10

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SOLUTIONS

LECTURE NOTES
This chapter is coverable in two lectures. The first deals with concentration units, the second with colligative
properties. Note that molarity first appeared in Chapter 4. Discussed here for the first time are the calculations
necessary for preparing a solution of known molarity from a more concentrated solution. The important point
to get across is that the number of moles of solute stays the same.
Perhaps the most difficult topic in this chapter is the conversion from one concentration unit to another.
Students typically dont know where to start. The unnumbered table on p. 262 of the text should be helpful.

Lecture 1
I. Concentration Units
A. Molarity
no. of moles solute
no. of liters solution

M=

1. How to prepare 35.0 mL of 0.200 M Al(NO3 )3 from solid?

molar mass = (26.98 + 42.03 + 144.00) g/mol = 213.01 g/mol
moles needed = 0.0350 L 0.200 mol/L = 7.00 103 mol
mass needed = 7.00 103 mol 213.01 g/mol = 1.49 g
Thus, dissolve 1.49 g of aluminum nitrate in enough water to form 35.0 mL of solution.
2. How to prepare 35.0 mL of 0.200 M aluminum nitrate from a 0.500 M solution?
Note that the number of moles of solute stays the same.
0.0350 L 0.200 mol/L = (0.500 mol/L) V

V = 0.0140 L = 14.0 mL.

Hence, dilute to 35.0 mL with water.

B. Mole fraction

XA =

no. of moles of A
total no. of moles

Dissolve 12.0 g CH3 OH in 100.0 g water. What are the mole fractions of CH3 OH and H2 O?
mol CH3 OH = 12.0 g 1 mol/32.0 g = 0.375

XCH3 OH = 0.375/(0.375 + 5.549) = 0.0633;

mol H2 O = 100.0 g 1 mol/18.02 g = 5.549

XH2 O = 1 0.0633 = 0.9367.

105

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Chapter 10
C. Mass percent solute
mass % =

mass solute
100
total mass solution

ppm (liquid solution) = mass % 104

ppb (liquid solution) = mass % 107

D. Molality

m=

no. of moles solute

no. of kg solvent

What is the molality of the solution of CH3 OH in (B) above?

m = 0.375/0.1000 = 3.75 mol/kg solvent.

E. Conversions between units
What is the molality of a 0.200 M aluminum nitrate solution (d = 1.012 g/mL)?
Work with one liter of solution:

mass = 1012 g.

mass Al(NO3 )3 = 0.200 mol 213.01 g/mol = 42.6 g;

Thus, molality =

mass water = 1012 g 43 g = 969 g

0.200 mol
= 0.206 mol/kg.
0.969 kg

II. Principles of Solubility

A. Nature of solute and solvent
Most nonelectrolytes that are appreciably soluble in water are hydrogen bonded (methyl alcohol,
hydrogen peroxide, ...). Other types of nonelectrolytes are generally more soluble in nonpolar or
slightly polar solvents such as benzene or toluene.

Lecture 2
II. Principles of Solubility (continued from Lecture 1)
B. Effect of temperature
Increase in T favors an endothermic process.
solid + water solution

H usually positive, so solubility usually increases with T.

gas + water solution

H usually negative, so solubility usually decreases with T.

C. Effect of pressure
This is negligible, except for gases for which solubility is directly proportional to the partial
pressure of the gas. Beverages are carbonated under high pressure.
III. Colligative Properties of Nonelectrolytes
These depend primarily upon the concentration of solute particles rather than type.

