Chapter 15

Homework 15A

Page 484: 48-53, 66-68

48. 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.

49. What determines whether a solute will be soluble in a given solvent?

the polarity of the solute and solvent (like dissolves like), temperature, and (for gases) pressure.

50. 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.

51. What does it mean if two liquids are said to be miscible?

Two liquids are miscible when they are soluble in each other in any proportion (ratio).

52. What are three ways to increase the rate of solvation?

increase the temperature of the solvent, increase the surface area of the solute, agitation (stirring)

53. Why are gases less soluble at higher temperatures?

An increase in temperature increases the kinetic energy of the gas particles. More gas particles escape the surface of the solution.

66. The solubility of a gas is 2.0 g/L at 50.0 kPa of pressure. How much gas will dissolve in 1 L at a pressure of 10.0 kPa?

67. The solubility of a gas is 4.5 g/L at a pressure of 1.0 atm. At what pressure will there be 45 g of gas in 1.0 L of solution?

0.40g/L 50.0kPa

g/L) 2.0(10.0kPa)(

P

SP S

P

S

P

S

1

122

2

2

1

1

atm 10. g/L 4.5

g/L) atm)(45 (1.0

S

SP P

P

S

P

S

1

212

2

2

1

1

68. 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 x 10-4 atm. What is the solubility of CO2 in the open drink? Express your answer in grams per liter.

P

SP S

P

S

P

S

1

122

2

2

1

1

mol/L 6-9.0E atm 4.0

mol/L) 4atm)(0.12-(3.0E

P

SP S

1

122

4g/L-4.0E mol 1.0

g 44.0

L

mol6-E .09 x

CO2 = 44.0 g/mol

Chapter 15

Homework 15B

Page 484: 69-75

Page 485: 76-79 (part “a”)

69. Calculate the percent by mass of 3.55 g NaCl dissolved in 88 g water.

percent by mass = (g solute/g solution) x 100

70. Calculate the percent by mass of benzene in a solution containing 14.2 g of benzene in 28.0 g of carbon tetrachloride.

3.9%m/m 100x OH 88g NaCl 3.55g

NaCl 3.55g

2

33.6%m/m 100x CCl 28g HC 14.2g

Benzene 14.2g

466

71. What is the percent by volume of 25 mL of methanol (MeOH) in 75 mL of water?

72. A solution is made by adding 1.23 mol KCl to 1000.0 g of water. What is the percent by mass of KCl in this solution?

25%v/v 100x OH 75mL MeOH 25mL

MeOH 25mL

2

KCl 91.8g 1mol

g 74.6 x KCl .23mol1

8.41%m/m 100x OH 1000.0g KCl 91.8g

KCl 91.8g

2

KCl = 39.1 + 35.5 = 74.6 g/mol

73. What mass of water must be added to 255.0 g NaCl to make a 15.00 percent by mass aqueous solution?

solution 100.g

NaCl 15.00g m/m 5.00%1

solution 1700.g NaCl g 255.0 x NaCl 15.00g

solution 100.g

Invert to cancel “NaCl”

1700.g solution – 255.0g NaCl = 1,445g H2O

74. The label on a 250-mL stock bottle reads “21.5% alcohol by volume.” What volume of alcohol does it contain?

solution 100.g

alcohol 21.5mL v/v1.5%2

alcohol 54mL solution mL 250 x solution 100.mL

alcohol 21.5mL

2 sig figs!

75. A 14.0 percent by mass solution of potassium iodide dissolved in water has a density of 1.208 g/mL. How many grams of KI are in 25.0 mL of the solution?

