol

L

� �

(a)0:25 mol

75:0 mL

� �1000 mL

1 L

� �¼ 3:3M

(b)1:75 mol

0:75 L

� �¼ 2:3M KBr

(c)35:0 g

1:25 L

� �1 mol

82:03 g

� �¼ 0:341M NaC2H3O2

(d)75 g CuSo4�5 H2O

1:0 L

� �1 mol

249:7 g

� �¼ 0:30M CuSO4

14. Molarity problems M ¼ mol

L

� �

(a)0:50 mol

125 mL

� �1000 mL

1 L

� �¼ 4:0M

(b)2:25 mol

1:50 L

� �¼ 1:50M CaCl2

(c)275 g

775 mL

� �1 mol

180:2 g

� �1000 mL

1 L

� �¼ 1:97M C6H12O6

(d)125 gMgSO4�7 H2O

2:50 L

� �1 mol

246:5 g

� �¼ 0:203MMgSO4

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- Chapter 14 -

15. Molarity ¼ mol solute

L solutionor mol solute ¼ L solutionð Þ Molarityð Þ

(a) 1:5 Lð Þ 1:20 mol H2SO4

L

� �¼ 1:8 mol H2SO4

(b) 25:0 mLð Þ 1 L

1000 mL

� �0:0015 mol BaCl2

L

� �¼ 3:8� 10�5 mol BaCl2

(c) 125 mLð Þ 1 L

1000 mL

� �0:35 mol K3PO4

L

� �¼ 0:044 mol K3PO4

16. Molarity ¼ mol solute

L solutionor mol solute ¼ L solutionð Þ Molarityð Þ

(a) 0:75 Lð Þ 1:50 mol HNO3

L

� �¼ 1:1 mol HNO3

(b) 10:0 mLð Þ 1 L

1000 mL

� �0:75 mol NaClO3

L

� �¼ 7:5� 10�3mol NaClO3

(c) 175 mLð Þ 1 L

1000 mL

� �0:50 mol LiBr

L

� �¼ 0:088 mol LiBr

17. (a) 2:5 Lð Þ 0:75 mol K2CrO4

L

� �194:2 g

1 mol

� �¼ 360 g K2CrO4

(b) 75:2 mLð Þ 1 L

1000 mL

� �0:050 mol HC2H3O2

L

� �60:05 g

1 mol

� �¼ 0:226 g HC2H3O2

(c) 250 mLð Þ 1 L

1000 mL

� �16 mol HNO3

L

� �63:02 g

1 mol

� �¼ 250 g HNO3

18. (a) 1:20 Lð Þ 18 mol H2SO4

L

� �98:09 g

1 mol

� �¼ 2:1� 103g H2SO4

(b) 27:5 mLð Þ 1 L

1000 mL

� �1:5 mol KMnO4

L

� �158:0 g

1 mol

� �¼ 6:52 g KMnO4

(c) 120 mLð Þ 1 L

1000 mL

� �0:025 mol Fe2 SO4ð Þ3

L

� �399:9 g

1 mol

� �¼ 1:2 g Fe2 SO4ð Þ3

19. (a) 0:15 mol H3PO4ð Þ 1 L

0:750 mol H3PO4

� �1000 mL

1 L

� �¼ 2:0� 102 mL

(b) 35:5 g H3PO4ð Þ 1 mol

97:99 g

� �1 L

0:750 mol H3PO4

� �1000 mL

1 L

� �¼ 483 mL

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- Chapter 14 -

20. (a) 0:85 mol NH4Clð Þ 1 L

0:250 mol NH4Cl

� �1000 mL

1 L

� �¼ 3:4� 103 mL

(b) 25:2 gð ÞNH4Cl1 mol

53:49 g

� �1 L

0:250 mol NH4Cl

� �1000 mL

L

� �¼ 1:88� 103 mL

21. Dilution problem V1M1 ¼ V2M2

(a) V1 ¼ 125 mL V2 ¼ 125 mLþ 775 mLð Þ ¼ 900:mL

M1 ¼ 5:0M M2 ¼ M2

125 mLð Þ 5:0Mð Þ ¼ 900:mLð Þ M2ð Þ

M2 ¼ 125 mLð Þ 5:0Mð Þ900:mL

¼ 0:694M

(b) V1 ¼ 250 mL V2 ¼ 250 mLþ 750 mLð Þ ¼ 1:00� 103 mL

M1 ¼ 0:25M M2 ¼ M2

250 mLð Þ 0:25Mð Þ ¼ 1:00� 103 mL� �

M2ð Þ

M2 ¼ 250 mLð Þ 0:25Mð Þ1:00� 103 mL

¼ 0:063M

(c) First calculate the moles of HNO3 in each solution. Then calculate the molarity.