Solutions

107

A. Vapor pressure lowering (Raoults law)

P1 = X1 P1 ; P = X2 P1 , where X1 = the mole fraction of the solvent, X2 = the mole fraction of the
solute, P1 = the vapor pressure of the solution and P1 = the vapor pressure of pure solvent.
B. Boiling point elevation, freezing point lowering
1. Results from vapor pressure lowering (see Figure 10.8 in Section 10.3 of the text).
Tf = 1.86 C m,

2. For water solutions:

Tb = 0.052 C m

Determine the freezing point and boiling point of a solution containing 20.0 g of ethylene glycol
(MM = 62.0 g/mol) in 50.0 g water.
Since m =

20.0/62.0
= 6.45, it follows that
0.0500
Tf = 6.45 1.86 C = 12.0 C;

Tf = 12.0 C

Tb = 6.45 0.52 C = 3.4 C;

Tb = 103.4 C

3. Use in determining molar mass.

Suppose a solution is prepared by dissolving 0.100 g of nonelectrolyte in 1.00 g of water. If its
freezing point is found to be 1.00 C, then what is its molar mass?
molality =

1.00 0.100/MM
=
;
1.86
0.00100

MM =

0.186
= 186 g/mol
0.00100

C. Osmosis, osmotic pressure

Water moves through a semipermeable membrane from a region of high vapor pressure (pure
water) to one of low vapor pressure (solution).
= MRT;

a 1 M solution at 25 C has an osmotic pressure of 24.5 atm.

IV. Colligative Properties of Electrolytes

Colligative effects are larger because of the increased number of solute particles.
Tf = 1.86 C m i,
where i is approximately equal to the number of moles of ions per mole of solute.
NaCl Na+ (aq) + Cl (aq)

i=2

CaCl2 (s) Ca2+ (aq) + 2Cl (aq)

i=3

Actually, i is usually less than predicted because of ion atmosphere effects.

108

Chapter 10

DEMONSTRATIONS
1. Gas solubility: SHAK 2 205, J. Chem. Educ. 69 573 (1992)
2. Effect of temperature on solubility: GILB F 10, SHAK 3 280
3. Raoults law: GILB F 37, F 38, SHAK 3 242, 254
4. Boiling point elevation: SHAK 3 297
5. Freezing point lowering: SHAK 3 290, J. Chem. Educ. 68 1038 (1991)
6. Osmosis: GILB F 41, SHAK 3 283
7. Osmotic pressure: GILB F 42, SHAK 3 286

SUMMARY PROBLEM
(a) mass C6 H6 = 725 mL 0.879 g/mL = 637 g
mass % palmitic acid =
(b) mol C16 H32 O2 =

M=

Vsolution =

112 g + 637 g
= 830 mL
0.902 g/mL

637 g
= 8.16
78.1 g/mol

P1 =

8.16
1.00 102 = 95 mm Hg
8.16 0.438

0.438 mol
= 0.688
0.637 kg

Tb =
Tf =
(e) M =

112 g
= 0.438
256.4 g/mol

0.438 mol
= 0.527 mol/L
0.830 L

(c) mol C6 H6 =
(d) m =

112 g
100 = 15.0%
112 g + 637 g

2.53 C

m
5.10 C

RT

(0.688 m) = 1.74 C;

Tb = 80.10 + 1.74 = 81.84 C

(0.688 m) = 3.51 C;

Tf = 5.50 3.51 = 1.99 C

1.45 atm
= 0.0589 mol/L
(0.0821 L atm/mol K)(300 K)

mol cholesterol = 0.0589 mol/L 0.525 L = 0.0309

MM =

12.0 g
= 388 g/mol
0.0309 mol

Solutions

109

PROBLEMS
1. (a)

12.15 g
100% = 6.49%
12.15 g + 175 g

(b) mol Ni(NO3 )2 =

12.15 g
= 0.06650;
182.7 g/mol

0.06650
= 0.00671
0.06650 + 0.1333 + 9.71

3. Assume 100.0 g of vinegar (solution);

mol HC2 H3 O2 =

175 g
= 9.71
18.02 g/mol

mol NO3 = 2(mol Ni(NO3 )2 ) = 0.1333

mol Ni2+ = mol Ni(NO3 )2 = 0.06650;

