Mixtures and Solutions - Weebly...15. Challenge If 18 mL of methanol are used to make an aqueous solution that is 15% methanol by volume, how many milliliters of solution are produced?

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SOLUTIONS MANUALCHAPTER 14

Mixtures and Solutions

Solutions Manual Chemistry: Matter and Change • Chapter 14 277

Section 14.1 Types of Mixturespages 476 – 479

Section Assessment 14.1page 479

1. Explain Use the properties of seawater to

describe the characteristics of mixtures.

Answers will vary but might include that

seawater is a heterogeneous mixture with dirt

and mud particles, and it is a homogeneous

mixture with dissolved substances.

2. Distinguish between suspensions and colloids.

Suspension particles are larger than colloidal

particles. Suspension particles settle out of the

mixture, whereas colloidal particles do not.

3. Identify the various types of solutions. Describe

the characteristics of each type of solution.

All solutions are homogeneous mixtures

containing two or more substances. Solutions

may be liquid, solid, or gas. Solution types are

identified in Table 14.2.

4. Explain Use the Tyndall effect to explain why

it is more difficult to drive through fog using

high beams than using low beams.

High beams are aimed farther down the road

than low beams. Because the fog scatters

light, there is less light from the high beams to

illuminate the road than from the low beams.

Also, because the high beams are aimed more

directly into the fog, more of their light is

reflected back toward the driver, making it more

difficult to see.

5. Describe different types of colloids.

See Table 14.1 for descriptions of colloid types.

6. Explain Why do dispersed colloid particles

stay dispersed?

The particles do not settle out because they have

polar or charged layers surrounding them. These

layers repel each other and prevent the particles

from settling or separating.

7. Summarize What causes Brownian motion?

Collisions of particles of the dispersion medium

with the dispersed particles results in Brownian

motion.

8. Compare and Contrast Make a table that

compares the properties of solutions, suspen-

sions, and colloids.

Student tables will vary, but should include

particle size, if the particles settle out, and if the

particles display the Tyndall effect. A sample table

follows.

Suspensions, Colloids, and Solutions

Particle sizeParticles settle?

Tyndall effect?

Suspensions Large (wide variation)

Yes Yes

Colloids 1 nm–1000 nm No Yes

Solutions Atomic scale (atoms, ions, and molecules)

No No

Section 14.2 Solution Concentrationpages 480–488

Practice Problemspages 481–488

9. What is the percent by mass of NaHCO3 in a

solution containing 20.0 g NaHCO3 dissolved

in 600.0 mL H2O?

600.0 mL H2O 3 1.0 g/mL 5 600.0 g H2O

20 g NaHCO3

___ 600 g H2O 1 20 g NaHCO3

3 100 5 3%

10. You have 1500.0 g of a bleach solution. The

percent by mass of the solute sodium

hypochlorite, NaOCl, is 3.62%. How many

grams of NaOCl are in the solution?

3.62% 5 100 3 mass NaOCI

__ 1500.0 g

mass NaOCl 5 54.3 g

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278 Chemistry: Matter and Change • Chapter 14 Solutions Manual


11. In question 10, how many grams of solvent are

in the solution?

1500.0 g 2 54.3 g 5 1445.7 g solvent

12. Challenge The percent by mass of calcium

chloride in a solution is found to be 2.65%. If

50.0 grams of calcium chloride is used, what is

the mass of the solution?

2.65% 5 100 3 50 g CaCl2

__ mass of solution

mass of solution 5 1886.79 g

13. What is the percent by volume of ethanol

in a solution that contains 35 mL of ethanol

dissolved in 155 mL of water?

35 mL

__ 155 mL 1 35 mL

3 100 5 18%

14. What is the percent by volume of isopropyl

alcohol in a solution that contains 24 mL of

isopropyl alcohol in 1.1 L of water?

24 mL

__ 24 mL 1 1100 mL

3 100 5 2.1%

15. Challenge If 18 mL of methanol are used to

make an aqueous solution that is 15% methanol

by volume, how many milliliters of solution are

produced?

15% 5 18 mL __ x mL solution

3 100

x 5 120 mL

16. What is the molarity of an aqueous solution

containing 40.0 g of glucose (C6H12O6) in 1.5 L

of solution?

mol C6H12O6 5 40.0 g 3 1 mol

_ 180.16 g

5 0.222 mol

molarity 5 mol C6H12O6

__ 1.5 L solution

5 0.222 mol

_ 1.5 L

5 0.15M

17. Calculate the molarity of 1.60 L of a solution

containing 1.55 g of dissolved KBr.

mol KBr 5 1.55 g 3 1 mol

_ 119.0 g

5 0.0130 mol KBr

molarity 5 mol KBr

__ 1.60 L solution

5 0.0130 mol

__ 01.60 L

5 8.13 3 1023M

18. What is the molarity of a bleach solution

containing 9.5 g of NaOCl per liter of bleach?

mol NaOCl 5 9.5 g 3 1 mol

_ 74.44 g

5 0.13 mol

molarity 5 mol NaOCI

__ 1.00 L solution

5 0.128 mol

_ 1.00 L

5 0.13M

19. Challenge How much calcium hydroxide

(Ca(OH)2), in grams, is needed to produce

1.5 L of a 0.25M solution?

0.25M 5 x mol Ca(OH)2

__ 1.5 L solution

x 5 0.38 mol Ca(OH)2

0.38 mol Ca(OH)2 3 74.08 g

_ mol

5 28 g Ca(OH)2

20. How many grams of CaCl2 would be dissolved

in 1.0 L of a 0.10M solution of CaCl2?

mol CaCl2 5 (0.10M)(1.0 L) 5 (0.10 mol/L)(1.0 L)

5 0.10 mol CaCl2

mass CaCl2 5 0.10 mol CaCl2 3 110.98 g

_ 1 mol

mass of CaCl2 5 11 g

21. How many grams of CaCl2 should be dissolved

in 500.0 mL of water to make a 0.20M solution

of CaCl2?

mol CaCl2 5 500.0 mL 3 1 L _

1000 mL 3 0.20M

5 0.10 mol

mass CaCl2 5 0.10 mol CaCl2 3 110.98 g

_ 1 mol


22. How many grams of NaOH are in 250 mL of a

3.0M NaOH solution?

mol NaOH 5 250 mL 3 1 L _

1000 mL 3 3.0M

5 250 mL 3 1 L _

1000 mL 3

3.0 mol _

1 L

5 0.75 mol

mass NaOH 5 0.75 mol NaOH 3 40.00 g

_ 1 mol

mass of NaOH 5 3.0 3 101 g

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23. Challenge What volume of ethanol (C2H3OH)

is in 100.0 mL of 0.15M solution? The density

of ethanol is 0.7893 g/mL.

100 mL 3 1 L _

1000 mL 3

0.15 mol ethanol __

1 L solution

3 46 g ethanol

__ 1 mol ethanol

3 1 mL ethanol

__ 0.7893 g ethanol

5 0.87 mL

24. What volume of a 3.00M KI stock solution

would you use to make 0.300 L of a 1.25M KI

solution?

