CHAPTER 14 SOLUTIONS - Chemeketa Community Collegefaculty.chemeketa.edu/lemme/CH 121/solutions/Hein9thCh14.pdf · CHAPTER 14 SOLUTIONS SOLUTIONS TO REVIEW QUESTIONS 1. These diagrams

CHAPTER 14

SOLUTIONS

SOLUTIONS TO REVIEW QUESTIONS

1.

These diagrams are intended to illustrate the orientation of the water molecules about theions, not the number of water molecules.

2. From Table 14.2, approximately 4.5 g of NaF would be soluble in 100 g of water at 50 C.

3. From Figure 14.4, solubilities in water at 25 C are:(a) KCl(b)(c)

4. Potassium fluoride has a relatively high solubility when compared to lithium or sodiumfluoride. For lithium and sodium halides, the order of solubility (in order of increasingsolubilities) is:

For potassium halides, the order of increasing solubilities is:

5. (a) at 60 C, 25 g (c) at 80 C, 30 g(b) HCl at 20 C, 72 g (d) at 0 C, 14 g

6. KNO3

°KNO3°°Li2SO4°KClO3

Cl- Br - F- I-

F- Cl- Br - I-

39 g>100 g H2OKNO3

9 g>100 g H2OKClO3

35 g>100 g H2O°

°

Na+ Cl–HH

H

O

H

H

O

H

H H

H

O

H

H

O

H

H

O

HH

O

H

O

O

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HEINS14-180-206v3.qxd 12/22/06 5:05 AM Page 180

7. A one molal solution in camphor will show a greater freezing point depression than a 2 molal solution in benzene.

8. Cube 1 cm 0.01 cm

Volume

Number 1 cm cube 1

Area of face

Total surface area

9. From Figure 14.4, the solubility of in water is

approximately at 30 C, at 40 C. Therefore, the solutionof of water would be saturated at 10 C, 20 C, and 30 C. The solution wouldbe unsaturated at 40 C and 50 C.

10. The dissolving process involves solvent molecules attaching to the solute ions ormolecules. This rate decreases as more of the solvent molecules are already attached tosolute molecules. As the solution becomes more saturated, the number of unused solventmolecules decreases. Also, the rate of recrystallization increases as the concentration ofdissolved solute increases.

11. A supersaturated solution of may be prepared in the following sequence:

(a) Determine the mass of necessary to saturate a specific amount of waterat room temperature.

(b) Place a bit more in the water than the amount needed to saturate thesolution.

(c) Heat the solution until all the solid dissolves.

(d) Cover the container and allow it to cool undisturbed. The cooled solution, whichshould contain no solid is supersaturated.

To test for supersaturation, add one small crystal of to the solution. Immediatecrystallization is an indication that the solution was supersaturated.

NaC2H3O2

NaC2H3O2,

NaC2H3O2

NaC2H3O2

NaC2H3O2

°°°°°63 g>150 g

°H2O46 g>100 g°H2O42 g>100 g

NH4Cl63 g NH4Cl

150 g H2O=

42 g NH4Cl

100 g H2O

11 * 106 cubes216 faces>cube211 * 10-4 cm2>face2 = 6 * 102 cm2

6 * 102 cm26 cm2

1 * 10-4 cm21 cm2

106 311 cm32>11 * 10-6 cm32 = 106 cubes4>1 * 10-6 cm31 cm3

¢tf = a2 mol solute

kg benzeneb a 5.1°C kg benzene

mol soluteb = 10.2°C (freezing point depression)

¢tf = a1 mol solute

kg camphorb a 40°C kg camphor

mol soluteb = 40°C (freezing point depression)

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12. Because the concentration of water is greater in the thistle tube, the water will flowthrough the membrane from the thistle tube to the urea solution in the beaker. Thesolution level in the thistle tube will fall.

13. A true solution is one in which the size of the particles of solute are between True solutions are homogeneous and the ratio of solute to solvent can be varied. They canbe colored or colorless but are transparent. The solute remains distributed evenly in thesolution, it will not settle out.

14. The two components of a solution are the solute and the solvent. The solute is dissolvedinto the solvent or is the least abundant component. The solvent is the dissolving agent orthe most abundant component.

15. It is not always apparent which component in a solution is the solute. For example, in asolution composed of equal volumes of two liquids, the designation of solute and solventwould 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 individualparticles are so small that the force of molecular collisions is large compared to the forceof gravity.

17. Yes. It is possible to have one solid dissolved in another solid. Metal alloys are of thistype. Atoms of one metal are dissolved among atoms of another metal.

18. Orange. The three reference solutions are KCl, and They all containions in solution. The different colors must result from the different anions dissolved in

the solutions: (purple) and (orange). Therefore, it is predictable that theion present in an aqueous solution of will impart an orange color to

the solution.

19. Hexane and benzene are both nonpolar molecules. There are no strong intermolecularforces 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, sosodium chloride is insoluble in hexane.

20. Coca Cola has two main characteristics, taste and fizz (carbonation). The carbonation isdue to a dissolved gas, carbon dioxide. Since dissolved gases become less soluble astemperature 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 gaseoussubstances and does not have a fixed composition.

Na2Cr2O7Cr2O7

2-Cr2 O7

2-MnO4

-K+

K2Cr2O7.KMnO4,

0.1 - 1nm.

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- Chapter 14 -


22. A teaspoon of sugar would definitely dissolve more rapidly in 200 mL of hot coffee thanin 200 mL of iced tea. The much greater thermal agitation of the hot coffee will helpbreak the sugar molecules away from the undissolved solid and disperse them throughoutthe solution. Other solutes in coffee and tea would have no significant effect. Thetemperature difference is the critical factor.

23. The solubility of gases in liquids is greatly affected by the pressure of a gas above theliquid. The greater the pressure, the more soluble the gas. There is very little effect ofpressure regarding the dissolution of solids in liquids.

24. For a given mass of solute, the smaller the particles, the faster the dissolution of thesolute. This is due to the smaller particles having a greater surface area exposed to thedissolving action of the solvent.