XNi(NO3 )2 =

mol H2 O =

5.00% HC2 H3 O2 = 5.00 g HC2 H3 O2 in solution

5.00 g
= 0.0833;
60.05 g/mol

Vsol0 n =

100.0 g
= 99.4 mL
1.006 g/mL

M = 0.0833 mol/0.0994 L = 0.838 M

5. (a) 8 glasses of H2 O
(b) 1.00 g Ag

8 oz
1 lb
453.6 g 0.028 g Ag+
1000 mg

= 0.051 mg
6
1 glass 16 oz
1 lb
1g
10 g H2 O

106 g H2 O
1 L H2 O

= 3.6 104 L
0.028 g Ag 1000 g H2 O

7. Molar mass of CuSO4 = 159.62 g/mol

(a) M =

(12.50/159.62) mol
= 0.164 mol/L
0.478 L

(b) mass = (0.299 mol/L)(0.283 L)(159.62 g/mol) = 13.5 g

(c) V =

(4.163 g)(1 mol/159.62 g)

= 0.03099 L = 30.99 mL
0.8415 mol/L

9. Molar mass of caffeine = 194.2 g/mol

(a) Assume 1.000 mole of solution.

Thus solution has 0.900 mol H2 O, and 0.100 mol caffeine

mass H2 O = 0.900 mol(18.02 g/mol) = 16.2 g;

m=

0.100 mol
= 6.17
0.0162 kg

ppm solute =

16.2 g
106 = 4.55 105
(16.2 + 19.4) g

mass caffeine = 0.100 mol(194.2 g/mol) = 19.4 g

mass % =

16.2 g
100 = 45.5%
(16.2 + 19.4) g

110

Chapter 10
(b) Assume 1.000 106 g of solution.
mol H2 O =

m=

9.987 105 g
= 5.542 104 ;
18.02 g/mol

mol caffeine =

6.535 mol
= 6.543 103
998.7 kg

XH2 O =

mol H2 O =

mass % =

85.5 g
= 4.74 ;
18.02 g/mol

mol caffeine =

14.5 g
= 0.0747
194.2 g/mol
ppm =

14.5 g
106 = 1.45 105
100.0 g

4.74 mol
= 0.984
(4.74 + 0.0747) mol

(d) Assume 1000.00 g of solvent (H2 O).

Thus solution has 0.2560 mol caffeine.

mol H2 O =

1000.00 g
= 55.49 ;
18.02 g/mol

mass % =

1000.00 g
100 = 95.26%
(1000.00 + 49.72) g

XH2 O =

9.987 105 g
100 = 99.87%
106 g

Thus solution has 85.5 g H2 O, and 14.5 g caffeine.

0.047 mol
= 0.873
0.0855 kg

XH2 O =

1269 g
= 6.535
194.2 g/mol

5.542 104 mol

= 0.9999
(5.542 104 + 6.535) mol

(c) Assume 100.0 g of solution.

m=

Thus solution has 9.987 105 g H2 O, and 1269 g caffeine.

mass caffeine = (0.2560 mol)(194.2 g/mol) = 49.72 g

ppm =

49.72 g
106 = 4.737 104
(1000.00 + 49.72) g

55.49 mol
= 0.9954
(55.49 + 0.2560) mol

11. (a) MM K2 Cr2 O7 = 294.20 g/mol;

Mass K2 Cr2 O7 needed = (0.465 L)(0.3550 mol/L)(294.2 g/mol) = 48.6 g

Weigh out 48.6 g, dissolve in enough water to form 465 mL of solution.

(b) V(0.750) = (0.465)(0.3550);

V = 0.220 L;

Dilute 0.220 L of 0.750 M K2 Cr2 O7 to 0.465 L.