(3.00M)V1 5 (1.25M)(0.300 L)

V1 5 (1.25M)(0.300 L)

__ 3.00M

5 0.125 L 5 125 mL

25. How many milliliters of a 5.0M H2SO4 stock

solution would you need to prepare 100.0 mL

of 0.25M H2SO4?

(5.0M)V1 5 (0.25M)(100.0 mL)

V1 5 (0.25M)(100.0 mL)

__ 5.0 M

5 5.0 mL

26. Challenge If 0.5 L of 5M stock solution of

HCl is diluted to make 2 L of solution, how

much HCl, in grams was in the solution?

mol HCl 5 5M 3 0.5 L 5 2.5 mol HCl

mass of HCl 5 36.45 g HCl

__ 1 mol

3 2.5 mol

mass of HCl 5 91.15 g

27. What is the molality of a solution containing

10.0 g Na2SO4 dissolved in 1000.0 g of water?

mol Na2SO4 5 10.0 g Na2SO4 3 1 mol

_ 142.04 g

5 0.0704 mol Na2SO4

molality 5 0.0704 mol Na2SO4

__ 1.0000 Kg H2O

5 0.0704 m

28. Challenge How much (Ba(OH)2), in grams, is

needed to make a 1.00m aqueous solution?

mol Ba(OH)2 5 1 mol __

1 kg solvent

molar mass of Ba(OH)2 5 171 g Ba(OH)2

__ 1 mol

mass of Ba(OH)2 5 171 g

29. What is the mole fraction of NaOH in an aqueous

solution that contains 22.8% NaOH by mass?

Assume 100.0 g sample.

Then, mass NaOH 5 22.8 g

mass H2O 5 100.0 g 2 (mass NaOH) 5 77.2 g

mol NaOH 5 22.8 g 3 1 mol

_ 40.00 g

5 0.570 mol NaOH

mol H2O 5 77.2 g 3 1 mol

_ 18.02 g

5 4.28 mol H2O

mol fraction NaOH 5 mol NaOH

___ mol NaOH 1 mol H2O

5 0.570 mol NaOH

____ 0.570 mol NaOH 1 4.28 mol H2O

5 0.570

_ 4.85

5 0.118

The mole fraction of NaOH is 0.118.

30. Challenge If the mole fraction of sulfuric acid

(H2SO4) in an aqueous solution is 0.125, what

is the percent by mass of H2SO4?

0.125 5 mole fraction of H2SO4

1 2 0.125 5 0.875 mole fraction of water

Assume a sample of the solution totals 100.0 moles.

By definition there would be 87.5 moles of water

and 12.5 moles of sulfuric acid in the sample.

87.5 mol of H2O 3 18.02 g

_ 1 mol

5 1580 g H2O

12.5 mol H2SO4 3 98.08 g

_ 1 mol

5 1230 g H2SO4

percent by mass H2SO4

5 1230 g H2SO4

___ (1580 1 1230) g solution

3 100

5 43.8% H2SO4 by mass


31. Compare and contrast five quantitative

ways to describe the composition of solutions.

molarity, molality, and mole fraction are based

on moles of solute per some other quantity;

percent by volume and molarity are defined on a

per volume of solution basis; molality and mole

fraction are based on a per quantity of solvent

basis; percent by mass and percent by volume are

the only ratios involving percentages

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32. Explain the similarities and differences

between a 1M solution of NaOH and a 1m

solution of NaOH.

Both solutions contain NaOH (solute) dissolved in

water (solvent). The 1 m solution contains 1 mole

of NaOH per kilogram of water; the 1M solution

contains 1 mole of NaOH per liter of solution.

33. Calculate A can of chicken broth contains

450 mg of sodium chloride in 240.0 g of broth.

What is the percent by mass of sodium chloride

in the broth?

450 mg NaCl 3 1 g _

1000 mg 5 0.45 g NaCl

percent by mass 5 0.45 g

_ 240.0 g

3 100 5 0.19%

34. Solve How much ammonium chloride

(NH4Cl), in grams, is needed to produce 2.5 L

of a 0.5M aqueous solution?

mol of NH4Cl 5 0.5M

_ 1 L

3 2.5 L

5 1.25 mol of NH4Cl

mass of NH4Cl 5 1.25 mol NH4Cl 3 53.49 g NH4Cl

__ 1 mol

mass of NH4Cl 5 66.86 g

35. Outline the laboratory procedure for preparing

a specific volume of a dilute solution from a

concentrated stock solution.

Calculate the volume of stock solution needed

and add it to a volumetric flask. Add water up to

the flask’s calibration line.

Section 14.3 Factors Affecting Solvationpages 489–497

Practice Problemspage 497

36. If 0.55 g of a gas dissolves in 1.0 L of water at

20.0 kPa of pressure, how much will dissolve at

110.0 kPa of pressure?

S1 5 0.55 g

_ 1.0 L

5 0.55 g/L

S2 5 S1 3 P2

_ P1

5 0.55 g/L 3 110.0 kPa

_ 20.0 kPa

5 3.0 g/L

37. A gas has a solubility of 0.66 g/L at 10.0 atm

of pressure. What is the pressure on a 1.0-L

sample that contains 1.5 g of gas?

S2 5 1.5 g _ 1.0 L

5 1.5 g/L

P2 5 P1 3 S2

_ S1

5 10.0 atm 3 1.5 g/L

_ 0.66 g/L

5 23 atm

38. Challenge The solubility of a gas at 7.0 atm

of pressure is 0.52 g/L. How many grams of the

gas would be dissolved per 1 L if the pressure

increased 40.0 percent?

P2 5 P1 1 (P1)(0.400)

5 (7.0 atm) 1 (7.0 atm)(0.400)

5 9.8 atm

S2 5 S1 3 P2

_ P1

S2 5 (0.52 g/L) 3 9.8 atm

_ 7.0 atm

S2 5 0.73 g/L


39. Describe factors that affect the formation of

solutions.

Surface area, temperature, and pressure affect the

formation of solutions.

40. Define solubility.

Solubility refers to the maximum amount of

solute that can dissolve in a given amount of

solvent at a particular temperature and pressure.

41. Describe how intermolecular forces affect

solvation?

The attractive forces between solute and solvent

particles overcome the forces holding the solute

particles together, thus, pulling the solute

particles apart.

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42. Explain on a particle basis why the vapor pres-

sure of a solution is lower than a pure solvent.

When a solvent contains a solute, fewer solvent

particles occupy the surface. Fewer particles

escape into the gaseous state.

43. Sumarize If a seed crystal was added to a

supersaturated solution, how would you charac-

terize the resulting solution?

After the excess solute particles crystallize out of

solution, the solution is saturated.

44. Make and Use Graphs Use the informa-

tion in Table 14.4 to graph the solubilities of

aluminum sulfate, lithium sulfate, and potas-

sium chloride at 0°C, 20°C, 60°C, and 100°C.