25. In a saturated solution, the net rate of dissolution is zero. There is no further increase inthe amount of dissolved solute, even though undissolved solute is continuouslydissolving, because dissolved solute is continuously coming out of solution, crystallizingat a rate equal to the rate of dissolving.

26. When crystals of and NaCl are mixed, the contact between the individual ions isnot intimate enough for the double displacement reaction to occur. When solutions of thetwo chemicals are mixed, the ions are free to move and come into intimate contact witheach 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 icyroads 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 per liter ofsolution.

29. The two solutions contain the same number of chloride ions. One liter of 1 M NaClcontains 1 mole of NaCl, therefore 1 mole of chloride ions. 0.5 liter of contains 0.5 mol of and 1 mole of chloride ions.

30. The champagne would spray out of the bottle all over the place. The rise in temperatureand the increase in kinetic energy of the molecules by shaking both act to decrease thesolubility of gas within the liquid. The pressure inside the bottle would be great. As thecork is popped, much of the gas would escape from the liquid very rapidly, causing the champagne to spray.

10.5 L2¢1 mol MgCl2

L≤ ¢ 2 mol Cl-

1 mol MgCl2≤ = 1 mol Cl-

MgCl2

1 M MgCl2

HNO3

AgNO3

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31. The number of grams of NaCl in 750 mL of 5.0 molar solution is

Dissolve the 220 g of NaCl in a minimum amount of water, then dilute the resultingsolution to a final volume of 750 mL (0.75 L).

32. A semipermeable membrane will allow water molecules to pass through in bothdirections. If it has pure water on one side and 10% sugar solutions on the other side ofthe 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 ofthe membrane. The net result is more water molecules pass from the pure water to thesugar solution. Osmotic pressure effect.

33. The urea solution will have the greater osmotic pressure because it has 1.67 mol solute kg while the glucose solution has only 0.83 mol solute kg

34. A lettuce leaf immersed in salad dressing containing salt and vinegar will become limpand wilted as a result of osmosis. As the water inside the leaf flows into the dressingwhere 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 highfluid content and crisp leaf.

35. The concentration of solutes (such as salts) is higher in seawater than in body fluids. Thesurvivors who drank seawater suffered further dehydration from the transfer of water byosmosis from body tissues to the intestinal tract.

36. Ranking of the specified bases in descending order of the volume of each required toreact with 1 liter of 1 M HCl. The volume of each required to yield 1 mole of ion isshown.

(a) 1 M NaOH 1 liter

(b) 0.83 liter

(c) 2 M KOH 0.50 liter

(d) 0.33 liter

37. The boiling point of a liquid or solution is the temperature at which the vapor pressureof the liquid equals the pressure of the atmosphere. Since a solution containing anonvolatile solute has a lower vapor pressure than the pure solvent, the boiling point ofthe solution must be at a higher temperature than for the pure solvent. At the higherboiling temperature the vapor pressure of the solution equals the atmospheric pressure.

1.5 M Ca(OH)2

0.6 M Ba(OH)2

OH-

H2O.>H2O,>

10.75 L2a5.0 mol NaCl

Lb a58.44 g

1 molb = 2.2 * 102 g NaCl

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38. The freezing point is the temperature at which a liquid changes to a solid. The vaporpressure of a solution is lower than that of a pure solvent. Therefore, the vapor pressurecurve of the solution intersects the vapor pressure curve of the pure solvent, at atemperature lower than the freezing point of the pure solvent. (See Figure 14.8b) At thispoint of intersection, the vapor pressure of the solution equals the vapor pressure of thepure solvent.

39. Water and ice are different phases of the same substance in equilibrium at the freezingpoint of water, 0 C. The presence of the methanol lowers the vapor pressure and hencethe freezing point of water. If the ratio of alcohol to water is high, the freezing point canbe 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) ofsucrose.

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

(c) 20.0 g (0.625 mol) of methyl alcohol is more effective than 20.0 g (0.435 mol) ofethyl alcohol.

41. Both molarity and molality describe the concentration of a solution. However, molarity isthe ratio of moles of solute per liter of solution, and molality is the ratio of moles ofsolute per kilogram of solvent.

42. of solution. Thevolume of the 5 molal solution will be larger than 1 liter Thevolume of the 5 molar solution is exactly 1 L to produce1 L of solution). The molarity of a 5 molal solution is therefore, less than 5 molar.

(5 mol NaCl + sufficient H2O(1 L H2O + 5 mol NaCl).

5 molar NaCl = 5 mol NaCl>L5 molal NaCl = 5 mol NaCl>kg H2O;

°

°

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CHAPTER 14

SOLUTIONS TO EXERCISES

1. Reasonably soluble: (a) KOH (b) (d) (e)

Insoluble: (c) ZnS

2. Reasonably soluble: (c) (d)

Insoluble: (a) (b) (e)

3. Mass percent calculations

(a)

(b) solution

(c)

solution

(d)

solution

4. Mass percent calculations

(a)

a 25.0 g

150.0 gb11002 = 16.7% NaNO3

25.0 g NaNO3 in 125.0 g H2O = 150.0

¢60.0 g

655 g≤11002 = 9.16% NaOH

60.0 g NaOH + 595 g H2O = 655 g

(33.0 mol H2O)a18.02 g

1 molb = 595 g H2O

(1.50 mol NaOH)a40.00 g

1 molb = 60.0 g NaOH

¢ 15 g

140. g≤11002 = 11% NH4C2H3O2

15 g NH4C2H3O2 + 125 g H2O = 140. g

(0.20 mol NH4C2H3O2)a 77.09 g

1 molb = 15 g NH4C2H3O2

¢2.50 g

12.5 g≤11002 = 20.0% Na3PO4

2.50 g Na3PO4 + 10.0 g H2O = 12.5 g

a 15.0 g

115.0 gb11002 = 13.0% KCl

15.0 g KCl + 100.0 g H2O = 115.0 g solution

BaSO4MgCO3PbI2

Fe(NO3)3CaCl2

Na2CrO4AgC2H3O2NiCl2

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(b) solution

(c)

solution

(d)