75 mLð Þ 1 L

1000 mL

� �0:50 mol

1 L

� �¼ 0:038 mol HNO3

75 mLð Þ 1 L

1000 mL

� �1:5 mol

1 L

� �¼ 0:11 mol HNO3

Total mol¼ 0.15 mol

Total volume¼ 75 mLþ 75 mL¼ 150. mL¼ 0.150 L

0:150 mol

0:150 L¼ 1:00M

22. Dilution problem V1M1 ¼ V2M2

(a) V1 ¼ 175 mL V2 ¼ 175 mLþ 275 mLð Þ ¼ 450:mL

M1 ¼ 3:0M M2 ¼ M2

175 mLð Þ 3:0Mð Þ ¼ 450:mLð Þ M2ð Þ

M2 ¼ 175 mLð Þ 3:0Mð Þ450:mL

¼ 1:2M

(b) V1 ¼ 350 mL V2 ¼ 350 mLþ 150 mLð Þ ¼ 5:0� 102 mL

M1 ¼ 0:10M M2 ¼ M2

350 mLð Þ 0:10Mð Þ ¼ 5:0� 102 mL� �

M2ð Þ

M2 ¼ 350 mLð Þ 0:10Mð Þ5:0� 102 mL

¼ 0:070M

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- Chapter 14 -

(c) First calculate the moles of HCl in each solution. Then calculate the molarity.