Which substance’s solubility is most affected by

increasing temperature?

600 10020

So

lub

ilit

y (

g/1

00 g

H2O

)

10

30

20

40

50

60

70

80

90

Temperature (°C)

Solubility v. Temperature

Li2SO4

Al2(SO4)3

KCl

Aluminum sulfate shows the greatest change in

solubility over the temperature range.

Section 14.4 Colligative Properties of Solutionspages 498–504

Practice Problemspage 503

45. What are the boiling point and freezing point of

a 0.625m aqueous solution of any nonvolatile,

nonelectrolyte solute?

DTb 5 0.512°C/m 3 0.625m 5 0.320°C

Tb 5 100°C 1 0.320°C 5 100.320°C

DTf 5 1.86°C/m 3 0.625m 5 1.16°C

Tf 5 0.0°C 2 1.16°C 5 21.16°C

46. What are the boiling point and freezing point of

a 0.40m solution of sucrose in ethanol?

DTb 5 1.22°C/m 3 0.40m 5 0.49°C

Tb 5 78.5°C 1 0.49°C 5 79.0°C

DTf 5 1.99°C/m 3 0.40m 5 0.80°C

Tf 5 2114.1°C 2 0.80°C 5 2114.9°C

47. Challenge If a 0.045m solution (consisting of a

nonvolatile, nonelectrolyte solute) is experimen-

tally found to have a freezing point depression of

0.080ºC. What is the freezing point depression

constant (Kf)? Which is most likely to be the

solvent—water, ethanol, or chloroform?

Kf 5 DTf

_ m

5 0.080ºC

_ 0.045m

5 1.8ºC/m

It is most likely water because the calculated

value is closest to 1.86°C/m


48. Explain the nature of colligative properties.

Colligative properties depend on the number of

solute particles in a solution.

49. Describe four colligative properties of solutions.

vapor pressure lowering: the decrease in vapor

pressure with increasing solute particles in

solution; boiling point elevation: the increase in

boiling point with increasing solute particles in

solution; freezing point depression: the decrease

in freezing point with increasing solute particles

in solution; osmotic pressure: the change in

osmotic pressure with increasing solute particles

in solution

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50. Explain why a solution has a higher boiling

point than the pure solvent.

Solute particles in solution decrease the vapor

pressure above the solution. Because a solution

boils when its vapor pressure equals the external

pressure, this decrease in vapor results in the

need for a higher temperature in order for the

solution to boil.

51. Solve An aqueous solution of calcium chloride

(CaCl2) boils at 101.3ºC. How many kilograms

of calcium chloride were dissolved in 1000 grams

of the solvent?

m 5 DTb _ Kb

5 1.3ºC _

0.512ºC/m

5 2.53m 5 2.53 moles solute particles/1 kg

solvent

2.53 mol particles 3 1 mol CaCl2

__ 3 mol particles

3 110.98 g _ 1 mol

3 1 kg _

1000 g 5 0.0936 kg

52. Calculate the boiling point elevation of a solu-

tion containing 50.0 g of glucose (C6H12O6)

dissolved in 500.0 g of water. Calculate the

freezing point depression for the same solution.

50.0 g glucose 3 1 mol

_ 180.15 g

5 0.278 mol glucose

molality 5 0.278 mol glucose

__ 0.5000 kg H2O

5 0.556m

DTb 5 (0.512°C/m)(0.556m) 5 0.285°C

Tb 5 100.000°C 1 0.285°C 5 100.285°C

DTf 5 (1.86°C/m)(0.556m) 5 1.03°C

Tf 5 0.00°C 2 1.03°C 5 21.03°C

53. Investigate A lab technician determines the

boiling point elevation of an aqueous solution

of a nonvolatile, nonelectrolyte to be 1.12°C.

What is the solution’s molality?

1.12°C 5 0.512°C/m 3 m

m 5 2.19m

Chapter 14 Assessment pages 508–511

Section 14.1

Mastering Concepts

54. Explain what is meant by the statement “not

all mixtures are solutions.”

Solutions are homogeneous mixtures that are

uniform in composition with a single phase.

Mixtures can also be heterogeneous, where the

substances that make them up remain distinct.

55. What is the difference between solute and

solvent?

A solute is the substance being dissolved.

The solvent is the substance in which the

solute dissolves.

56. What is a suspension and how does it differ

from a colloid?

A suspension is a heterogeneous mixture that

settles out if left undisturbed. The particles

dispersed in a colloid are much smaller than those

in a suspension and do not settle out.

57. How can the Tyndall effect be used to distin-

guish between a colloid and a solution? Why?

A beam of light is visible in a colloid but not in

a solution. Dispersed colloid particles are large

enough to scatter light (Tyndall effect).

58. Name a colloid formed from a gas dispersed in

a liquid.

Student answers may include whipped cream or

beaten egg whites.

59. Salad dressing What type of heterogeneous

mixture is shown in Figure 14.24 on page 508?

What characteristic is most useful in classifying

the mixture?

The mixture is a suspension. Left undisturbed, the

mixture components settle out.

60. What causes Brownian motion observed in

liquid colloids?

The random particle movements in liquid colloids

result from collisions between particles in the

mixture.

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61. Aerosol sprays are categorized as colloids.

Identify the phases of an aerosol spray.

The most abundant mixture component is in

the gas phase. The dispersed particles are in the

liquid phase.

Section 14.2

Mastering Concepts

62. What is the difference between percent by mass

and percent by volume?

Percent by mass is a comparison between the

mass of solute and the total mass of the solution.

Percent by volume is a comparison between the

volume of the solute and the total volume of the

solution.

63. What is the difference between molarity and

molality?

Molarity is solution concentration expressed

as the moles of solute per volume of solution.

Molality expresses concentration as moles of

solute per kilogram of solvent. Molality does not

depend upon the temperature of the solution.

64. What factors must be considered when creating

a dilute solution from a stock solution?

The molarity and volume of both stock solution

and dilute solution are required in the formula

M1V1 5 M2V2.

65. How do 0.5M and 2.0M aqueous solutions of

NaCl differ?

The 2M solution contains more moles of NaCl per

volume of water than the 0.5M solution.

66. Under what conditions might a chemist describe

a solution in terms of molality? Why?

Under conditions of changing temperature.

Because molality is based on mass, it does not

change with temperature.

Mastering Problems

67. According to lab procedure, you stir 25.0 g

of MgCl2 into 550 mL of water. What is the

percent by mass of MgCl2 in the solution?

percent by mass of MgCl2

5 25.0 g MgCl

___ 25.0 g MgCl 1 550 g H2O

3 100 5 4.3%

68. How many grams of LiCl are in 275 g of a 15%

aqueous solution of LiCl?

mass of LiCl 5 275 g 3 15

__ 100

5 41 g

69. You need to make a large quantity of a 5%

solution of HCl but have only 25 mL HCl.