solution 30.2g 1mL

1.208g solution x 5.0mL2

solution 100.g

KI 14.0g m/m % 4.01

KI 4.23g solution 100.g

KI 14.0g solution x 0.2g3

76. What is the molarity of the following solutions?a. 2.5 mol KCl in 1.0 L of solution

77. What is the molarity of the following solutions?a. 0.96 g MgCl2 in 500 mL of solution

2.5M solution 1.0L

KCl 2.5mol

0.02M solution 0.5L

MgCl 0.010mol 2

22 MgCl 0.010mol 95.3g

1mol x MgCl .96g0

MgCl2 = 24.3 + 2(35.5) = 95.3 g/mol

0.5L 1000mL

1L x 00mL5

78. How many moles of solute are contained in the following solutions?a. 15.25 mL 2.10M CaCl2

2CaCl 0.0320mol L 1

2.10mol x 0.01525L

0.01525L 1000mL

1L x 5.25mL1

2.10M = 2.10mol/L

79. How many grams of solute are contained in the following solutions?a. 64.3 mL 0.0238M KOH

KOH 0.0859g 1mol

56.1g x

L 1

0.0238mol x 0.0643L

0.0643L 1000mL

1L x 4.3mL6

0.0238M = 0.0238mol/L

KOH = 39.1 + 16.0 + 1.0 = 56.1 g/mol

Chapter 15

Homework 15C

Page 485: 80-85

80. How many milliliters of 2.55M NaOH is needed to make 125 mL 0.75M NaOH solution?

C1 = 2.55M; V1 = ???; C2 = 0.75M; V2 = 125mL

81. How many milliliters of 0.400M HBr solution can be made from 50.0 mL of 8.00M HBr solution?

C1 = 8.00M; V1 = 50.0; C2 = .400M; V2 = ???mL

37mL 2.55M

5mL)(0.75M)(12

C

VC V ;VCVC Since

1

2212211

0mL010 0.400M

.0mL)(8.00M)(50

C

VC V ;VCVC Since

2

1122211

(Or, 1.00 x 103 mL)

82. What is the molarity of each resulting solution when the following mixtures are prepared? a. 500.0 mL H2O is added to 20.0 mL 6.00M HNO3

500.0mL + 20.0mL = 520.0mL

C1 = 6.00M; V1 = 20.0mL; C2 = ???M; V2 = 520.0mL

0.231M 520.0mL

.0mL)(6.00M)(20

V

VC C ;VCVC Since

2

1122211

82b. 30.0 mL 1.75M HCl is added to 80.0 mL 0.450M HCl(tricky! Mixing two solutions of different concentrations!)

0.0300L + 0.0800L = 0.1100L

0.0525mol 1L

(1.75mol) x0.0300L

0.0360mol 1L

(0.450mol) x0.0800L

0.805M 0.110L

0.0525mol) (0.0360mol

0.0300L 1000mL

1L x30.0mL 0.0800L

1000mL

1L x80.0mL

83. Calculate the molality of the following solutions.a. 15.7 g NaCl in 100.0 g H2O

Molality – moles solute/Kg solvent: NaCl = 58.5 g/mol

NaCl 0.268mol 58.5g

1mol xNaCl 15.7g

OH 0.1000Kg 1000g

1Kg xOH 100.0g 22

m2.68 0.1000Kg

mol 0.268

83. Calculate the molality of the following solutions.b. 20.0 g CaCl2 in 700.0 g H2O

Molality – moles solute/Kg solvent: CaCl2 = 111.1g/mol

22 CaCl 0.180mol 111.1g

1mol xCaCl 20.0g

OH 0.7000Kg 1000g

1Kg xOH 700.0g 22

m0.257 0.7000Kg

mol 0.180

83. Calculate the molality of the following solutions. c. 3.76 g NaOH in 0.850 L H2O

Molality – moles solute/Kg solvent: NaOH = 40.0g/mol

NaOH 0.0940mol 40.0g

1mol xNaOH 3.76g

OH 0.850Kg 1L

1.00Kg xOH 0.850L 22

m0.111 0.850Kg

mol 0.0940

Since the density of water = 1.00g/mL, this is the same ratio as 1.00Kg/L - (multiplied top and bottom by 1000)

84. Calculate the mole fraction of NaCl, CaCl2, and NaOH in the solutions listed in the previous problem.a. 15.7 g NaCl in 100.0 g H2O

Mole fraction = moles solute/total moles

NaCl 0.268mol 58.5g

1mol xNaCl 15.7g

OH mol 5.56 18.0g

1mol xOH 100.0g 22

0.0460 mol 5.56 mol 0.268

mol 0.268 X

(mole fraction has NO units!)

84. Calculate the mole fraction of NaCl, CaCl2, and NaOH in the solutions listed in the previous problem. b. 20.0 g CaCl2 in 700.0 g H2O

Mole fraction = moles solute/total moles

22 CaCl 0.180mol 111.1g

1mol xCaCl 20.0g

OH mol 38.9 18.0g

1mol xOH 700.0g 22

0.00461 mol 38.9 mol 0.180

mol 0.180 X

84. Calculate the mole fraction of NaCl, CaCl2, and NaOH in the solutions listed in the previous problem.c. 3.76 g NaOH in 0.850 L H2O

Mole fraction = moles solute/total moles

NaOH 0.0940mol 40.0g

1mol xNaOH 3.76g

OH 47.2mol 18.0g

1mol x

1mL

1.00g x

1L

1000mL xOH 0.850L 22

.001990 47.2mol 0.0940mol

mol 0.0940 X

(Density of water = 1.00g/mL)

85. What are the molality and mole fraction of solute in a 35.5 percent by mass aqueous solution of formic acid (HCOOH)?

“%” = “parts out of 100”, so:

100g solution – 35.5g solute = 64.5g water!

Molality = moles solute /Kg solvent.