50:0 mLð Þ 1 L

1000 mL

� �0:250 mol

1 L

� �¼ 0:0125 mol HCl

25:0 mLð Þ 1 L

1000 mL

� �0:500 mol

1 L

� �¼ 0:0125 mol HCl

Total mol¼ 0.0250 mol

Total volume¼ 50.0 mLþ 25.0 mL¼ 75.0 mL¼ 0.0750 L

0:250 mol

0:0750 L¼ 0:333M

23. V1M1 ¼ V2M2

(a) V1ð Þ 15Mð Þ ¼ 750 mLð Þ 3:0Mð Þ

V1 ¼ 750 mLð Þ 3:0Mð Þ15M

¼ 150 mL 15M H3PO4

(b) V1ð Þ 16Mð Þ ¼ 250 mLð Þ 0:50Mð Þ

V1 ¼ 250 mLð Þ 0:50Mð Þ16M

¼ 7:8 mL 16M HNO3

24. V1M1 ¼ V2M2

(a) V1ð Þ 18Mð Þ ¼ 225 mLð Þ 2:0Mð Þ

V1 ¼ 225 mLð Þ 2:0Mð Þ18M

¼ 25 mL 18M H2SO4

(b) V1ð Þ 15Mð Þ ¼ 75 mLð Þ 1:0Mð Þ

V1 ¼ 75 mLð Þ 1:0Mð Þ15M

¼ 5:0 mL 15M NH3

25. 0:125 Lð Þ 6:0 mol HC2H3O2

L

� �¼ 0:75 mol HC2H3O2

(a) Final volume after mixing

125 mLþ 525 mL ¼ 650:mL ¼ 0:650 L0:75 mol HC2H3O2

0:650 L¼ 1:2M HC2H3O2

(b) 175 mLð Þ 1 L

1000 mL

� �1:5 mol HC2H3O2

L

� �¼ 0:26 mol HC2H3O2

Total moles ¼ 0:75 molþ 0:26 mol ¼ 1:01 mol HC2H3O2

Finfal volume ¼ 125 mLþ 175 mL ¼ 300:mL ¼ 0:300 L

1:01 mol HC2H3O2

0:300 L¼ 3:37M HC2H3O2

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- Chapter 14 -

26. 0:175 Lð Þ 3:0 mol HCl

L

� �¼ 0:53 mol HCl

(a) Final volume after mixing

175 mLþ 250 mL ¼ 425 mL ¼ 0:425 L0:53 mol HCl

0:425 L¼ 1:2M HCl

(b) 115 mLð Þ 1 L

1000 mL

� �6:0 mol HCl

L

� �¼ 0:69 mol HCl

Total moles ¼ 0:53 molþ 0:69 mol ¼ 1:22 mol HCl

Final volume ¼ 175 mLþ 115 mL ¼ 290:mL ¼ 0:290 L

1:22 mol HCl

0:290 L¼ 4:21M HCl

27. 3 Ca NO3ð Þ2 aqð Þ þ 2 Na3PO4 aqð Þ ���! Ca3 PO4ð Þ2 sð Þ þ 6 NaNO3 aqð Þ,

(a) 2:7 mol Na3PO4ð Þ 1 mol Ca3 PO4ð Þ22 mol Na3PO3

� �¼ 1:4 mol Ca3 PO4ð Þ2

(b) 0:75 mol Ca NO3ð Þ2� � 6 mol NaNO3

3 mol Ca NO3ð Þ2

� �¼ 1:5 mol NaNO3

(c) L Ca NO3ð Þ2 ���! mol Ca NO3ð Þ2 ���! mol Na3PO4

1:45 L Ca NO3ð Þ2� � 0:225 mol

L

� �2 mol Na3PO4

3 mol Ca NO3ð Þ2

� �¼ 0:218 mol Na3PO4

(d) mL Ca NO3ð Þ2 ���! mol Ca NO3ð Þ2 ���! mol Ca3 PO4ð Þ2 ���! g Ca3 PO4ð Þ2125 mL Ca NO3ð Þ2� � 0:500 mol

1000 mL

� �1 mol Ca3 PO4ð Þ23 mol Ca NO3ð Þ2

� �310:18 g

mol

� �¼ 6:46 g Ca3 PO4ð Þ2

(e) mL Ca NO3ð Þ2 ���! mol Ca NO3ð Þ2 ���! mol Na3PO4 ���! mLNa3PO4

15:0 mL Ca NO3ð Þ2� � 0:50 mol

1000 mL

� �2 mol Na3PO4

3 mol Ca NO3ð Þ2

� �1000 mL

0:25 mol

� �¼ 20:mLNa3PO4

(f) Find mol Cal NO3ð Þ2 mL Na3PO4 ���! mol Na3PO4 ���! mol Ca NO3ð Þ250:0 mLNa3PO4ð Þ 2:0 mol

1000 mL

� �3 mol Ca NO3ð Þ22 mol Na3PO4

� �¼ 0:15 mol Ca NO3ð Þ2

M ¼ mol

LM ¼ 0:15 mol Ca NO3ð Þ2

0:0500 L

� �¼ 3:0M Ca NO3ð Þ2

28. 2 NaOH aqð Þ þ H2SO4 aqð Þ ���! Na2SO4 aqð Þ þ 2 H2O lð Þ,

(a) 3:6 mol H2SO4ð Þ 1 mol Na2SO4

1 mol H2SO4

� �¼ 3:6 mol Na2SO4

(b) 0:025 mol NaOHð Þ 2 mol H2O

2mol NaOH

� �¼ 0:025 mol H2O

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- Chapter 14 -

(c) L H2SO4 ���! mol H2SO4 ���! mol NaOH

2:50 L H2SO4ð Þ 0:125 mol

1 L

� �2 mol NaOH

1mol H2SO4

� �¼ 0:625 mol NaOH

(d) mLNaOH���! mol NaOH���! mol Na2SO4 ���! g Na2SO4

25 mLNaOHð Þ 0:050 mol

1000 mL

� �1 mol Na2SO4

2 mol NaOH

� �142:05 g

mol

� �¼ 0:089 g Na2SO4

(e) mL NaOH���! mol NaOH���! mol H2SO4 ���! mLH2SO4

25:5 mLNaOHð Þ 0:750 mol

1000 mL

� �1 mol H2SO4

2 mol NaOH

� �1000 mL

0:250 mol

� �¼ 38:25 mLH2SO4

(f) Find mol NaOH mLH2SO4 ���! mol H2SO4 ���! mol NaOH

35:72 mLH2SO4ð Þ 0:125 mol

1000 mL

� �2 mol NaOH

1mol H2SO4

� �¼ 8:93� 10�3 mol NaOH

M ¼ mol

LM ¼ 8:93� 10�3 mol NaOH

0:04820 L

� �¼ 0:185M NaOH

29. 2 KMnO4 aqð Þ þ 16 HCl aqð Þ ���! 2MnCl2 aqð Þ þ 5 Cl2 gð Þ þ 8 H2O lð Þ þ 2 KCl aqð Þ(a) 15:0 mLHClð Þ 0:250 mol

1000 mL

� �8 mol H2O

16 mol HCl

� �¼ 1:88� 10�3 mol H2O

(b) 1:85 mol MnCl2ð Þ 2 mol KMnO4

2 mol MnCl2

� �1 L

0:150 mol KMnO4

� �¼ 12:3 L KMnO4

(c) 125 mLKClð Þ 0:525 mol

1000 mL

� �16 mol HCl

2 mol KCl

� �1000 mL

2:50 mol

� �¼ 210:mLHCl

(d) 15:60 mLKMnO4ð Þ 0:250 mol

1000 mL

� �16 mol HCl

2 mol KMnO4

� �¼ 0:0312 mol HCl

M ¼ 0:0312 mol HCl

0:02220 L

� �¼ 1:41M HCl

(e) mL HCl���! mol HCl���! mol Cl2 ���! L Cl2 gas at STPð Þ

125 mLHClð Þ 2:5 mol

1000 mL

� �5 mol Cl2

� �22:4 L

mol

� �¼ 2:2 L Cl2

(f) Limiting reactant problem. Convert volume of both reactants to liters of Cl2 gas.