What volume of 5% solution can be made from

this volume of HCl?

volume of solution 5 25 mL HCl

_ 5 3 100 5 500 mL

70. Calculate the percentage by volume of a solu-

tion created by adding 75 mL of acetic acid to

725 mL of water.

percent volume

5 75 mL CH3COOH

____ 75 mL CH3OOH 1 725 mL solution

3 100

5 9.4%

71. Calculate the molarity of a solution that

contains 15.7 g of CaCO3 dissolved in 275 mL

of water.

mol of CaCO3 5 15.7 g CaCO3

3 1 mol CaCO3

__ 100.01 g CaCO3

5 0.157 mol CaCO3

275 mL 3 1 L _

1000 mL 5 0.275 L

molarity 5 0.157 mol CaCO3

__ 0.275 L solution

5 0.571M

72. What is the volume of a 3.00M solution made

with 122 g of LiF?

mol of LiF 5 122 g LiF 3 1 mol LiF

_ 25.9 g LiF

5 4.71 mol LiF

volume of solution 5 4.71 mol

_ 3.00M

5 1.57 L

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73. How many moles of BaS would be used to

make 1.5 3 103 mL of a 10.0M solution?

1.5 3 103 mL 3 1 L _

1000 mL 5 1.5 L

mol BaS 5 10.0 mol

_ 1 L

3 1.5 L

mol BaS 5 15 mol

74. How many grams of CaCl2 are needed to make

2.0 L of a 3.5M solution?

mol of CaCl2 5 3.5 mol

_ 1 L

3 2.0 L 5 7.0 mol CaCl2

mass of CaCl2 5 7.0 mol CaCl2 3 110.1 g CaCl2

__ 1 mol CaCl2


75. Stock solutions of HCl with various molarities

are frequently prepared. Complete Table 14.7

by calculating the volume of concentrated, or

12M, hydrochloric acid that should be used to

make 1.0 L of HCl solution with each molarity

listed.

Molarity of HCl desired

Volume of 12M HCl stocksolution needed (mL)

0.50 42 mL

1.0 83 mL

1.5 130 mL

2.0 170 mL

5.0 420 mL

V1 5 0.50 mol/L 3 1 L

__ 12 mol/L

5 0.042 L HCl

0.042 L HCl 3 1000 mL

__ 1 L

5 42 mL HCl

V1 5 1.0 mol/L 3 1 L

__ 12 mol/L

5 0.083 L HCl

0.083 L HCl 3 1000 mL

_ 1 L

5 83 mL HCl

V1 5 1.5 mol/L 3 1 L

__ 12 mol/L

5 0.13 L HCl

0.13 L HCl 3 1000 mL

_ 1 L

5 130 mL HCl

V1 5 2.0 mol/L 3 1 L

__ 12 mol/L

5 0.17 L HCl

0.17 L HCl 3 1000 mL

_ 1 L

5 170 mL HCl

V1 5 5.0 mol/L 3 1 L

__ 12 mol/L

5 0.42 L HCl

0.42 L HCl 3 1000 mL

_ 1 L

5 420 mL HCl

76. How much 5.0M nitric acid (HNO3), in

milliliters, is needed to make 225 mL of

1.0M HNO3?

V1 5 1.0 M 3 225 mL

__ 5.0 M

volume of HNO3 5 45 mL

77. Experiment In the lab, you dilute 55 mL of

a 4.0M solution to make 250 mL of solution.

Calculate the molarity of the new solution.

M2 5 4.0M 3 55 mL

__ 250 mL

5 0.88M

78. How many milliliters of 3.0M phosphoric acid

(H3PO4) can be made from 95 mL of a 5.0M

H3PO4 solution?

V2 5 5.0 M 3 95 mL

__ 3.0 M

5 160 mL

79. If you dilute 20.0 mL of a 3.5M solution

to make 100.0 mL of solution, what is the

molarity of the dilute solution?

M2 5 3.5M 3 20 mL

__ 100 mL

5 0.70M


75.3 grams of KCl dissolved in 95.0 grams of

water?

mol KCl 5 75.3 g KCl 3 1 mol KCl _

74.6 g KCl

5 1.01 mol KCl

95.0 g H2O 3 1 kg _

1000 g 5 0.0950 kg H2O

m 5 1.01 mol KCl

__ 0.0950 kg H2O

5 10.6 mol/kg

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81. How many grams of Na2CO3 must be dissolved

into 155 grams of water to create a solution

with a molality of 8.20 mol/kg?

155 g H2O 3 1 kg _

1000 g 5 0.155 kg H2O

mol NaCO3 5 8.20 mol/kg 3 0.155 kg 5 1.27 mol

mass of NaCO3 5 1.27 mol NaCO3 3 83.00 g NaCO3 __

mol NaCO3

5 105 g


30.0 g of naphthalene (C10H8) dissolved in

500.0 g of toluene?

30.0 g C10H8 3 1 mol C10H8

__ 128 g C10H8

5 0.234 mol C10H8

500.0 g toluene 3 1 kg _

1000 g 5 0.5000 kg

m 5 0.234 mol C10H8

__ 0.5000 kg toluene

5 0.468m

83. What are the molality and mole fraction of

solute in a 35.5 percent by mass aqueous

solution of formic acid (HCOOH)?

35.5% means 35.5 g HCOOH

__ 100.0 g solution

35.5 g HCOOH 3 1 mol HCOOH

__ 46.03 g HCOOH

5 0.771 mol HCOOH

mass of water 5 100.0 g 2 35.5 g 5 64.5 g

5 6.45 3 1022 kg

moles of water 5 64.5 g H2O 3 1 mol H2O

__ 18.02 g H2O

5 3.58 mol H2O

molality 5 0.771 mol HCOOH

__ 6.45 3 1022 kg H2O

5 12.0m

mole fraction 5 0.771 mol

___ 0.771 mol 1 3.58 mol

5 0.177

84. What is the mole fraction of H2SO4 in a

solution containing the percentage of sulfuric

acid and water shown in Figure 14.25?

H2O72.7%

H2SO427.3%

27.3 g H 2 S O 4 3 1 mol H 2 S O 4

__ 97.1 g H 2 S O 4

5 0.281 mol H 2 S O 4

72.7 g H 2 O 3 1 mol H 2 O

__ 18.02 g H 2 O

5 4.034 mol H2O

X H 2 S O 4 5 0.281 mol H 2 S O 4

____ 0.281 mol H 2 S O 4 1 4.034 mol H 2 O

5 0.0650

85. Calculate the mole fraction of MgCl2 in a

solution created by dissolving 132.1 g MgCl2

into 175 mL of water.

132.1 g MgC l 2 3 1 mol MgC l 2

__ 95.21 g

5 1.387 mol MgCl2

175 mL H 2 O 3 1.0 g H 2 O

_ 1 mL H 2 O

3 1 mol H 2 O

__ 18.0 g H 2 O

5 9.72 mol H2O

XMgCl2 5

1.387 mol MgC l 2

____ 1.387 mol MgCl2 1 9.72 mol H2O

5 0.125

Section 14.3

Mastering Concepts

86. Describe the process of solvation.

A solute introduced into a solvent is surrounded

by solvent particles. Due to the attraction

between solute and solvent particles, solute

particles are pulled apart and surrounded by

solvent particles. Once separated, solute particles

disperse into solution.