13. (a) mol Al2 (SO4 )3 = 0.225 L 0.1885 mol/L = 0.0424

(b) MAl2 (SO4 )3 = 0.0424 mol/1.450 L = 0.0292 mol/L

MAl3+ = 2(MAl2 (SO4 )3 ) = 2(0.0292 M) = 0.0584 M;

MSO 2 = 3(MAl2 (SO4 )3 ) = 3(0.0292 M) = 0.0876 M

4

15. Consider one liter of solution weighing 1689 g.

mass H3 PO4 = (0.850)(1689 g) = 1436 g;

MH3 PO4 =
XH3 PO4 =

1436 g
1 mol

= 14.7 mol/L;
1.00 L
97.99 g
14.7
= 0.511
14.7 + 14.0

mass H2 O = 1689 1436 = 253 g

mH3 PO4 = (14.7 mol)/(0.253 kg H2 O) = 57.9

Solutions

111

17. MMKOH = 56.11 g/mol

(a) Assume one liter of solution;

mass of solution = 1.050 g/mL 1000 ml = 1050 g

KOH:

n = 1.13 mol;

H2 O:

mass = 1050 g 63.4 g = 987 g

mass = 1.13 mol 56.11 g/mol = 63.4 g

m = 1.13 mol/0.987 kg = 1.14

% KOH = (63.4 g/1050 g) 100 = 6.04%

(b) Assume one hundred grams of solution;

n=

volume of solution =

100.0 g
= 77.5 mL
1.29 g/mL

30.0 g
= 0.535 mol
56.11 g/mol

KOH:

mass = 30.0 g;

H2 O:

mass = 100.0 g 30.0 g = 70.0 g

m = 0.535 mol/0.070 kg = 7.64

M = 0.535 mol/0.0775 L = 6.90 mol/L

(c) Assume one thousand grams of solvent (H2 O)

KOH:

n = 14.2 mol;

solution:

mass = 1000.0 + 797 = 1797 g ;

mass = (14.2 mol)(56.11 g/mol) = 797 g

M = 14.2 mol/1.257 L = 11.3 mol/L

volume =

1797 g
= 1257 mL
1.43 g/mL

% KOH = (797 g/1797 g) 100 = 44.4%

19. Assume that the sucrose amount in 30 L of maple sap is negligible.

30 L sap 30 L H2 O 30 kg H2 O
(1/4)(30) = 7.5 kg;
In syrup:

mass of H2 O 22.5 kg

1 kg syrup 66% sucrose = 660 g;

mol sucrose =

660 g
= 1.93
342 g/mol

m = 1.93 mol/22.5 kg = 0.086

21. (a) CCl4 ;
(b) C6 H14 ;

it is nonpolar like benzene

it is nonpolar like benzene

(c) heptanoic acid;

dispersion forces similar to those of benzene

(d) propyl chloride;

dispersion forces similar to those of benzene

23. (a) H2 O2 ; hydrogen bonding

(c) HCl; ionic vs molecular

(b) NaOH; ionic versus network covalent

(d) CH3 OH; hydrogen bonding

25. (a) H = Hf Pb2+ (aq) + 2 Hf Cl (aq) Hf PbCl2 (s) = ( 1.7 334.3 + 359.4) kJ = +23.3 kJ
(b) yes; solution process is endothermic

112

Chapter 10

27. (a) k = 3.8 104

M
atm

(b) C = 5.0 107

1 atm
= 5.0 107 M/mm Hg
760 mm Hg

M
mm Hg

293 mm Hg = 1.5 104 M

(c) mol He = (1.5 104 mol/L)(10.00 L) = 1.5 103

V=

(1.5 103 mol)(0.0821 L atm/mol K)(298 K)

= 0.095 L = 95 mL
(293/760 atm)

29. (a) C = (0.0769 M/atm)(3.0 atm) = 0.23 M

(b) C = (0.0313 M/atm)(3.4 104 atm) = 1.1 105 M
31. Assume one hundred mL of vodka. Mass of vodka = 100.0 g. Volume C2 H5 OH in vodka = 40.0 mL
C2 H5 OH:

mass = 40.0 mL 0.789 g/mL = 31.56 g;

H2 O:

mass = 100.0 - 31.56 = 68.4 g = 0.0684 kg

n=

31.56 g
= 0.685 mol
46.07 g/mol

m = 0.685 mol/0.0684 kg = 10.0

Tf = 1.86

m
C

10.0 m = 18.6 C;

Tf = 18.6 C

33. (a) XH2 O = 1.000 0.288 = 0.712;