15:0 mLHClð Þ 0:750 mol

1000 mL

� �5 mol Cl2

16 mol HCl

� �22:4 L

mol

� �¼ 0:0788 L Cl2

12:0 mLKMnO4ð Þ 0:550 mol

1000 mL

� �5 mol Cl2

2 mol KMnO4

� �22:4 L

mol

� �¼ 0:373 L Cl2

HCL is limiting reactant. 0.0788 L of Cl2 are produced.

30. K2CO3 aqð Þ þ 2 HC2H3O2 aqð Þ ���! 2 KC2H3O2 aqð Þ þ H2O lð Þ þ CO2 gð Þ

(a) 25:0 mLHC2H3O2ð Þ 0:150 mol

1000 mL

� �1 mol H2O

2mol HC2H3O2

� �¼ 1:88� 10�3 mol H2O

- 170 -

- Chapter 14 -

16 mol HCl

(b) 17:5 mol KC2H3O2ð Þ 1 mol K2CO3

2 mol KC2H3O2

� �1 L

0:210 mol K2CO3

� �¼ 41:7 L K2CO3

(c) 75:2 mLK2CO3ð Þ 0:750 mol

1000 mL

� �2 mol HC2H3O2

1 mol K2CO3

� �1000 mL

1:25 mol HC2H3O2

� �¼ 90:2 mLHC2H3O2

(d) 18:50 mLK2CO3ð Þ 0:250 mol

1000 mL

� �2 mol HC2H3O2

1 mol K2CO3

� �¼ 9:25� 10�3 mol HC2H3O2

M ¼ 9:25� 10�3 mol HC2H3O2

0:01015 L

� �¼ 0:911M HC2H3O2

(e) mL HCL���! mol HCl���! mol Cl2 ���! L Cl2 gas at STPð Þ

105 mL of HC2H3O2ð Þ 1:5 mol

1000 mL

� �1 mol CO2

2 mol HC2H3O2

� �22:4 L

mol

� �¼ 1:8 L CO2

(f) Limiting reactant problem. Convert volume of both reactant to liters of CO2 gas.

25:0 mLK2CO3ð Þ 0:350 mol

1000 mL

� �1 mol CO2

1 mol K2CO3

� �22:4 L

mol

� �¼ 0:196 L CO2

25:0 mLHC2H3O2ð Þ 0:250 mol

1000 mL

� �1 mol CO2

2 mol HC2H3O2

� �22:4 L

mol

� �¼ 0:0700 L CO2

HC2H3O2 is the limiting reactant. 0.0700 L of CO2 are produced.