87. What are three ways to increase the rate of

solvation?

increase the temperature of the solvent, increase

the surface area of the solute, agitation

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88. Explain the difference between saturated and

unsaturated solutions.

A saturated solution contains the maximum

amount of solute under a given set of conditions.

An unsaturated solution contains less than the

maximum amount.

Mastering Problems

89. At a pressure of 1.5 atm, the solubility of a gas

is 0.54 g/L. Calculate the solubility when the

pressure is doubled.

S 2 5 0.54 g/L 3 3.0 atm

__ 1.5 atm

S 2 5 1.08 g/L

90. At 4.5 atm of pressure, the solubility of a gas is

9.5 g/L. How many grams of gas will dissolve

in 1L if the pressure is reduced by 3.5 atm?

S 2 5 9.5 g/L 3 1.0 atm

__ 4.5 atm

S 2 5 2.1 g/L

91. Using Figure 14.26, compare the solubility of

potassium bromide (KBr) and potassium nitrate

(KNO3) at 80°Celsius.

So

lub

ilit

y (

g/1

00 g

of

wa

ter)

240

220

200

180

160

140

120

100

80

60

40

20

0

Temperature (ºC)

20 40 60 80 100 120

NaCl

NaClO2

KNO3

KBr

Solubility v. Temperature

The solubility of KBr is 95 g/100 g H 2 O. The

solubility of KN O 3 is nearly twice as high at the

same temperature, at nearly 170 g/100 g H 2 O.

92. The solubility of a gas at 37.0 kPa is 1.80 g/L.

At what pressure will the solubility reach

9.00 g/L?

P 2 5 37.0 kPa 3 9.00 g/L

__ 1.8 g/L

P 2 5 185 kPa

93. Use Henry’s Law to complete the Table 14.8.

Solubility and Pressure

Solubility (g/L) Pressure (kPa)

2.9 25

3.7 32

4.5 39

P 2 5 32 kPa 3 2.9 g/L

__ 3.7 g/L

P 2 5 25 kPa

S 2 5 3.7 g/L 3 39 kPa

__ 32 kPa

S 2 5 4.5 g/L

94. Soft Drinks The partial pressure of CO2

inside a bottle of soft drink is 4.0 atm at 25°C.

The solubility of CO2 is 0.12 mol/L. When the

bottle is opened, the partial pressure drops

to 3.0 3 1024 atm. What is the solubility of

CO2 in the open drink? Express your answer in

grams per liter.

S 5 (0.12 mol/L)(3.0 3 10–4 atm)

___ 4.0 atm

5 9.0 3 10–6 mol/L CO2

9.0 3 10–6 mol C O 2

__ 1 L

3 44.01 g CO2

__ 1 mol CO2

5 4.0 3 10–4 g/L CO2

Section 14.4

Mastering Concepts

95. Define the term colligative property.

A physical property of a solution that is affected

by the number of solute particles but not their

nature.

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96. Use the terms dilute and concentrated to

compare the solution on both sides of a

membrane.

If there is a concentration gradient, the solution

is more dilute on one side of the membrane

and more concentrated on the other side of the

membrane.

97. Identify each variable in the following formula.

D T b 5 K b m

D T b represents the difference between the boiling

points of a solution and the pure solvent; K b is

the molal boiling point elevation constant; m

represents the solution molality.

98. Define the term osmotic pressure, and explain

why it is considered a colligative property.

Osmotic pressure is the pressure exerted by

water molecules that move into solution through

osmosis. Osmotic pressure is a colligative property

because it depends on the number of solute

particles dissolved in solution.

Mastering Problems

99. Calculate the freezing point of a solution of

12.1 grams of naphthalene ( C 10 H 8 ) dissolved

into 0.175 kg of benzene ( C 6 H 6 ). Refer to

Table 14.6 for the necessary constant.

mol C 10 H 8 5 12.1 g C10H8 3 1 mol C 10 H 8

__ 128.08 g C 10 H 8

5 0.0945 mol C 10 H 8

m 5 0.0945 mol C 10 H 8

__ 0.175 kg C 6 H 6

5 0.540m

D T f 5 5.12°C/m 3 0.540m 5 2.76°C

T f 5 5.5°C 2 2.76°C 5 2.74°C

100. In the lab, you dissolve 179 grams of MgC l 2

into 1.00 liter of water. Use Table 14.6 to find

the freezing point of the solution.

mol MgC l 2 5 179 g MgC l 2

__ 95.3 g/mo l

5 1.88 mol MgCl2

kg H 2 O 5 1.00 L H 2 O 3 1000 mL

_ 1 L

3 1 g H 2 O

_ 1 mL H 2 O

3 1 kg _

1000 g 5 1.00 kg H2O

m 5 1.88 mol MgC l 2

__ 1 kg H 2 O

5 1.88m

particle m 5 1.88m 3 3 5 5.64m

DTf 5 1.86°C/m 3 5.64m 5 10.5°C

Tf 5 0.0°C 2 10.5°C 5 210.5°C

101. Cooking A cook prepares a solution for

boiling by adding 12.5 grams of NaCl to a pot

holding 0.750 liters of water. At what tempera-

ture should the solution in the pot boil? Use

Table 14.5 for the necessary constant.

mol NaCl 5 12.5 g NaCl

__ 58.44 g/mol

5 0.214 mol NaCl

kg H2O 5 0.750 L H2O 3 1000 mL

_ 1 L

3 1 g H 2 O

_ 1 mL H 2 O

3 1 kg _

1000 g 5 0.750 kg H2O

solution molality 5 0.214 mol NaCl

__ 0.750 kg H 2 O

5 0.285m

particle molality 5 0.285m 3 2 5 0.570m

DTb 5 0.512°C/ m 3 0.570 m 5 0.292°C

Tb 5 100.00°C 1 0.292°C 5 100.29°C

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102. The boiling point of ethanol ( C 2 H 5 OH)

changes from 78.5°C to 85.2°C when an

amount of naphthalene ( C 10 H 8 ) is added to

1.00 kg of ethanol. How much naphthalene, in

grams, is required to cause this change? Refer

to Table 14.5 for needed data.

D T b 5 85.2°C 2 78.5°C 5 6.70°C

solution molality 5 6.70°C

_ 1.22°C/m

5 5.49m

moles of solute 5 5.49 mol C 10 H 8

__ 1 kg C 2 H 5 OH

3 1.00 kg C 2 H 5 OH 5 5.49 mol C10H8

grams of solute 5 5.49 mol C 10 H 8 3 128 g C 10 H 8

__ 1 mol C 10 H 8

5 703 g C10H8

103. Ice Cream A rock salt (NaCl), ice, and water

mixture is used to cool milk and cream to

make homemade ice cream. How many grams

of rock salt must be added to water to lower

the freezing point by 10.0°C?