5. solution

6. solution contains 10.0 g per 100. g solution

solution

7. (a)

(b) (25 g solution)a100. g solution - 7.5 g CaSO4

100. g solutionb = 23 g solvent

(25 g solution)a 7.5 g CaSO4

100. g solutionb = 1.9 g CaSO4

(25.0 g NaCl)¢100. g solution

10.0 g NaCl≤ = 250. g

NaClA 10.0% NaCl

(25.2 g AgNO3)a100. g solution

15.5 g AgNO3b = 163 g solution

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

a 118 g

1.43 * 103 gb11002 = 8.25% H2SO4

118 g H2SO4 + 1.31 * 103 g H2O = 1.43 * 103 g solution

(72.5 mol H2O)a18.02 g

1 molb = 1.31 * 103 g H2O

(1.20 mol H2SO4)a98.09 g

1 molb = 118 g H2SO4

a150 g

380 gb11002 = 39% K2CrO4

150 g K2CrO4 + 225 g H2O = 380 g

(0.75 mol K2CrO4)a 194.2 g

1 molb = 150 g K2CrO4

¢1.25 g

36.3 g≤11002 = 3.44% CaCl2

1.25 g CaCl2 in 35.0 g H2O = 36.3 g

- Chapter 14 -

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8. (a)

(b)

9. Mass/volume percent.

10. Mass/volume percent.

11. Volume percent.

12. Volume percent.

13. Molarity problems

(a)

(b)

(c)

(d) ¢75 g CuSO4• 5 H2O

1.0 L≤ a 1 mol

249.7 gb = 0.30 M CuSO4

¢ 35.0 g

1.25 L≤ a 1 mol

82.03 gb = 0.341 M NaC2H3O2

a1.75 mol

0.75 Lb = 2.3 M KBr

a0.25 mol

75.0 mLb a1000 mL

1 Lb = 3.3 M

aM =mol

Lb

¢ 2.0 mL C6H14

9.0 mL solution≤11002 = 22% C6H14

¢10.0 mL CH3OH

40.0 mL solution≤11002 = 25.0% CH3OH

a 4.20 g NaCl

12.5 mL solutionb11002 = 33.6% NaCl

¢ 22.0 g CH3OH

100. mL solution≤11002 = 22.0% CH3OH

(75 g solution)a100. g solution - 12.0 g BaCl2100. g solution

b = 66 g solvent

(75 g solution)a 12.0 g BaCl2100. g solution

b = 9.0 g BaCl2

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- Chapter 14 -


14. Molarity problems

(a)

(b)

(c)

(d)

15.

(a)

(b)

(c)

16.

(a)

(b)

(c)

17. (a)

(b)

(c) 1250 mL2a 1 L

1000 mLb a 16 mol HNO3

Lb a 63.02 g

1 molb = 250 g HNO3

175.2 mL2a 1 L

1000 mLb a 0.050 mol HC2H3O2

Lb a60.05 g

1 molb = 0.226 g HC2H3O2

12.5 L2a0.75 mol K2CrO4

Lb a194.2 g

1 molb = 360 g K2CrO4

1175 mL2a 1 L

1000 mLb a 0.50 mol LiBr

Lb = 0.088 mol LiBr

110.0 mL2a 1 L

1000 mLb a 0.75 mol NaClO3

Lb = 7.5 * 10-3 mol NaClO3

10.75 L2a1.50 mol HNO3

Lb = 1.1 mol HNO3

Molarity =mol solute

L solution or mol solute = 1L solution21Molarity2

1125 mL2a 1 L

1000 mLb a 0.35 mol K3PO4

Lb = 0.044 mol K3PO4

125.0 mL2a 1 L

1000 mLb a 0.0015 mol BaCl2

Lb = 3.8 * 10-5 mol BaCl2

11.5 L2a1.20 mol H2SO4

Lb = 1.8 mol H2SO4

Molarity =mol solute

L solution or mol solute = 1L solution21Molarity2

¢125 g MgSO4• 7 H2O

2.50 L≤ a 1 mol

246.5 gb = 0.203 M MgSO4

¢ 275 g

775 mL≤ a 1 mol

180.2 gb a 1000 mL

1 Lb = 1.97 M C6H12O6

a2.25 mol

1.50 Lb = 1.50 M CaCl2

a0.50 mol

125 mLb a1000 mL

1 Lb = 4.0 M

aM =mol

Lb

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18. (a)

(b)

(c)

19. (a)

(b)

20. (a)

(b)

21. Dilution problem

(a)

(b)

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

Total mol = 0.15 mol

(75 mL)a 1 L

1000 mLb a 1.5 mol

1 Lb = 0.11 mol HNO3

(75 mL)a 1 L

1000 mLb a0.50 mol

1 Lb = 0.038 mol HNO3

HNO3

M2 =1250 mL210.25 M2

1.00 * 103 mL= 0.063 M

1250 mL210.25 M2 = 11.00 * 103 mL21M22 M1 = 0.25 M M2 = M2

V1 = 250 mL V2 = (250 mL + 750 mL) = 1.00 * 103 mL

M2 =1125 mL215.0 M2

900. mL= 0.694 M

1125 mL215.0 M2 = 1900. mL21M22 M1 = 5.0 M M2 = M2

V1 = 125 mL V2 = (125 mL + 775 mL) = 900. mL

V1M1 = V2M2

125.2 g) NH4Cla 1 mol

53.49 gb a 1 L

0.250 mol NH4Clb a1000 mL

Lb = 1.88 * 103 mL

10.85 mol NH4Cl2a 1 L

0.250 mol NH4Clb a 1000 mL

1 Lb = 3.4 * 103 mL

135.5 g H3PO42a 1 mol

97.99 gb

a 1 L

0.750 mol H3PO4 b a1000 mL

1 Lb = 483 mL

10.15 mol H3PO42a 1 L

0.750 mol H3PO4 b a 1000 mL

1 Lb = 2.0 * 102 mL

1120 mL2¢ 1 L

1000 mL≤ a 0.025 mol Fe2(SO4)3

Lb a 399.9 g

1 molb = 1.2 g Fe2(SO4)3

127.5 mL2¢ 1 L

1000 mL≤ a 1.5 mol KMnO4

Lb a 158.0 g

1 molb = 6.52 g KMnO4

11.20 L2a18 mol H2SO4

Lb a 98.09 g

1 molb = 2.1 * 103 g H2SO4

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- Chapter 14 -


22. Dilution problem

(a)

(b)

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

23.