31. Molality ¼ m ¼ mol solute

kg solvent

(a)2:0 mol HCl

175 g H2O

� �1000 g

kg

� �¼ 11mHCl

(b)14:5 g C12H22O11

550:0 g H2O

� �1000 g

kg

� �1 mol

342:3 g

� �¼ 0:0770mC12H22O11

(c)25:2 mL CH3OH

595 g CH3CH2OH

� �0:791 g

mL

� �1000 g

kg

� �1 mol

32:04 g

� �¼ 0:985m CH3OH

32. Molality ¼ m ¼ mol solute

kg solvent

(a)125 mol CaCl2

750:0 g H2O

� �1000 g

kg

� �¼ 1:67mCaCl2

(b)2:5 g C6H12O6

525 g H2O

� �1000 g

kg

� �1 mol

180:2 g

� �¼ 0:026mC6H12O6

(c)17:5 mL CH3ð Þ2CHOH

35:5 mLH2O

� �0:785 g

mL

� �1 mL

1:00 g

� �1000 g

kg

� �1 mol

60:09 g

� �¼ 6:44m CH3ð Þ2CHOH

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- Chapter 14 -

33. (a)2:68 g C10H8

38:4 g C6H6

� �1 mol

128:2 g C10H8

� �1000 g C6H6

kg

� �¼ 0:544m

(b) Kf (for benzene)¼ 5:1�Cm

Freezing point of benzene¼ 5.5�C

Dtf ¼ 0:544mð Þ 5:1�Cm

� �¼ 2:8�C

Freezing point of solution ¼ 5:5�C� 2:8�C ¼ 2:7�C

(c) Kb (for benzene) ¼ 2:53�Cm

Boiling point of benzene¼ 80.1�C

Dtb ¼ 0:544mð Þ 2:53�Cm

� �¼ 1:38�C

Boiling point of solution ¼ 80:1�Cþ 1:38�C ¼ 81:5�C

34. (a)100:0 g C2H6O2

150:0 g H2O

� �1 mol

62:07 g

� �1000 g

kg

� �¼ 10:74m

(b) Dtb ¼ mKb ¼ 10:74mð Þ 0:512�Cm

� �¼ 5:50�C Increase in boiling pointð Þ

Boiling point ¼ 100:00�Cþ 5:50�C ¼ 105:50�C

(c) Dtf ¼ mKf ¼ 10:74mð Þ 1:86�Cm

� �¼ 20:0�C Decrease in freezing pointð Þ

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�Cm

Dtf ¼ 16:6�C� 13:2�C ¼ 3:4�C

Dtf ¼ mKf

m ¼ 3:4�C3:90�C=m

¼ 0:87m

Convert 8.00 g unknown/60.0 g HC2H3O2 to g/mol (molar mass)

Conversion:g unknown

g HC2H3O2

���! g unknown

kg HC2H3O2

���! g

mol

8:00 g unknown

60:0 g HC2H3O2

� �1000 g

kg

� �1 kg CH2H3O2

0:87 mol unknown

� �¼ 153 g=mol

36. Dtf ¼ 2:50�C Kf for H2Oð Þ ¼ 1:86�Cm

Dtf ¼ mKf

m ¼ 2:50�C1:86�C=m

¼ 1:34m

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- Chapter 14 -

Convert 4.80 g unknown=22.0 g H2O to g=mol (molar mass)

4:80 g unknown

22:0 g H2O

� �1000 g

kg

� �1 kg H2O

1:34 mol unknown

� �¼ 163 g=mol

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 bywatermolecules). The sodiumand chloride ions are separated fromone 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

4� 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.

43. Molarity ¼ M ¼ mol solute

L solution;

13:5 g C6H12O6

L

� �1 mol C6H12O6

180:2 g

� �¼ 0:0749M C6H12O6

44. 1:0mHCl ¼ 1 mol HCl

1 kg H2O¼ 36:46 g HCl

1000 g H2O

Total mass of solution ¼ 1000 gþ 36:46 g ¼ 1036:46 g

Therefore, 1:0mHCl ¼ 1 mol HCl

1036:46 g HCl solution

NaOHþ HCl���! NaClþ H2O

Calculate the grams NaOH to neutralize HCl

250:0 g solutionð Þ 1 mol HCl

1036:46 g solution

� �1 mol NaOH

1mol HCl

� �40:00 g

mol

� �¼ 9:648 g NaOH

- 173 -

- Chapter 14 -

has not been evenly distributed throughout that solvent yet. The MnO

Calculate the grams of 10.0% NaOH solution that contains 9.648 g NaOH.

9:648 g NaOH

x¼ 10:0 g NaOH

100:0 g 10:0%NaOH solution

x ¼ 96:5 g 10%NaOH solution

45. (a) 1:0 L syrupð Þ 1000 mL

L

� �1:06 g

mL

� �15:0 g sugar

100: g syrup

� �¼ 1:6� 102 g sugar

(b)1:6� 102g C12H22O11

L

� �1 mol

342:3 g

� �¼ 0:47M

(c) m ¼ mol sugar

kg H2O15% sugar by mass ¼ 15:0 g C12H22O11 þ 85:0 g H2O

15:0 g C12H22O11

85:0 g H2O

� �1000 g H2O

1 kg H2O

� �1 mol

342:3 g C12H22O11

� �¼ 0:516m

46. Kf ¼ 5:1�Cm

Dtf ¼ 0:614�C

3:84 g C4H2N

250:0 g C6H6

� �1000 g

kg

� �¼ 15:4 g C4H2N

kg C6H6

Dtf ¼ mKf

m ¼ 0:614�C5:1�C=m

¼ 0:12m ¼ 0:12 mol C4H2N

kg C6H6

15:4 g C4H2N

kg C6H6

� �1 kg C6H6

0:12 mol C4H2N

� �¼ 128 g=mol ¼ 1:3� 102 g=mol

Empirical mass C4H2Nð Þ ¼ 64:07 g

130 g

64:07 g¼ 2:0 (number of empirical formulas per molecular formula)

Therefore, the molecular formula is twice the empirical formula, or C8H4N2.