D T f 5 K f m

m 5 D T f _ K f

5 10.0°C

_ 1.86°C/m

5 5.38m ions of Na1 and Cl2

molality 5 moles of solute

__ kilograms of solvent

5 2.69 mol NaCl

__ 1 kg solvent

2.69 mol NaCl

__ 1 kg H2O

3 58.44 g NaCl

__ 1 mol NaCl

5 157 g NaCl per 1 kg H 2 O

Mixed Review

104. Apply your knowledge of polarity and solu-

bility to predict whether solvation is possible

in each situation shown in Table 14.9. Explain

your answers.

MgCl2(s) in H2O(l): Yes. NH3(l) in C6H6(l): No.

H2(g) in H2O(l): No. I2(l) in Br2(l): Yes. Predictions

are based on the general rule “like dissolves

like.” A polar solvent like water will dissolve

a polar solute like magnesium chloride, and

a nonpolar solvent like liquid bromine will

dissolve a nonpolar solute like liquid iodine.

Ammonia is a polar molecule, while benzene

is nonpolar. Water is a polar molecule while

diatomic hydrogen is nonpolar.

105. Household Paint Some types of paint

are colloids composed of pigment particles

dispersed in oil. Based on what you know

about colloids, recommend an appropriate

location for storing cans of leftover household

paint. Justify your recommendation.

When a colloid is exposed to heat, suspended

particles can settle out. Paint should be stored

in a cool location where it cannot freeze, and

away from direct sunlight and objects like water

heaters or furnaces that generate heat.

106. Which solute has the greatest effect on the

boiling point of 1.00 kg of water: 50.0 grams

of strontium chloride (SrC l 2 ) or 150.0 grams

of carbon tetrachloride (CC l 4 )? Justify your

answer.

50.0 grams SrC l 2 has the greatest effect.

mol SrC l 2 5 50.0 g mol SrCl2

__ 158.6 g mol SrCl2

5 0.315 mol SrC l 2

solution molality 5 0.315 mol

_ 1.00 kg

5 0.315m


D T b 5 0.512°C/m 3 0.945m 5 0.484°C

T b SrC l 2 solution 5 100.0°C 1 0.484°C

5 100.484°C

mol CC l 4 5 150.0 g mol CCl4

__ 154 g mol

5 0.974 mol CC l 4

solution molality 5 0.974 mol

_ 1.00 kg

5 0.974m


D T b 5 0.512°C/m 3 0.974m 5 0.310°C

T b CC l 4 solution 5 100.0°C 1 0.310°C 5 100.31°C

107. Study Table 14.4. Analyze solubility and

temperature data to determine the general trend

followed by the gases (NH3, CO2, O2) in the

chart. Compare this trend to the trend followed

by most of the solids in the chart. Identify the

solids listed that do not follow the general trend

followed by most of the solids in the chart.

For the gases, solubility decreases as

temperature increases. For most solids, solubility

increases as temperature increases. Ca(OH)2 and

Li2SO4 do not follow the general trend for solids.

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108. An air sample yields the percent composition

shown in Figure 14.27. Calculate the mole

fraction of each gas present in the sample.

Oxygen21.0%

Nitrogen78.0%

Argon1.00%

78.0 g N 2 3 1 mol N2

_ 28.0 g N2

5 2.79 mol N 2

21.0 g O 2 3 1 mol O 2

_ 32.0 g O 2

5 0.656 mol O 2

1.00 g Ar 3 1 mol Ar

_ 39.9 g Ar

5 0.0251 mol Ar

XN2 5

2.79 mol N2

_____ 2.79 mol N 2 1 0.656 mol O 2 1 0.0251 mol Ar

5 0.804

XO2 5

0.656 mol O2 _____

2.79 mol N 2 1 0.656 mol O 2 1 0.0251 mol Ar

5 0.189

X Ar 5

0.0251 mol N 2

_____ 2.79 mol N 2 1 0.656 mol O 2 1 0.0251 mol Ar

5 0.00723

109. If you prepared a saturated aqueous solution

of potassium chloride at 25°C and then heated

it to 50°C, would you describe the solution

as unsaturated, saturated, or supersaturated?

Explain.

unsaturated; the solubility of KCl in water

increases with temperature. A solution at 50°C

holds more solute than one at 25°C.

110. How many grams of calcium nitrate

(Ca(NO3)2) would you need to prepare 3.00 L

of a 0.500M solution?

3.00 L 3 0.500 mol Ca(N O 3 ) 2 __

1 L 3

164.09 g Ca(NO3)2

__ 1 mol Ca(N O 3 ) 2

5 246 g Ca(NO3)2

111. What would be the molality of the solu-

tion described in the previous problem? The

density of the Ca(N O 3 ) 2 solution is 1.08 kg/L.

mass of solution 5 3.00 L 3 1.08 kg solution

__ 1 L solution

5 3.24 kg solution

mass of solute 5 246 g 3 1 kg _

1000 g 5 0.246 kg

mass of solvent 5 3.24 kg solution 2 0.246 kg

solute

moles of Ca(NO3)2 5 3.00 L 3 0.500 mol Ca(N O 3 ) 2

__ 1 L

5 1.50 mol Ca(NO3)2

molality, m 5 1.50 mol Ca(N O 3 ) 2

____ 3.24 kg solution 2 0.246 kg solute

5 0.501m

Think Critically

112. Develop a plan for making 1000 mL of a 5%

by volume solution of hydrochloric acid in

water. Your plan should describe the amounts

of solute and solvent necessary, as well as the

steps involved in making the solution.

% by volume 5 volume solute

__ volume solution

3 100

5% 5 volume solute

__ 1000 mL

3 100

vol solute 5 50 mL

50 mL HCl needed. Subtract the volume of HCl

from the total solution volume to determine

a volume of 950 mL H2O needed. Dissolve

50 mL HCl in somewhat less than 950 mL H2O. Add

water until the volume of the solution is

1000 mL.

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113. Compare and infer Study the phase

diagram in Figure 14.21. Compare the dotted

lines surrounding DTf and DTb and describe

the differences you observe. How might these

lines be positioned differently for solutions of

electrolytes and nonelectrolytes? Why?

Pure solventSolution

LIQUIDSOLID

GAS

Increasing Temperature

Freezingpoint ofsolution

1 atm

Boilingpoint ofsolution

Normalfreezing pointof water

Normalboiling pointof water

Incr

easi

ng

Pre

ssure

Tf

∆ Tb

∆

P∆

Phase Diagram

The freezing point of the solution is below the

normal freezing point of water, while the boiling

point of the solution is above the normal boiling

point of water. DTf and DTb would be larger for

electrolytes than nonelectrolytes. Electrolytes

dissociate in water, resulting in a larger number

of particles in solution.

114. Extrapolate The solubility of argon in water at various pressures is shown in Figure 14.28. Extrapolate the data to 15 atm. Use Henry’s law to verify the solubility deter-mined by your extrapolation.