(a)

(b)

V1 =1250 mL210.50 M2

16 M= 7.8 mL 16 M HNO3

1V12116 M2 = 1250 mL210.50 M2V1 =

1750 mL213.0 M215 M

= 150 mL 15 M H3PO4

1V12115 M2 = 1750 mL213.0 M2V1M1 = V2M2

0.250 mol

0.0750 L= 0.333 M

Total volume = 50.0 mL + 25.0 mL = 75.0 mL = 0.0750 L Total mol = 0.0250 mol

(25.0 mL)a 1 L

1000 mLb a 0.500 mol

1 Lb = 0.0125 mol HCl

(50.0 mL)a 1 L

1000 mLb a 0.250 mol

1 Lb = 0.0125 mol HCl

HCl

M2 =1350 mL210.10 M2

5.0 * 102 mL= 0.070 M

1350 mL210.10 M2 = 15.0 * 102 mL21M22 M1 = 0.10 M M2 = M2

V1 = 350 mL V2 = (350 mL + 150 mL) = 5.0 * 102 mL

M2 =1175 mL213.0 M2

450. mL= 1.2 M

1175 mL213.0 M2 = 1450. mL21M22 M1 = 3.0 M M2 = M2

V1 = 175 mL V2 = (175 mL + 275 mL) = 450. mL

V1M1 = V2M2

0.150 mol

0.150 L= 1.00 M

Total volume = 75 mL + 75 mL = 150. mL = 0.150 L

- Chapter 14 -

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24.

(a)

(b)

25.

(a) Final volume after mixing

(b)

26.

(a) Final volume after mixing

(b)

1.22 mol HCl

0.290 L= 4.21 M HCl

Final volume = 175 mL + 115 mL = 290. mL = 0.290 L

Total moles = 0.53 mol + 0.69 mol = 1.22 mol HCll

1115 mL2a 1 L

1000 mLb ¢ 6.0 mol HCl

L≤ = 0.69 mol HCl

0.53 mol HCl

0.425 L= 1.2 M HCl

175 mL + 250 mL = 425 mL = 0.425 L

10.175 L2a3.0 mol HCl

Lb = 0.53 mol HCl

1.01 mol HC2H3O2

0.300 L= 3.37 M HC2H3O2

Final volume = 125 mL + 175 mL = 300. mL = 0.300 L

Total moles = 0.75 mol + 0.26 mol = 1.01 mol HC2H3O2

1175 mL2a 1 L

1000 mLb ¢ 1.5 mol HC2H3O2

L≤ = 0.26 mol HC2H3O2

0.75 mol HC2H3O2

0.650 L= 1.2 M HC2H3O2

125 mL + 525 mL = 650. mL = 0.650 L

10.125 L2a6.0 mol HC2H3O2

Lb = 0.75 mol HC2H3O2

V1 =175 mL211.0 M2

15 M= 5.0 mL 15 M NH3

1V12115 M2 = 175 mL211.0 M2V1 =

1225 mL212.0 M218 M

= 25 mL 18 M H2SO4

1V12118 M2 = 1225 mL212.0 M2V1M1 = V2M2

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- Chapter 14 -


27.

(a)

(b)

(c)

(d)

(e)

(f)

28.

(a)

(b)

(c)

12.50 L H2SO42¢0.125 mol

1 L≤ a 2 mol NaOH

1 mol H2SO4b = 0.625 mol NaOH

L H2SO4 ¡ mol H2SO4 ¡ mol NaOH

10.025 mol NaOH2¢ 2 mol H2O

2 mol NaOH≤ = 0.025 mol H2O

13.6 mol H2SO42¢1 mol Na2SO4

1 mol H2SO4≤ = 3.6 mol Na2SO4

2 NaOH(aq) + H2SO4(aq) ¡ Na2SO4(aq) + 2 H2O(l),

M =mol

L M = a 0.15 mol Ca(NO3)2

0.0500 Lb = 3.0 M Ca(NO3)2

150.0 mL Na3PO4)¢ 2.0 mol

1000 mL≤ a3 mol Ca(NO3)2

2 mol Na3PO4b = 0.15 mol Ca(NO3)2

Find mol Ca(NO3)2 mL Na3PO4 ¡ mol Na3PO4 ¡ mol Ca(NO3)2

115.0 mL Ca(NO3)22¢ 0.50 mol

1000 mL≤ a 2 mol Na3PO4

3 mol Ca(NO3)2b a 1000 mL

0.25 molb = 20. mL Na3PO4

mL Ca(NO3)2 ¡ mol Ca(NO3)2 ¡ mol Na3PO4 ¡ mL Na3PO4

1125 mL Ca(NO3)22¢ 0.500 mol

1000 mL≤ a1 mol Ca3(PO4)2

3 mol Ca(NO3)2b a310.18 g

molb = 6.46 g Ca3(PO4)2

mL Ca(NO3)2 ¡ mol Ca(NO3)2 ¡ mol Ca3(PO4)2 ¡ g Ca3(PO4)2

11.45 L Ca(NO3)22¢0.225 mol

L≤ a 2 mol Na3PO4

3 mol Ca(NO3)2b = 0.218 mol Na3PO4

L Ca(NO3)2 ¡ mol Ca(NO3)2 ¡ mol Na3PO4

10.75 mol Ca(NO3)22¢ 6 mol NaNO3

3 mol Ca(NO3)2≤ = 1.5 mol NaNO3

12.7 mol Na3PO42¢ 1 mol Ca3 (PO4)2

2 mol Na3PO4≤ = 1.4 mol Ca3(PO4)2

3 Ca(NO3)2(aq) + 2 Na3PO4(aq) ¡ Ca3(PO4)2(s) + 6 NaNO3(aq),

- Chapter 14 -

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(d)

(e)

(f)

29.