47. 12:0 mol HClð Þ 36:46 g

mol

� �¼ 438 g HCl in 1:00 L solution

1:00 Lð Þ 1:18 g solution

mL

� �1000 mL

L

� �¼ 1180 g solution

1180 g solution� 438 g HCl ¼ 742 g H2O 0:742 kg H2Oð Þ

Since molality ¼ mol HCl

kg H2O¼ 12:0 mol HCl

0:742 kg H2O¼ 16:2mHCl

- 174 -

- Chapter 14 -

48. First calculate the g KNO3 in the solution.

The conversion is:mgKþ

mL���! g Kþ

mL���! g KNO3

mL���! g KNO3

5:5 mg Kþ

mL

� �1 g

1000 mg

� �101:1 g KNO3

39:10 g Kþ

� �450 mLð Þ ¼ 6:4 g KNO3

Now calculate the mol KNO3 and the molarity.

6:4 g KNO3ð Þ 1 mol

101:1 g

� �¼ 0:063 mol KNO3

0:063 mol KNO3

0:450 L¼ 0:14M

49. 16 fl: oz witch hazelð Þ 1 qt

32 oz

� �946:1 mL

qt

� �14 mL ethyl alcohol

100 mLwitch hazel

� �¼ 66 mL ethyl alcohol

50. Verification of Kb for water

Dtb ¼ mKb Dtb ¼ 101:62�C� 100�C ¼ 1:62�C Kb ¼ Dtbm

First calculate the molality of the solution.

m ¼ 16:10 g C2H6O2

62:07 g=molð Þ 0:0820 kg H2Oð Þ ¼3:16 mol C2H6O2

kg H2O

Kb ¼ Dtbm¼ 1:62�C

3:16 mol=kg H2O¼ 0:513�C kg H2O

mol

51. (a) 500:0 mL solutionð Þ 0:90 g NaCl

100:mL solution

� �¼ 4:5 g NaCl

(b)4:5 g NaCl

xmL

� �100ð Þ ¼ 9:0% x ¼ volume of 9:0% solution

x ¼ 4:5 g NaCl

9:0%¼ 50:mL 4:5 g NaCl in solutionð Þ

500:mL� 50:mL ¼ 450:mLH2Omust evaporate

52. From Figure 14.4, the solubility of KNO3 in H2O at 20�C is 32 g per 100. g H2O.

50:0 g KNO3ð Þ 100: g H2O

32:0 g KNO3

� �¼ 156 g H2O to produce a saturated solution:

175 g H2O� 156 g H2O ¼ 19 g H2Omust be evaporated:

53. 150 mL alcoholð Þ 100:mL solution

70:0 mL alcohol

� �¼ 210 mL solution

- 175 -

- Chapter 14 -

54. (a) 1:00 L solutionð Þ 1000 mL solution

L solution

� �1:21 g

mL

� �35:0 g HNO3

100: g solution

� �¼ 424 g HNO3

(b) 500: g HNO3ð Þ 1000 mL solution

424 g HNO3

� �1:00 L

1000 mL

� �¼ 1:18 L solution

55. Molarity ¼ M ¼ mol solute

L solution

85 g H3PO4

100 g solution

� �1:7 g solution

mL solution

� �1000 mL

L

� �1 mol H3PO4

97:99 g

� �¼ 15M H3PO4

56. First calculate the molarity of the solution

80:0 g H2SO4

500:mL

� �1000 mL

L

� �1 mol

98:09 g

� �¼ 1:63M H2SO4

M1V1 ¼ M2V2

1:63Mð Þ 500:mLð Þ ¼ 0:10Mð Þ V2ð Þ

V2 ¼ 1:63Mð Þ 500:mLð Þ0:10M

¼ 8:2� 103mL ¼ 8:2 L

57. 4:0 galð Þ 4 qt

1 gal

� �1 L

1:057 qt

� �5:25 mol HOCH2CH2OH

L

� �62:07 g

1 mol HOCH2CH2OH

� �

¼ 4:9� 103 g HOCH2CH2OH

58. Mgþ 2 HCl���! MgCl2 þ H2 gð Þ(a) mL HCl���! mol HCl���! mol H2

200:0 mLHClð Þ 3:00 mol

1000 mL

� �1 mol H2

2 mol HCl

� �¼ 0:300 mol H2

(b) PV ¼ nRT

P ¼ 720 torrð Þ 1 atm

760 torr

� �¼ 0:95 atm

T ¼ 27�C ¼ 300:K

n ¼ 0:300 mol

V ¼ nRT

P¼ 0:300 molð Þ 0:0821 L atm=mol Kð Þ 300:Kð Þ

0:95 atm¼ 7:8 L H2

59. Mg OHð Þ2 þ 2 HCl���! MgCl2 þ 2 H2O

Al OHð Þ3 þ 3 HCl���! AlCl3 þ 3 H2O

Calculate the moles of HCl neutralized by each base.