So

lub

ilit

y (

mg

ga

s/100 g

wa

ter)

70

60

50

40

30

20

10

0

Gas pressure (atm)

2.0 4.0 6.0 8.0 10.0

H2N2

CH4

O2

NO

Ar

Solubility v. Gas Pressure

S1

_ P1

5 S2

_ P2

S2 5 (55 mg Ar/100 g H 2 O)(15 atm)

___ (10.0 atm)

5 82 mg Ar/100 g H2O

115. Infer Dehydration occurs when more fluid

is lost from the body than is taken in. Scuba

divers are advised to hydrate their bodies

before diving. Use your knowledge of the

relationship between pressure and gas solubility

to explain the importance of hydration prior

to a dive.

As pressure increases with water depth during

a dive, gas concentration in the blood increases.

If blood (solvent) volume is low, the gas (solute)

concentration will be higher than normal levels

at specific depths. A well-hydrated diver has a

greater amount of solvent in which gases can be

dissolved.

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116. Graph Table 14.10 shows solubility data was

collected in an experiment. Plot a graph of the

molarity of KI versus temperature. What is the

solubility of KI at 55°C?

0 20 40 60 80 1000

8

9

10

11

12

13Solubility vs. Temperature

Temperature (°C)

So

lub

ilit

y (

M)

Molarity equals 8.67M, 9.76M, 10.6M, 11.6M,

and 12.4M at 20°C, 40°C, 60°C, 80°C, and 100°C,

respectively. The solubility of KI at 55°C is about

10.4M.

117. Design an Experiment You are given a

sample of a solid solute and three aqueous

solutions containing that solute. How would

you determine which solution is saturated,

unsaturated, and supersaturated?

Add a pinch of solute to each container. If

the solution is supersaturated, crystallization

will occur; saturated, no solute will dissolve;

unsaturated, solute will dissolve.

118. Compare Which of the following solutions

has the highest concentration? Rank the solu-

tions from the greatest to the smallest boiling

point depression. Explain your answer.

a. 0.10 mol NaBr in 100.0 mL solution

b. 2.1 mol KOH in 1.00 L solution

c. 1.2 mol KMnO4 in 3.00 L solution

The molarities are 1.0M NaBr, 2.1M KOH, and

0.40M KMnO4. Because the KOH solution

contributes the greatest concentration of

particles to solution, it has the greatest

boiling point elevation; KMnO4 has the lowest

concentration of particles and the smallest

boiling point depression. Boiling point elevation

depends only upon concentration.

0.10 mol NaBr

__ 0.1000 L

5 1.0M NaBr

2.1 mol KOH

__ 1.00 L

5 2.1M KOH

1.2 mol KMnO4

__ 3.00 L

5 0.40M KMnO4

Challenge

Measurements of Solubility of a Gas

Measurement Solubility

1 0.225

2 0.45

3 0.9

4 1.8

5 3.6

119. Interpret the solubility data in Table 14.11 using the concept of Henry’s Law.

In Henry’s law, solubility is directly proportional

to pressure. In this example, each measurement

indicates a doubling of the solubility

value. This indicates that the pressure is

also doubling between measurements. An

additional observation might include that from

Measurement 1 to Measurement 5, the solubility

has increased by a factor of 16. Therefore, the

pressure would do the same.

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120. You have a solution containing 135.2 grams

of dissolved KBr in 2.3 liters of water. What

volume of this solution, in mL, would you use

to make 1.5 liters of a 0.1M KBr? What is the

boiling point of this new solution?

Step1: Calculate molarity of original solution

135.2 g KBr 3 1 mol KBr

_ 119 g KBr

5 1.14 mol KBr

M 5 1.14 mol KBr

__ 2.3 L H2O

5 0.496 M

Step 2: Dilute the solution – Calculate required

volume

V1 5 0.10M 3 1.5 L

__ 0.496M

5 0.30 L 3 1000 mL _ 1 L

5 300 mL

Step 3: Calculate boiling point of new solution

ΔTb 5 Kbm

m 5 0.10 mol KBr

__ 1 L H2O

3 1 L H2O __

1000 mL H2O 3

1 mL H2O

_ 1 g H2O

1 1000 g H2O

__ 1 kg H2O

5 0.10m


ΔTb 5 0.512°C/m 3 0.20m 5 0.10°C

Tb 5 100.0°C 1 0.10°C 5 100.1°C

Cumulative Review

121. The radius of an argon atom is 94 pm.

Assuming the atom is spherical, what is the

volume of an argon atom in nm3? V 5 4/3πr3.

(Chapter 3)

94 pm 3 1 nm _

1000 pm 5 0.094 nm

V 5 (4/3)(3.14)(0.094 nm)3 5 3.5 3 1023 nm3

122. Identify which of the following molecules is

polar. (Chapter 8)

a. SiH4

nonpolar

b. NO2

polar

c. H2S

polar

d. NCl3

polar

123. Name the following compounds. (Chapter 7)

a. NaBr

sodium bromide

b. Pb(CH3COO)2

lead(II) acetate

c. (NH4)2CO3

ammonium carbonate

124. A 12.0-g sample of an element contains

5.94 3 1022 atoms. What is the unknown

element? (Chapter 10)

5.94 3 1022 atoms 3 1 mol

__ 6.02 3 1023 atoms

5 0.0987 mol

12.0 g __

0.0987 mol 5 122 g/mol

The atomic mass is 122 amu. The element is

antimony.

125. Pure bismuth can be produced by the reaction

of bismuth oxide with carbon at high

temperatures.

2Bi2O3 1 3C 0 4Bi 1 3CO2

How many moles of Bi2O3 reacted to produce

12.6 mol of CO2? (Chapter 11)

12.6 mol CO2 3 2 mol Bi2O3

__ 3 mol CO2

5 8.40 mol Bi2O3

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Additional Assessment

Writing in Chemistry

126. Homogenized Milk The first homogenized

milk was sold in the United States around 1919.

Today, almost all milk sold in this country is

homogenized, in the form of a colloidal

emulsion. Research the homogenization

process. Write a brief article describing the

process. The article may include a flowchart or

diagram of the process, as well as a discussion

of the reputed benefits and/or drawbacks

associated with drinking homogenized milk.

Student answers will vary. Students should note

that raw milk contains fat dispersed throughout.

If left to stand, the fat separates out, leaving a

cream layer and a skim milk layer. The process

of homogenization breaks the fat globules into

smaller sizes and reduces their tendency to form

a cream layer.

Document-Based Questions

127. Are dissolved oxygen values most closely

related to latitude or longitude? Why do you

think this is true?

Dissolved oxygen values are most closely related

to latitude. Surface land and water temperatures

are more closely correlated to latitude than

longitude.

128. At what latitude are average dissolved oxygen

values the lowest?

Values are lowest near the equator.

129. Describe the general trend defined by the data.

Relate the trend to the relationship between

gas solubility and temperature.

In general, dissolved oxygen in surface

ocean waters increases as latitude increases

towards both north and south. Surface water

temperatures are greatest near the equator.

Surface water temperature decreases toward the

poles. As temperature decreases, gas solubility

generally increases.