(a)

(b)

(c)

(d)

(e)

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

115.0 mL HCl)a0.750 mol

1000 mLb a 5 mol Cl2

16 mol HClb a22.4 L

molb = 0.0788 L Cl2

1125 mL HCl2¢ 2.5 mol

1000 mL≤ a 5 mol Cl2

16 mol HClb a 22.4 L

molb = 2.2 L Cl2

mL HCl ¡ mol HCl ¡ mol Cl2 ¡ L Cl2 (gas at STP)

M = a0.0312 mol HCl

0.02220 Lb = 1.41 M HCl

(15.60 mL KMnO4)a0.250 mol

1000 mLb a 16 mol HCl

2 mol KMnO4b = 0.0312 mol HCl

1125 mL KCl2¢0.525 mol

1000 mL≤ a 16 mol HCl

2 mol KClb a 1000 mL

2.50 molb = 210. mL HCl

11.85 mol MnCl22a2 mol KMnO4

2 mol MnCl2b a 1 L

0.150 mol KMnO4b = 12.3 L KMnO4

115.0 mL HCl2¢0.250 mol

1000 mL≤ a 8 mol H2O

16 mol HClb = 1.88 * 10-3 mol H2O

2 KMnO4(aq) + 16 HCl(aq) ¡ 2 MnCl2(aq) + 5 Cl2(g) + 8 H2O(l) + 2 KCl(aq)

M =mol

L M = a8.93 * 10-3 mol NaOH

0.04820 Lb = 0.185 M NaOH

135.72 mL H2SO4)¢0.125 mol

1000 mL≤ a 2 mol NaOH

1 mol H2SO4b = 8.93 * 10-3 mol NaOH

Find mol NaOH mL H2SO4 ¡ mol H2SO4 ¡ mol NaOH

125.5 mL NaOH2¢0.750 mol

1000 mL≤ a 1 mol H2SO4

2 mol NaOHb a 1000 mL

0.250 molb = 38.25 mL H2SO4

mL NaOH ¡ mol NaOH ¡ mol H2SO4 ¡ mL H2SO4

125 mL NaOH2¢0.050 mol

1000 mL≤ a 1 mol Na2SO4

2 mol NaOHb a 142.05 g

molb = 0.089 g Na2SO4

mL NaOH ¡ mol NaOH ¡ mol Na2SO4 ¡ g Na2SO4

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- Chapter 14 -


HCl is the limiting reactant. 0.0788 L of Cl2 are produced.

30.

(a)

(b)

(c)

(d)

(e)

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

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

31.

(a)

(b) ¢2.50 mol C6H6

250 g C6H14≤ a 1000 g

kgb = ¢10. mol C6H6

kg C6H14≤ = 10. m C6H6

¢14.0 g CH3OH

100. g H2O≤ a 1000 g

kgb a 1 mol

32.04 gb = ¢4.37 mol CH3OH

kg H2O≤ = 4.37 m CH3OH

Molality = m =mol solute

kg solvent

HC2H3O2

125.0 mL HC2H3O2)a0.250 mol

1000 mLb a 1 mol CO2

2 mol HC2H3O2b a22.4 L

molb = 0.0700 L CO2

125.0 mL K2CO3)a 0.350 mol

1000 mLb a 1 mol CO2

1 mol K2CO3b a 22.4 L

molb = 0.196 L CO2

1105 mL of HC2H3O22¢ 1.5 mol

1000 mL≤ a 1 mol CO2

2 mol HC2H3O2b a22.4 L

molb = 1.8 L CO2

mL HCl ¡ mol HCl ¡ mol Cl2 ¡ L Cl2 (gas at STP)

M = a9.25 * 10-3 mol HC2H3O2

0.01015 Lb = 0.911 M HC2H3O2

(18.50 mL K2CO3)a0.250 mol

1000 mLb a2 mol HC2H3O2

1 mol K2CO3b = 9.25 * 10-3 mol HC2H3O2

= 90.2 mL HC2H3O2

175.2 mL K2CO32¢0.750 mol

1000 mL≤ a2 mol HC2H3O2

1 mol K2CO3b a 1000 mL

1.25 mol HC2H3O2b

117.5 mol KC2H3O22a 1 mol K2CO3

2 mol KC2H3O2b a 1 L

0.210 mol K2CO3b = 41.7 L K2CO3

125.0 mL HC2H3O22¢ 0.150 mol

1000 mL≤ a 1 mol H2O

2 mol HC2H3O2b = 1.88 * 10-3 mol H2O

K2CO3(aq) + 2 HC2H3O2(aq) ¡ 2 KC2H3O2(aq) + H2O(l) + CO2(g)

112.0 mL KMnO4)a0.550 mol

1000 mLb a 5 mol Cl2

2 mol KMnO4b a 22.4 L

molb = 0.373 L Cl2

- Chapter 14 -

- 195 -


32.