PH2 O = (19.83 mm Hg)(0.712) = 14.1 mm Hg

(b) Assume one hundred g of solution;

XH2 O =

(61.0/18.02) mol
= 0.843;
(61.0/18.02) mol + (39.0/62.07) mol

(c) Assume one hundred g of solution;

XH2 O =

mass H2 O = 100.0 39.0 = 61.0 g

mass H2 O = 100.0 39.0 = 61.0 g

(1000.0/18.02) mol
= 0.9582;
(1000.0/18.02) mol + 2.42 mol

35. XCCl4 = P1 /P = 483 mm Hg/504 mm Hg = 0.042;

XCCl4 =

mol CCl4
;
mol CCl4 + mol C10 H8

P = 0.843 19.83 mm Hg = 16.7 mm Hg

0.042 =

P = 0.9582 19.83 mm Hg = 19.00 mm Hg

mol CCl4 =

0.163
;
0.163 + mol C10 H8

mass C10 H8 = 0.00715 mol 128.2 g/mol = 0.92 g

37. (a) = (0.217 mol/L)(0.0821 L atm/mol K)(295 K) = 5.26 atm
(b) M =

25.00 g
= 0.163
153.8 g/mol

(25.0/60.06) mol
= 0.608 mol/L
0.685 L

= (0.608 mol/L)(0.0821 L atm/mol K)(295 K) = 14.7 atm

mol C10 H8 = 0.00715

Solutions

113

(c) Assume one liter of solution;

mol urea =

mass of solution = 1000.0 mL 1.12 g/mL = 1120 g

(1120 0.150) g
= 2.80;
60.06 g/mol

= (2.80 mol/L)(0.0821 L atm/mol K)(295 K) = 67.8 atm

39. (a) Assume one hundred grams of solution

mass glycerine = 25.0 g;
mol glycerine =

mass H2 O = 100.0 25.0 = 75.0 g = 0.075 kg

25.0 g
= 0.272;
92.1 g/mol

m = 0.272 mol/0.0750 kg = 3.62

Tf = 3.62 m (1.86 C/m) = 6.73 C;

Tf = 6.73 C

Tb = 3.62 m (0.52 C/m) = 1.9 C;

(b) mol C3 H8 O2 =

28.0 g
= 0.368;
76.1 g/mol

Tb = 101.9 C
m = 0.368 mol/0.325 kg = 1.13

Tf = 1.13 m (1.86 C/m) = 2.10 C;

Tf = 2.10 C

Tb = 1.13 m (0.52 C/m) = 0.59 C;

Tb = 100.59 C

(c) mass C2 H5 OH = (25.0 mL)(0.780 g/mL) = 19.5 g;

mol C2 H5 OH =

19.5 g
= 0.423
46.07 g/mol

m = 0.423 mol/0.735 kg = 0.576

Tf = 0.576 m (1.86 C/m) = 1.07 C;

Tf = 1.07 C

Tb = 0.576 m (0.52 C/m) = 0.30 C;

Tb = 100.30 C

41. mass acetone = (39 mL)(0.790 g/mL) = 31 g;

mol acetone =

31 g
= 0.53
58.1 g/mol

m = 0.53 mol/0.225 kg = 2.36

Tf = 2.36 m (1.86 C/m) = 4.4 C;

Tf = 4.4 C

Tb = 2.36 m (0.52 C/m) = 1.2 C;

Tb = 101.2 C

43. mol lactic acid =

kf =

45. M =

Tf

RT

13.66 g
= 0.152;
90.1 g/mol

m = 0.152 mol/0.115 kg stearic acid = 1.32

(69.4 62.7) C
= 5.1 C/m
1.32 m

(2.5/760) atm
= 1.34 104 mol/L
0.0821 L atm/mol K 298 K

n = (1.34 104 mol/L)(0.125 L) = 1.68 105 mol;