- 176 -

- Chapter 14 -

1:20 gMg OHð Þ2� � 1 mol

58:33 g

� �2 mol HCl

1 mol Mg OHð Þ2

� �¼ 0:0411 mol HCl

1:00 g Al OHð Þ3� � 1 mol

78:00 g

� �3 mol HCl

1 mol Al OHð Þ3

� �¼ 0:0385 mol HCl

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) With equal masses of CH3OH and C2H5OH, the substance with the lower molar mass will repre-

sent more moles of solute in solution. Therefore, the CH3OH will be more effective than C2H5OH

as an antifreeze.

(b) 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.

1000 mL solutionð Þ 1:29 g

mL

� �¼ 1:29� 103 g solution

1:29� 103 g solution� � 38 g H2SO4

100 g solution

� �¼ 4:9� 102 g H2SO4

1:29� 103 g solution � 4:9� 102 g H2SO4 ¼ 8:0� 102 g H2O in the solution

m ¼ 490 g H2SO4

8:0� 102 g H2O

� �1000 g

kg

� �1 mol

98:09 g

� �¼ 6:2mH2SO4

M ¼ 4:9� 102 g H2SO4

L

� �1 mol

98:09 g

� �¼ 5:0M H2SO4

62. Freezing point depression is 5.4�C(a) Dtf ¼ mKf

m ¼ DtfKf

¼ 5:4�C1:86�C kg solvent=mol solute

¼ 2:9m

(b) Kb for H2Oð Þ ¼ 0:512�C kg solvent

mol solute¼ 0:512�C

m

Dtb ¼ mKb ¼ 2:9mð Þ 0:512�Cm

� �¼ 1:5�C

Boiling point ¼ 100�Cþ 1:5�C ¼ 101:5�C

63. Freezing point depression ¼ 0:372�C Kf ¼ 1:86�Cm

Dtf ¼ mKf

m ¼ 0:372�C1:86�C=m

¼ 0:200m

- 177 -

- Chapter 14 -

6:20 g C2H6O2ð Þ 1 mol

62:07 g

� �¼ 0:100 mol C2H6O2

0:100 mol C2H6O2ð Þ 1 kg H2O

0:200 mol C2H6O2

� �1000 g H2O

kg H2O

� �¼ 500: g H2O

64. (a) Freezing point depression¼ 20.0�C

12:0 L H2O1000 mL

L

� �1:00 g

mL

� �¼ 1:20� 104 g H2O

Dtf ¼ mKf

m ¼ 20:0�C1:86�C=m

¼ 10:8m

1:20� 104 g H2O� � 10:8 mol C2H6O2

1000 g H2O

� �62:07 g

mol

� �¼ 8:04� 103 g C2H6O2

(b) 8:04� 103 g C2H6O2

� � 1:00 mL

1:11 g

� �¼ 7:24 � 103 mL C2H6O2

(c) �F ¼ 1:8 �Cð Þ þ 32 ¼ 1:8 �20:0ð Þ þ 32 ¼ �4:0�F

65. HNO3 þ NaHCO3 ���! NaNO3 þ H2Oþ CO2

First calculate the grams of NaHCO3 in the sample.