(a)

(b)

33. (a)

(b)

(c)

34. (a)

(b)

(c)

Freezing point = 0.00°C - 20.0°C = -20.0°C

¢t f = mKf = 110.74 m2a1.86°Cmb = 20.0°C (Decrease in freezing point)

Boiling point = 100.00°C + 5.50°C = 105.50°C

¢t b = mKb = (10.74 m)a 0.512°Cm

b = 5.50°C (Increase in boiling point)

¢100.0 g C2H6O2

150.0 g H2O≤ a 1 mol

62.07 gb a 1000 g

kgb = 10.74 m

Boiling point of solution = 80.1°C + 1.38°C = 81.5°C

¢t b = (0.544 m)a2.53°Cmb = 1.38°C

Kb (for benzene) =2.53°C

m Boiling point of benzene = 80.1°C

Freezing point of solution = 5.5°C - 2.8°C = 2.7°C

¢t f = 10.544 m2a5.1°Cmb = 2.8°C

Kf (for benzene) =5.1°C

m Freezing point of benzene = 5.5°C

¢2.68 g C10H8

38.4 g C6H6≤ ¢ 1 mol

128.2 g C10H8≤ ¢ 1000 g C6H6

kg≤ = 0.544 m

¢ 0.250 g I2

1.0 kg H2O≤ a 1 mol I2

253.8 g I2b = 9.9 * 10-4 m I2

¢1.0 g C6H12O6

1.0 g H2O≤ a1000 g

kgb a 1 mol

180.2 gb = ¢5.5 mol C6H12O6

kg H2O≤ = 5.5 m C6H12O6

Molality = m =mol solute

kg solvent

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- Chapter 14 -


35. acetic acid

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

36.

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

37. Salt (NaCl) is an ionic compound. When it is dissolved in water the sodium andchloride ions separate or dissociate. The polar water molecules are attracted to thepolar sodium and chloride ions and are hydrated (surrounded by water molecules).The sodium and chloride ions are separated from one another and distributedthroughout the water in this way. and are hydrated in aqueous solution: SeeQuestion 1.

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 thesugar becomes hydrated. The water molecules help separate the sugar molecules fromeach other and distribute them throughout the water.

39. Sugar and salt behave differently when dissolved in water because salt is an ioniccompound and sugar is a molecular compound.

Cl-Na+

¢4.80 g unknown

22.0 g H2O≤ a1000 g

kgb ¢ 1 kg H2O

1.34 mol unknown≤ = 163 g>mol

>H2O> m =

2.50°C

1.86°C>m = 1.34 m

¢t f = mKf

¢t f = 2.50°C Kf (for H2O) =1.86°C

m

¢ 8.00 g unknown

60.0 g HC2H3O2≤ a1000 g

kgb ¢ 1 kg HC2H3O2

0.87 mol unknown≤ = 153 g>mol

Conversion: g unknown

g HC2H3O2¡

g unknown

kg HC2H3O2¡

g

mol

>HC2H3O2> m =

3.4°C

3.90°C>m = 0.87 m

¢t f = mKf

¢t f = 16.6°C - 13.2°C = 3.4°C

=3.90°C

mFreezing point of acetic acid is 16.6°C Kf

- Chapter 14 -

- 197 -


40. An isotonic sodium chloride solution has the same osmotic pressure as human bloodplasma. When blood cells are placed in an isotonic solution the osmotic pressure insidethe cells is equal to the osmotic pressure outside the cells so there is no change in theappearance of the blood cells.

41. The crystals give the solution its purple color. The purple streaks are formedbecause the solute has not been evenly distributed throughout that solvent yet. The

has a purple color in solution.

42. The line for slopes upward, because the solubility increases as the temperatureincreases. has the steepest slope of all the compounds given in the diagram. Itexhibits 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. First calculate the g NaOH to neutralize the HCl.

6.0 g NaOH required to neutralizethe acid

Now calculate the grams of 10% NaOH solution that contains 6.0 g NaOH

44.

Therefore,

Calculate the grams NaOH to neutralize HCl

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

x = 96.5 g 10% NaOH solution

9.648 g NaOH

x=

10.0 g NaOH

100.0 g 10.0% NaOH solution

1250.0 g solution2a 1 mol HCl

1036.46 solutionb a1 mol NaOH

1 mol HClb a 40.00 g

molb = 9.648 g NaOH

NaOH + HCl ¡ NaCl + H2O

1.0 m HCl =1 mol HCl

1036.46 g HCl solution

Total mass of solution = 1000 g + 36.46 g = 1036.46 g

1.0 m HCl =1 mol HCl

1 kg H2O=

36.46 g HCl

1000 g H2O

x = 60. g 10% NaOH solution

6.0 g NaOH

x=

10.0 g NaOH

100.0 g 10.0% NaOH solution

10.15 L HCl2a1.0 mol

Lb a1 mol NaOH

1 mol HClb a40.00 g

molb =

NaOH + HCl ¡ NaCl + H2O

KNO3

KNO3

MnO4

-

KMnO4

- 198 -

- Chapter 14 -


45. (a)

(b)

(c)

46.

Empirical mass

(number of empirical formulas per molecular formula)

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

47.

48. First calculate the in the solution.

The conversion is:mg K+

mL¡

g K+

mL¡

g KNO3

mL¡ g KNO3

g KNO3

Since molality =mol HCl

kg H2O=

12.0 mol HCl

0.742 kg H2O= 16.2 m HCl

1180 g solution - 438 g HCl = 742 g H2O (0.742 kg H2O)

11.00 L2a1.18 g solution

mLb a 1000 mL

Lb = 1180 g solution

112.0 mol HCl2a36.46 g

molb = 438 g HCl in 1.00 L solution

C8H4N2.

130 g

64.07 g= 2.0

(C4H2N) = 64.07 g

¢15.4 g C4H2N

kg C6H6≤ ¢ 1 kg C6H6

0.12 mol C4H2N≤ = 128 g>mol = 1.3 * 102 g>mol

m =0.614°C

5.1°C>m = 0.12 m =0.12 mol C4H2N

kg C6H6

¢t f = mKf

¢3.84 g C4H2N

250.0 g C6H6≤ a1000 g

kgb =

15.4 g C4H2N

kg C6H6

Kf =5.1°C

m ¢t f = 0.614°C

¢15.0 g C12H22O11

85.0 g H2O≤ ¢1000 g H2O

1 kg H2O≤ ¢ 1 mol

342.3 g C12H22O11≤ = 0.516 m

m =mol sugar

kg H2O 15% sugar by mass = 15.0 g C12H22O11 + 85.0 g H2O

¢1.6 * 102 g C12H22O11

L≤ a 1 mol

342.3 gb = 0.47 M

11.0 L syrup2a1000 mL

Lb a1.06 g

mLb a 15.0 g sugar

100. g syrupb = 1.6 * 102 g sugar

- Chapter 14 -

- 199 -


Now calculate the mol and the molarity.