Formic acid (F/A) = 2(1.0) + 2(16.0) + 12.0 = 46.0 g/mol

solution 100g

HCOOH 35.5g m/m 35.5%

F/A 0.772mol 46.0g

1mol F/A 35.5g x

OH 0.0645Kg 1000g

1Kg OH 64.5g 22 x

m12.0 0.0645Kg

0.772mol

85. What are the molality and mole fraction of solute in a 35.5 percent by mass aqueous solution of formic acid (HCOOH)?

“%” = “parts out of 100”, so:

100g solution – 35.5g solute = 64.5g water!

Mole fraction = moles solute/total moles

solution 100g

HCOOH 35.5g m/m 35.5%

F/A 0.772mol 46.0g

1mol F/A 35.5g x

OH 3.58mol 18.0g

1mol OH 64.5g 22 x

.1770 3.58mol 0.772mol

0.772mol

Chapter 15

Homework 15D

Page 485: 86-89

86. Using the information in Tables 15-4 and 15-5, calculate the freezing point and boiling point of 12.0 g of glucose (C6H12O6) in 50.0g H2O.

Find molality, then multiply by Kf and Kb of water:

Molality = moles solute /Kg solvent.

Glucose (G) = 6(12.0) + 12(1.0) + 6(16.0) = 180.0 g/mol G 0.0667mol

180.0g

1mol G 12.0g x

OH 0.0500Kg 1000g

1Kg OH 50.0g 22 x

m1.33 0.0500Kg

0.0667mol

C2.47 1.33 molal

C1.86 molality K OH T f2f

mxx

Freezing point = 0.00C - 2.47 = -2.47 C

C681.0 1.33 molal

C512.0 molality K OH T b2b

mxx

Boiling point = 100.000C + 0.681 = 100.681 C

Note: glucose is a non-electrolyte! (i = 1) Had this been an electrolyte, the molality would have been 2,3,4 or more times higher.

87. Using the information in Tables 15-4 and 15-5, calculate the freezing point and boiling point of each of the following solutions.a. 2.75m NaOH in water (NaOH is an electrolyte)

(i = 2): NaOH Na+1 + OH-1, so the particle molality = (2)(2.75m) = 5.50m

C2.01 5.50 molal

C1.86 molality K OH T f2f

mxx

Freezing point = 0.00C – 10.2 = -10.2C

C82.2 5.50 molal

C512.0 molality K OH T b2b

mxx

Boiling point = 100.00C + 2.82 = 102.82 C

87. Using the information in Tables 15-4 and 15-5, calculate the freezing point and boiling point of each of the following solutions.b. 0.586m of water in ethanol (EtOH)

(i = 1): water is NOT an electrolyte! so the particle molality =(1)(0.586m) = 0.586m

C17.1 0.586 molal

C1.99 molality K EtOH T ff

mxx

Freezing point = -114.1C – 1.17 = -115.3C

C715.0 0.586 molal

C1.22 molality K EtOH T bb

mxx

Boiling point = 78.5C + 0.715 = 79.2 C

87. Using the information in Tables 15-4 and 15-5, calculate the freezing point and boiling point of each of the following solutions. c. 1.26m of naphthalene (C10H8) in benzene (C6H6)

(i = 1): NOT an electrolyte: so the particle molality =(1)(1.26m) = 1.26m

C45.6 1.26 molal

C12.5 molality K HC T f66f

mxx

Freezing point = 5.5C – 6.45 = -1.0C

C19.3 1.26 molal

C53.2 molality K HC T b66b

mxx

Boiling point = 80.1C + 3.19 = 83.3 C

88. 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 10.0°C?Find the molality of particles, then the molality of the NaCl that is needed, then the grams of NaCl…

5.38 mol particles are needed per Kg water!

Since i for NaCl = 2,

m

m

mx 5.38 C1.86

C10.0

K

OH T m ; K OH T Since

f

2ff2f

NaCl of .692 particles 2

1NaCl xparticles m38.5 m

NaCl Kg

157g

mol 1

g 58.5 x

Kg 1

NaCl .69mol2 157g NaCl needed for

each Kg of water!

89. Calculate the freezing point and boiling point of a solution that contains 55.4 g NaCl and 42.3 g KBr dissolved in 750.3 mL H2O.

Find the moles of each, and the moles of particles of each!

1.89mol + 0.711mol = 2.60 total moles of particles

particles 1.89mol NaCl

particles 2 x

58.5g

1mol NaCl 55.4g x

particles 0.711mol KBr

particles 2 x

119.0g

1mol KBr 42.3g x

m3.47 Kg .7503

mol 2.60

7503Kg. 1000g

1Kg x

1mL

1.00g x 750.3mL

C45.6 3.47 molal

C1.86 molality K OH T f2f

mxx

Freezing point = 0.00C – 6.45 = -6.45 C

C78.1 3.47 molal

C512.0 molality K OH T b2b

mxx

Boiling point = 100.00C + 1.78 = 101.78 C