mL HNO3 ���! LHNO3 ���! mol HNO3 ���! mol NaHCO3 ���! g NaHCO3

150 mLHNO3ð Þ 1 L

1000 mL

� �0:055 mol

L

� �1 mol NaHCO3

1 mol HNO3

� �84:01 g

mol

� �¼ 0:69 g NaHCO3 in the sample

0:69 g NaHCO3

1:48 g sample

� �100ð Þ ¼ 47%NaHCO3

66. (a) Dilution problem:M1V1 ¼ M2V2

1:5Mð Þ 8:4 Lð Þ ¼ 17:8Mð Þ V2ð Þ

V2 ¼ 1:5Mð Þ 8:4 Lð Þ17:8M

¼ 0:71 L

0.71 L of 17.8M H2SO4 must be added (assume volumes are additive)

(b)17:8 mol

1000:mL

� �1:00 mLð Þ ¼ 0:0178 mol

(c)1:5 mol

1000:mL

� �1:00 mLð Þ ¼ 0:0015 mol H2SO4 in each mL

67. moles HNO3 total¼moles HNO3 from 3.00 Mþmoles HNO3 from 12.0 M

MTVT ¼ M3:00 MV3:00 M þM12:0 MV12:0 M

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- Chapter 14 -

Assume preparation of 1000. mL of 6 M solution

Let y¼ volume of 3.00 M solution; volume of 12.0 M¼ 1000. mL � y

6000. mL¼ 3.00 y mLþ 12,000 mL � 12.0 y

6000:mL ¼ 9:00 y y ¼ 6000:mL

9:00¼ 667 mL 3M

1000:mL� 667 mL ¼ 333 mL 12M

Mix together 667 mL 3.00 M HNO and 333 mL of 12.03 M HNO3

68. HBrþ NaOH���! NaBrþ H2O

First calculate the molarity of the diluted HBr solution.

The reaction is 1 mol HBr to 1 mol NaOH, so

MAVA ¼ MBVB

MAð Þ 100:0 mLð Þ ¼ 0:37Mð Þ 88:4 mLð Þ

MA ¼ 0:37Mð Þ 88:4 mLð Þ100:00 mL

¼ 0:33M HBr diluted solutionð Þ

Now calculate the molarity of the HBr before dilution.

M1V1 ¼ M2V2

M1ð Þ 20:0 mLð Þ ¼ 0:33Mð Þ 240:mLð Þ

M1 ¼ 0:33Mð Þ 240:mLð Þ20:0 mL

¼ 4:0M HBr original solutionð Þ

69. Ba NO3ð Þ2 þ 2 KOH���! Ba OHð Þ2 þ 2 KNO3

This is a limiting reactant problem. First calculate the moles of each reactant and determine the limiting

reactant.

M � L ¼ moles

L

� �Lð Þ ¼ moles

0:642 mol

L

� �0:0805 Lð Þ ¼ 0:0517 mol Ba NO3ð Þ2

0:743 mol

L

� �0:0445 Lð Þ ¼ 0:0331 mol KOH

According to the equation, twice as many moles of KOH as Ba(NO3)2 are needed, so KOH is the limiting

reactant.

70. (a)0:25 mol

L

� �0:0458 Lð Þ ¼ 0:011 mol Li2CO3

(b)0:25 mol

L

� �0:75 Lð Þ 73:89 g

mol

� �¼ 14 g Li2CO3

- 179 -

- Chapter 14 -

(6.00 M )(1000. mL)¼ (3.00M)(y)þ (12.0 )(1000. mL � y)M

Mto get 1000. mL of 6.00 3HNO .

(c) 6:0 g Li2CO3ð Þ 1 mol

73:89 g

� �1000:mL

0:25 mol

� �¼ 3:2� 102 mL solution

(d) Assume 1000. mL solution

1:22 g

mL

� �1000:mLð Þ ¼ 1220 g solution

0:25 mol

L

� �73:89 g Li2CO3

mol

� �¼ 18 g Li2CO3 per L solution

% ¼ g solute

g solution

� �100ð Þ ¼ 18 g

1220 g

� �100ð Þ ¼ 1:5% (mass percent)

71. The balanced equation is

2 HCl aqð Þ þ Na2SO3 aqð Þ ���! 2 NaCl aqð Þ þ H2O lð Þ þ SO2 gð Þ

125 mLHClð Þ 2:50 mol

1000 mL

� �1 mol SO2

2 mol HCl

� �¼ 0:156 mol SO2

75 mLNa2 SO3ð Þ 1:75 mol

1000 mL

� �1 mol SO2

1 mol Na2SO3

� �¼ 0:131 mol SO2

Na2SO3 is the limiting reactant 0.131 mol of SO2 gas will be produced

The gas is at non-standard conditions, so use PV¼ nRT to find the liters of SO2.

V ¼ nRT

PV ¼ 0:131 molð Þ 0:0821 L atm=mol Kð Þ 295 Kð Þ

775 torrð Þ 1 atm=760 torrð Þ ¼ 3:11 L SO2

72. mass of solute¼mass of container & solute � mass of container � mass of water

mass of water¼ (5.549 moles) (18.02 g=mol)¼ 100.0 g

mass of solute¼ 563 g � 375 g � 100.0 g¼ 88 g

solubility in water¼ g solute=100 g H2O at 20� CUsing this data, solubility is 88 g solute=100.0 g water¼NaNO3 (see Table 14.3).

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- Chapter 14 -