49.

Alternate solution:

50. Verification of for water

First calculate the molality of the solution.

51. (a)

(b)

must evaporate500. mL - 50. mL = 450. mL H2O

x =4.5 g NaCl

9.0%= 50. mL (4.5 g NaCl in solution)

a4.5 g NaCl

x mLb11002 = 9.0% x = volume of 9.0% solution

1500.0 mL solution2a 0.90 g NaCl

100. mL solutionb = 4.5 g NaCl

Kb =¢t b

m=

1.62°C

3.16 mol>kg H2O=

0.513°C kg H2O

mol

m =16.10 g C2H6O2

162.07 g>mol210.0820 kg H2O2 =3.16 mol C2H6O2

kg H2O

¢t b = mKb ¢t b = 101.62°C - 100°C = 1.62°C Kb =¢t b

m

Kb

x = 455 g solution

a25.0 g KClx

b = a 5.50 g KCl

100. g solutionb

125.0 g KCl2a100. g solution

5.50 g KClb = 455 g solution

0.063 mol KNO3

0.450 L= 0.14 M

16.4 g KNO32a 1 mol

101.1 gb = 0.063 mol KNO3

KNO3

¢5.5 mg K+

mL≤ a 1 g

1000 mgb ¢101.1 g KNO3

39.10 g K+ ≤1450 mL2 = 6.4 g KNO3

- 200 -

- Chapter 14 -


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

to produce a saturated solution.

must be evaporated.

53.

54. (a)

(b)

55. Assume 1.000 L (1000. mL) of solution

56. First calculate the molarity of the solution

57.

58.

(a)

1200.0 mL HCl2a 3.00 mol

1000 mLb ¢ 1 mol H2

2 mol HCI≤ = 0.300 mol H2

mL HCl ¡ mol HCl ¡ mol H2

Mg + 2 HCl ¡ MgCl2 + H2(g)

M2 =116 M2110.0 mL2

500.0 M= 0.32 M HNO3

116 M2110.0 mL2 = 1M221500. mL2M1V1 = M2V2

V2 =11.63 M21500. mL2

0.10 M= 8.2 * 103 mL = 8.2 L

11.63 M21500. mL2 = 10.10 M21V22M1V1 = M2V2

¢80.0 g H2SO4

500. mL≤ a1000 mL

Lb a 1 mol

98.09 gb = 1.63 M H2SO4

a1000. mL

Lb a1.21 g solution

mLb ¢ 35.0 g HNO3

100. g solution≤ a 1 mol

63.02 gb = 6.72 M HNO3

1500. g HNO32¢1000 mL solution

424 g HNO3≤ a 1.00 L

1000 mLb = 1.18 L solution

11.00 L solution2a1000 mL solution

L solutionb a 1.21 g

mLb ¢ 35.0 g HNO3

100. g solution≤ = 424 g HNO3

1150 mL alcohol2a100. mL solution

70.0 mL alcoholb = 210 mL solution

175 g H2O - 156 g H2O = 19 g H2O

150.0 g KNO32a 100. g H2O

32.0 g KNO3b = 156 g H2O

100. g H2O.°H2OKNO3

- Chapter 14 -

- 201 -


(b)

59.

Calculate the moles of HCl neutralized by each base.

reacts with more HCl than Therefore, ismore effective in neutralizing stomach acid.

60. (a) With equal masses of and the substance with the lower molarmass will represent more moles of solute in solution. Therefore, the will bemore effective than as an antifreeze.

(b) Equal molal solutions will lower the freezing point of the solution by the sameamount.

61. Calculate molarity and molality. Assume 1000 mL of solution to calculate the amounts ofand in the solution.

M = ¢4.9 * 102 g H2SO4

L≤ a 1 mol

98.09 gb = 5.0 M H2SO4

m = ¢ 490 g H2SO4

8.0 * 102 g H2O≤ a1000 g

kgb a 1 mol

98.09 gb = 6.2 m H2SO4

1.29 * 103 g solution - 4.9 * 102 g H2SO4 = 8.0 * 102 g H2O in the solution

11.29 * 103 g solution2¢ 38 g H2SO4

100 g solution≤ = 4.9 * 102 g H2SO4

11000 mL solution2a1.29 g

mLb = 1.29 * 103 g solution

H2OH2SO4

C2H5OHCH3OH

C2H5OH,CH3OH

Mg(OH)21.00 g Al(OH)3.1.20 g Mg(OH)2

11.00 g Al(OH)32a 1 mol

78.00 gb ¢ 3 mol HCl

1 mol Al(OH)3≤ = 0.0385 mol HCl

11.20 g Mg(OH)22a 1 mol

58.33 gb ¢ 2 mol HCl

1 mol Mg(OH)2≤ = 0.0411 mol HCl

Al(OH)3 + 3 HCl ¡ AlCl3 + 3 H2O

Mg(OH)2 + 2 HCl ¡ MgCl2 + 2 H2O

V =nRT

P=10.300 mol210.0821 L atm>mol K21300. K2

0.95 atm= 7.8 L H2

n = 0.300 mol

T = 27°C = 300. K

P = 1720 torr2a 1 atm

760 torrb = 0.95 atm

PV = nRT

- 202 -

- Chapter 14 -


62. Freezing point depression is

(a)

(b)

63. Freezing point

64. (a) Freezing point

(b)

(c) °F = 1.8 (°C) + 32 = 1.8 (-20.0) + 32 = -4.0°F

18.04 * 103 g C2H6O22a1.00 mL

1.11 gb = 7.24 * 103 mL C2H6O2

11.20 * 104 g H2O2¢ 10.8 mol C2H6O2

1000 g H2O≤ a 62.07 g

molb = 8.04 * 103 g C2H6O2

m =20.0°C

1.86°C>m = 10.8 m

¢t f = mKf

12.0 L H2Oa1000 mL

Lb a1.00 g

mLb = 1.20 * 104 g H2O

depression = 20.0°C

10.100 mol C2H6O22¢ 1 kg H2O

0.200 mol C2H6O2≤ ¢ 1000 g H2O

kg H2O≤ = 500. g H2O

16.20 g C2H6O22a 1 mol

62.07 gb = 0.100 mol C2H6O2

m =0.372°C

1.86°C>m = 0.200 m

¢t f = mKf

depression = 0.372°C Kf =1.86°C

m

Boiling point = 100°C + 1.5°C = 101.5°C

¢t b = mKb = 12.9 m2a0.512°Cm

b = 1.5°C

Kb (for H2O) =0.512°C kg solvent

mol solute=

0.512°Cm

m =¢t f

Kf=

5.4°C

1.86°C kg solvent>mol solute= 2.9 m

¢t f = mKf

5.4°C

- Chapter 14 -

- 203 -


65.