MM = 0.100 g/1.68 105 mol = 6.0 103 g/mol

114

Chapter 10
Tb

47. m =

kb

(80.78 80.10) C
= 0.27;
2.53 C/m

MM = 5.00 g/0.027 mol;

49. Simplest formula:

n = (0.27 mol/kg)(0.100 kg) = 0.027 mol

formula mass = molar mass;

Molecular formula is C12 H26 O

Assume one hundred grams of compound.

mol C =

49.5 g
= 4.125;
12.0 g/mol

mol O =

16.5 g
= 1.031;
16.00 g/mol

mol H =

5.2 g
= 5.16
1.008 g/mol
18.9 g
= 2.061
14.01 g/mol

mol N =

Ratios of atoms: 2 N : 4 C : 5 H : 1 O;

Simplest formula is C4 H5 N2 O;

formula mass = 97 g/mol

Molecular formula:
mass solvent = (100.0 mL)(0.877 g/mL) = 87.7 g = 0.0877 kg
Tf

m =

kf

(5.50 3.03) C
= 0.484;
5.10 C/m

mol solute = 0.484 mol/kg 0.0877 kg = 0.0424

MM = 8.25 g/0.0424 mol = 194.4 g/mol;

194/97 = 2;

MM = 2(formula mass)

Molecular formula is C8 H10 N4 O2

51. M =

RT

(4.18/760) atm
= 2.25 104 mol/L
(0.0821 L atm/mol K)(298 K)

n = (2.25 104 mol/L)(0.0500 L) = 1.12 105 mol;

MM =

0.225 g
= 2.00 104 g/mol
1.12 105 mol

53. (a) Tf = 2(0.25)(1.86 C) = 0.93 C;

Tb = 2(0.25)(0.52 C) = 0.26 C;
(b) Tf = 1.9 C;

Tf = 1.9 C;

(c) Tf = 5(0.25)(1.86 C) = 2.3 C;

Tb = 5(0.25)(0.52 C) = 0.65 C;

55.

iRT

57. (a) i =
(b) (iii)

Tf = 0.93 C
Tb = 100.26 C
Tb = 100.52 C
Tf = 2.3 C
Tb = 100.65 C

7.7 atm
= 0.16 mol/L
(2)(0.0821 L atm/mol K)(298 K)
Tf

m kf

0.38 C
= 1.0
(0.20 m)(1.86 C/m)

Solutions
59. m =

Tf

i kf

115
=

1.81 C
= 0.487 mol/kg;
(2)(1.86 C/m)

mol LiX = (0.487 mol/kg)(0.283 kg) = 0.138

MM of X = 26.0 MM of Li+ = 26.0 6.94 = 19.1 g/mol

MM of LiX = 3.58 g/0.138 mol = 26.0 g/mol;

X is F
61. Assume one thousand mL (one L) of solution.
(a) mass sucrose = 1203 0.450 = 541 g;

mass of solution = (1000.0 ml)(1.203 g/mL) = 1203 g

mol sucrose =

541 g
= 1.58
343.2 g/mol

M = 1.58 mol/1 L soln = 1.58 mol/L

(b) mass H2 O = 1203 541 = 662 g = 0.662 kg;
(c) XH2 O =

m = 1.58 mol/0.662 kg = 2.39

(662/18.02) mol
= 0.959
(662/18.02) mol + 1.58 mol

(d) Tb = kf m = (0.52 C/m)(2.39 m) = 1.2 C;

P = XH2 O P = 0.959(23.76 mm Hg) = 22.8 mm Hg

Tb = 101.2 C

63. mass solution = (2.50 104 mL)(1.08 g/mL) = 2.70 104 g

mass KMnO4 = (0.150)(2.70 104 g) = 4.05 103 g;
Weigh 4.05 103 g KMnO4 and dilute with water to 25.0 L
mol KMnO4 =