Standardized Test Practicepages 512–513

Percent by mass

Percent by volume

Bromine (Br2) Concentration ofFour Aqueous Solutions

0.9000

0.8000

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000

Pe

rce

nt

1 32 4

Solution number

0.7

94

7

0.2

57

5

0.3

18

9

0.1

59

6

0.4

77

9

0.1

54

50.1

03

0

0.0

515

1. What is the volume of bromine (Br2) in 7.000 L

of Solution 1?

a. 55.63 mL

b. 8.808 mL

c. 18.03 mL

d. 27.18 mL

c

Volume of Br2 5 (7.000 L) 3 (0.002575) 5

0.01803 L 5 18.03 mL

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2. How many grams of Br2 are in 55.00 g of

Solution 4?

a. 3.560 g

b. 0.084 98 g

c. 1.151 g

d. 0.2628 g

d

Mass of Br2 5 (55.00 g) 3 (0.004779) 5 0.2628 g

3. Which one is an intensive physical property?

a. volume

b. length

c. hardness

d. mass

c

4. What is the product of this synthesis reaction?

Cl2(g) 1 2NO(g) 0?

a. NCl2b. 2NOCl

c. N2O2

d. 2ClO

b

5. If 1 mole of each of the solutes listed below

is dissolved in 1 L of water, which solute will

have the greatest effect on the vapor pressure of

its respective solution?

a. KBr

b. C6H12O6

c. MgCl2d. CaSO4

c

MgCl2 will produce the most number of particles

in solution: 1 mol Mg21, 2 mol Cl2

C

77.9%

H11.7%

O

10.4%

6. What is the empirical formula for this

substance?

a. CH2O

b. C8HO

c. C10 H18O

d. C7 H12O

c

Assume a 100.0 g sample.

Determine the number of moles.

77.9 g C 3 1 mol C

_ 12.01 g C

5 6.486 mol C

10.4 g O 3 1 mol O

_ 16.00 g O

5 0.65 mol O

11.7 g H 3 1 mol H

_ 1.008 g H

5 11.607 mol H

Calculate the simplest ratio of moles.

6.486 mol C

__ 0.65

5 9.98; 0.65 mol O

__ 0.65

5 1.00;

11.607 mol H

__ 0.65

5 17.86

The empirical formula is C10H18O.

7. What is the correct chemical formula for the

ionic compound formed by the calcium ion

(Ca21) and the acetate ion (C2H3O22)?

a. CaC2H3O2

b. CaC4H6O3

c. (Ca)2C2H3O2

d. Ca(C2H3O2)2

d

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8. If 16 moles of H2 are used, how many moles of Fe will be produced?

a. 6b. 3c. 12

d. 9

c

Fe3O4(s) 1 4 H2(g) → 3 Fe(s) 1 4 H2O(l)

16 mol H2 3 3 mol Fe

_ 4 mol H2

5 12 mol Fe

9. If 7 moles of Fe3O4 are mixed with 30 moles of

H2, what will be true?

a. There will be no reactants left.

b. 2 moles of hydrogen gas will be left over

c. 30 moles of water will be produced

d. 7 moles of Fe will be produced

b

Fe3O4(s) 1 4 H2(g) → 3 Fe(s) 1 4 H2O(l)

7 mol Fe3O4 3 4 mol H2

__ 1 mol Fe3O4

5 28 mol H2 used

30 mol H2 2 28 mol used 5 2 mol H2 remaining

10. What is the molar mass of Fe3O4?

a. 231.56 g/mol

b. 71.85 g/mol

c. 287.40 g/mol

d. 215.56 g/mol

a

3 mol Fe 3 55.847 g/mol 5 167.541 g Fe

4 mol O 3 15.999 g/mol 5 63.996 g O

molar mass = 167.541 g 1 63.996 g 5 231.537 g/mol

100 20 30 50 60 70 80 90 10040

So

lub

ilit

y (

g o

f so

lute

/100 g

H2O

)

100

90

80

70

60

50

40

30

20

10

0

Temperature (°C)

Solubilities as a Functionof Temperature

Ce2(SO

4)3

KClO3

KCl

CaCl2

NaCl

11. How many moles of KClO3 can be dissolved in

100 g of water at 60°C?

21 grams

12. Which can hold more solute at 208C: NaCl

or KCl? How does this compare to their

solubilities at 808C?

At 20°C, the NaCl solution can hold more solute.

At 80°C, the solubilities are reversed and KCl is

more soluble than NaCl.

13. How many moles of KClO3 would be required

to make 1 L of a saturated solution of KClO3 at

75°C?

(30 g/L)(1 mol/122.55 g KClO3) 5 0.245 mol KClO3

in 1 liter.

Use the information below to answer

Questions 14 and 15.

The electron configuration for silicon is 1s2 2s2 2p6

3s2 3p2.

14. Explain how this configuration demonstrates

the aufbau principle.

The aufbau principle dictates that electrons must

fill the lowest available energy levels before

filling any higher energy levels.

15. Draw the orbital diagram for silicon. Explain

how Hund’s rule and the Pauli exclusion

principle are used in constructing the orbital

diagram.

Hund’s rule mandates that the last two electrons

will be placed in separate p-orbitals. The Pauli

exclusion principal determines that shared elec-

trons in any given orbital must have opposite

spins, as shown by up and down arrows.

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑

1s 2s 2p 3s 3p

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16. What volume of a 0.125M NiCl2 solution

contains 3.25 g NiCl2?

a. 406 mL

b. 32.5 mL

c. 38.5 mL

d. 26.0 mL

e. 201 mL

e

3.25 g 3 1 mol

_ 129.6 g

3 1 L _

0.125 mol 3

1000 mL _

1 L

5 201 mL

17. Which is NOT a colligative property?

a. boiling point elevation

b. freezing point depression

c. vapor pressure increase

d. osmotic pressure

e. heat of solution

c

Use the data table below to answer

Questions 18 and 19.

Electronegativities of Selected Elements

H

2.20

Li Be B C N O F

0.98 1.57 2.04 2.55 3.04 3.44 3.93

Na Mg Al Si P S Cl

0.93 1.31 1.61 1.90 2.19 2.58 3.16

18. What is the electronegativity difference in the

compound Li2O?

a. 1.48

b. 2.46

c. 3.4

d. 4.42

e. 5.19

b

3.44 – 0.98 5 2.46

19. Which bond has the greatest polarity?

a. C2H

b. Si2O

c. Mg2Cl

d. Al2N

e. H2Cl

c

C-H: 2.55 2 2.20 5 0.35

Si-O: 3.44 2 1.90 5 1.54

Mg-Cl: 3.16 2 1.31 5 1.85

Al-N: 3.04 2 1.61 5 1.43

H-Cl: 3.16 2 2.20 5 0.96

Mg-Cl has the greatest polarity.

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Mixtures and Solutions - Weebly...15. Challenge If 18 mL of methanol are used to make an aqueous solution that is 15% methanol by volume, how many milliliters of solution are produced?