First calculate the grams of in the sample.

66. (a) Dilution problem:

0.71 L of is to be diluted to 8.4 L.

must be added (assume volumes are additive)

(b)

(c)

67.

Assume preparation of 1000. mL of 6 M solution

Let of 3.00 M solution; volume of

Mix together 667 mL 3.00 M and 333 mL of 12.0 M to get 1000. mL of 6.00 M HNO3.

HNO3HNO3

1000. mL - 667 mL = 333 mL 12 M

6000. mL = 9.00 y y =6000. mL

9.00= 667 mL 3 M

6000. mL = 3.00 y mL + 12,000 mL - 12.0 y

16.00 M211000. mL2 = 13.00 M21y2 + 112.0 M211000. mL - y212.0 M = 1000. mL - yy = volume

MTVT = M3.00 MV3.00 M + M12.0 MV12.0 M

moles HNO3 total = moles HNO3 from 3.00 M + moles HNO3 from 12.0 M

a 1.5 mol

1000. mLb11.00 mL2 = 0.0015 mol H2SO4 in each mL

a 17.8 mol

1000. mLb11.00 mL2 = 0.0178 mol

8.4 L - 0.71 L = 7.7 L H2O

17.8 M H2SO4

V2 =11.5 M218.4 L2

17.8 M= 0.71 L

11.5 M218.4 L2 = 117.8 M21V22M1V1 = M2V2

¢0.69 g NaHCO3

1.48 g sample≤11002 = 47% NaHCO3

= 0.69 g NaHCO3 in the sample

1150 mL HNO32a 1 L

1000 mLb a0.055 mol

Lb ¢ 1 mol NaHCO3

1 mol HNO3≤ a 84.01 g

molb

mL HNO3 ¡ L HNO3 ¡ mol HNO3 ¡ mol NaHCO3 ¡ g NaHCO3

NaHCO3

HNO3 + NaHCO3 ¡ NaNO3 + H2O + CO2

- 204 -

- Chapter 14 -


68.First calculate the molarity of the diluted HBr solution.

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

(diluted solution)

Now calculate the molarity of the HBr before dilution.

(original solution)

69.

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

According to the equation, twice as many moles of KOH as are needed, soKOH is the limiting reactant.

is formed

70. (a)

(b)

(c) 16.0 g Li2CO32a 1 mol

73.89 gb a1000. mL

0.25 molb = 3.2 * 102 mL solution

a0.25 mol

Lb10.75 L2a73.89 g

molb = 14 g Li2CO3

a0.25 mol

Lb10.0458 L2 = 0.011 mol Li2CO3

10.0331 mol KOH2¢1 mol Ba(OH)2

2 mol KOH≤ a 171.3 g

molb = 2.84 g Ba(OH)2

Ba(NO3)2

a0.743 mol

Lb10.0445 L2 = 0.0331 mol KOH

a0.642 mol

Lb10.0805 L2 = 0.0517 mol Ba(NO3)2

M * L = amoles

Lb1L2 = moles

Ba(NO3)2 + 2 KOH ¡ Ba(OH)2 + 2 KNO3

M1 =10.33 M21240. mL2

20.0 mL= 4.0 M HBr

1M12120.0 mL2 = 10.33 M21240. mL2M1V1 = M2V2

MA =10.37 M2188.4 mL2

100.00 mL= 0.33 M HBr

1MA21100.0 mL2 = 10.37 M2188.4 mL2MAVA = MBVB

HBr + NaOH ¡ NaBr + H2O

- Chapter 14 -

- 205 -


(d) Assume 1000. mL solution

71. The balanced equation is

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

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

72.

Using this data, solubility is (see Table 14.3).88 g solute>100.0 g water = NaNO3

solubility in water = g solute>100 g H2Omass of solute = 563 g - 375 g - 100.0 g = 88 gmass of water = 15.549 moles2118.02 g>mol2 = 100.0 gmass of solute = mass of container & solute - mass of container - mass of water

V =nRT

P V =

(0.131 mol)(0.0821 L atm/mol K)(295 K)

(775 torr)(1 atm/760 torr)= 3.11 L SO2

PV = nRT

Na2SO3

(75.0 mL Na2SO3)a 1.75 mol

1000 mLb a 1 mol SO2

1 mol Na2SO3b = 0.131 mol SO2

(125 mL HCl)a 2.50 mol

1000 mLb a 1 mol SO2

2 mol HClb = 0.156 mol SO2

2 HCl(aq) + Na2SO3(aq) ¡ 2 NaCL(aq) + H2O(l) + SO2(g)

% = a g solute

g solutionb11002 = a 18 g

1220 gb11002 = 1.5%

a0.25 mol

Lb ¢73.89 g Li2CO3

mol≤ = 18 g Li2CO3 per L solution

a1.22 g

mLb11000. mL2 = 1220 g solution

- 206 -

- Chapter 14 -


CHAPTER 14 SOLUTIONS - Chemeketa Community Collegefaculty.chemeketa.edu/lemme/CH 121/solutions/Hein9thCh14.pdf · CHAPTER 14 SOLUTIONS SOLUTIONS TO REVIEW QUESTIONS 1. These diagrams

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