4.05 103 g
= 25.6;
158.0 g/mol

M = 25.6 mol/25.0 L = 1.02 mol/L

65. (a) BaSO4

(b) Ba2+ (aq) + SO42 (aq) BaSO4 (s)
mass solution = (25.00 mL)(1.107 g/mL) = 27.68 g;
mol H2 SO4 = mol SO42 =

mass H2 SO4 = (27.68 g)(0.1525) = 4.220 g

4.220 g
= 0.04303;
98.0 g/mol

mol BaCl2 = mol Ba2+ = (2.45 mol/L)(0.050 L) = 0.122

SO42 is limiting;

0.0430 mol BaSO4 is obtained

mass BaSO4 = (0.04303 mol)(233.4 g/mol) = 10.04g

(c) mol Cl = 2(mol BaCl2 ) = 2(0.122) = 0.244;

MCl =

0.244 mol
= 3.25 mol/L
(0.050 + 0.025) L

116

Chapter 10

67. m =

Tb

kb

(81.20 80.10) C
= 0.435;
2.53 C/m

0.435 mol C10 H8 / thousand g of benzene

mass C10 H8 = (0.435 mol)(128 g/mol) = 55.8 g;

% benzene =

1000.0 g
100 = 94.7%
(1000.0 + 55.8) g

69. Presumably, solution inside the cell is more dilute than that outside. Water moves out of the cell by
osmosis. The cell shrinks.
6 M in Na+ ;

71.

73. 1 ppb =

3 M in S2

1g
106 g 103 g
1 g

=
1g
1 kg
1 kg
109 g

75. (a) greater

(b) less than

(c) greater

(d) less than

77. (a) more ions in CaCl2

(b) formation of solution usually absorbs heat to break down the crystal lattice
(c) If P > , water moves in reverse direction.
(d) i = 3 as opposed to i = 1
(e) One liter of solution contains approximately one kilogram of water.
79. (a) Ions present lower the freezing point.
(b) Sugar crystallizes out on cooling.
(c) If of solution = of blood, or water will move one way or another.
(d) more O2 present
(e) When pressure is released, bubbles of CO2 come out of solution.
81. (a) <

(b) <

(c) >

(d) >

(e) =

Solutions
82. m =

117

Tb

i kb

3.0 C
= 2.9;
(2)(0.52 C/m)

M =

i RT

122 atm
= 2.49
(2)(0.0821 L atm/mol K)(298 K)

Assume one thousand g of solvent; solution has 2.9 moles KNO3 = 291 g.
mass solution = (1000 + 291) g = 1291 g

M = mol solute/Vsol0 n ;

2.49 = 2.9/V;

V = 1.16 L = 1160 mL

density = mass/volume = 1291 g/1160 mL = 1.1 g/mL

83. mol KOH = 158.2 g

1 mol
= 2.819 in one liter, 0.2819 mol in 100 mL.
56.11 g

mass water = 0.2819/0.250 = 1.128 kg = 1128 g

mass water available = 113 g 15.82 g = 97 g; add 1.03 103 g of water
84. Assume one thousand mL of solution.
moles solute = molarity (M);
mass H2 O (in kg) =
molality (m) =

mass solute = M MM;

mass solution mass solute

1000d M MM
=
1000
1000

moles solute
=
mass H2 O (kg)

85. Let x = mass of X;

M
M MM
1000

moles X = x/410;

(x/410) + [(0.100 x)/342]

;
0.00100

86.

mass solution = 1000 d

moles sugar = (0.100 x)/342

x = 0.049 g;

49% X

142 g 0.300 0.15 2

= 0.0018 g/cm3
7.0 103 cm3

87. (a) MNaOH =

49.92 g
1 mol

= 2.08 mol/L
0.600 L 40.00 g

(b) If NaOH is limiting: 1.248 mol NaOH yields 1.872 mol H2

If Al is limiting: 1.530 mol Al yields 2.295 mol H2
Yield of H2 must be 1.872 mol
(c) V =

(1.872 mol)(0.0821 L atm/mol K)(298 K)

= 47.4 L
(734.8/760) atm

118

Chapter 10

88. V = nRT/P

(ideal gas law)

n/P = constant

(Henrys law)

V = constant RT;

V must be constant at constant temperature