CHM 1046. Chapter 12 Homework Solutions.
Problems: 2, 5, 8, 23, 32, 34, 42, 52, 64, 68, 70, 72, 78, 82, 87, 94
2) A solution is a homogeneous mixture, that is, a mixture that has the same composition throughout.
The solvent is the major component of a solution. The solutes are the minor components of a solution.
5) Entropy is a measure of the amount of disorder in a system. The more disordered the system is, the larger the value for entropy.
In general, a process that increases the amount of disorder in a system is more likely to occur than one that does not. Since forming a solution increases the amount of disorder, entropy favors the formation of a solution.
8) The statement "like dissolves like" means that polar solutes tend to dissolve well in polar solvents, and that nonpolar solutes tend to dissolve well in nonpolar solvents.
Polar solutes usually do not dissolve well in nonpolar solvents, and nonpolar solutes usually do not dissolve well in polar solvents.
23) Colligative properties are properties of solutions of nonvolatile solutes with volatile solvents. They are properties that at most depend only on the physical properties of the solvent and the concentration of solute particles.
There are four colligative properties - vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
32) Both CCl
4
and CH
2
Cl
2
have a tetrahedral molecular geometry. In both molecules the C - Cl bonds are polar (EN(C) = 2.5, EN(Cl) = 3.0). However, because of the geometry, the contributions of the four polar bonds cancel in CCl
4
, and the molecule is nonpolar. In CH
2
Cl
2
the contributions of the two polar C - Cl bonds do not cancel, and so the molecule is polar. Polar molecules dissolve better in polar solvents, like water, than nonpolar molecules, so CH
2
Cl
2
should be more soluble in water than CCl
4
.
34) Molecules a and c (toluene and isobutene) are both hydrocarbons (they contain only carbon and hydrogen atoms). The molecules are nonpolar. Therefore these two substances should dissolve well in hexane (a nonpolar solvent) but not dissolve well in water (a polar solvent). They will interact with the hexane by dispersion forces.
1

Molecules b and d (sucrose and ethylene glycol) contain several polar - O - H bonds, and so the molecules are polar. Therefore these two molecules should dissolve well in water (a polar solvent) but not dissolve well in hexane (a nonpolar solvent). They will interact with water by hydrogen bonding (a type of dipole-dipole interaction) and dispersion forces.
42) According to Figure 12.11, the solubility of potassium nitrate (KNO
3
) in water at
T = 25.

C is about 37 g per 100 g water. So a solution containing 32 g KNO
3
per 100 g water is unsaturated.
52) M(KNO
3
) = 101.1 g/mol M(H
2
O) = 18.02 g/mol g(KNO
3
) = 88.4 g mol(KNO g solution = 1500 mL 1.05 g = 1575. g
3
) = 88.4 g 1 mol = 0.8743 mol
101.1 g
1 mL
Since g solution = g water + g KNO
3
, g water = g solution - g KNO
3
= 1575. g - 88.4 g = 1486.6 g
So... molarity (M) = 0.8743 mol KNO
3
= 0.583 mol/L
1.5 L soln molality(m) = 0.8743 mol KNO
3
= 0.588 mol/kg
1.4866 kg water mass percent KNO
3
= 88.4 g KNO
3
.
100% = 5.6% KNO
3
by mass
1575. g soln
Comments:
1) I have not been that rigorous in significant figures, since I feel Tro has given some of the information in the problem with too few significant figures. On an exam I would be careful to give all of the information to an appropriate number of significant figures.
2) Tro seems not to like to give units for molality - but molality does have units.
The units are mol solute/kg solvent (or mol/kg).
3) For dilute aqueous solutions we expect that the numerical value for molarity and molality should be close to one another. This is due to the fact that the density of water is 1.00 g/mL, so 1. kg of water occupies 1. L of volume. In fact, the values here are close, as expected.
4) We have sufficient information in this problem to also calculate the mole fraction of KNO
3
in the solution.
2

64) M(CH
3
OH) = 32.04 g/mol M(H
2
O) = 18.02 g/mol g(CH
3
OH) = 20.2 ml 0.782 g = 15.80 g g(H
2
O) = 100.0 mL 1.00 g = 100.0 g
1 mL 1 mL mol(CH
3
OH) = 15.80 g 1 mol = 0.493 mol mol(H
2
O) = 100.0 g 1 mol = 5.549 mol
32.04 g 18.02 g a) molarity(M) = 0.493 mol CH
3
OH = 4.18 mol/L
0.118 L soln b) molality(m) = 0.493 mol CH
3
OH = 4.93 mol/kg
0.1000 kg water c) percent by mass methanol = 15.80 g .
100% = 13.6% methanol
(15.80 + 100.0)g d) mole fraction methanol = 0.493 = 0.0816
(0.493 + 5.549) mol e) mole percent methanol = (0.0816)
.
100% = 8.16% methanol by moles
68) Let N = naphthalene and H = hexane
M
N
(C
10
H
8
) = 128.2 g/mol M
H
(C
6
H
14
) = 86.17 g/mol
Assume 100.0 g of solution. Then there are 10.85 g naphthalene and (100.00 -
10.85) g = 89.15 g hexane.
To find the mole fraction of hexane (the solvent) we need to find the moles of each substance moles naphthalene = 10.85 g 1 mol = 0.0846 mol
128.2 g moles hexane = 89.15 g 1 mol = 1.0346 mol
86.17 g
So X
H
= 1.0346 mol = 0.9244
(1.0346 + 0.0846) mol
Assuming Raoult's law is obeyed, then p
H
= X
H
p
H

= (0.9244) (151 torr) = 140. torr
3

70) M(CaCl
2
) = 111.0 g/mol M(H
Assuming Raoult's law applies, p
W
= X
W
p
W

, or
2
O) = 18.02 g/mol
X
W
= p
W
= 81.6 torr = 0.881
p
W

92.6 torr
Now, assume that there is a total of 1.000 moles in the system. Then moles water = 0.881 mol g water = 0.881 mol 18.02 g = 15.88 g
1 mol
For the moles of CaCl
2
we must account for the fact that this is a strong electrolyte.
CaCl
2
(s)

Ca 2+ (aq) + 2 Cl (aq)
So moles solute particles = (1.000 - 0.881) mol = 0.119 mol
Moles CaCl
2
= 0.119 mol particle s 1 mol CaCl
2
= 0.0397 mol CaCl
2
3 mol particles g(CaCl
2
) = 0.0397 mol 110.0 g = 4.37 g
1 mol
Finally, to get the mass percent CaCl
2 mass percent CaCl
2
= 4.37 g
(4.37 + 15.88)g
.
100% = 21.6%
72) Let P = pentane and H = hexane, and assume Raoult's law applies. p total
= p
P
+ p
H
= X
P
p
P

+ X
H
p
H

We have one equation and two unknowns. But X
P
+ X
H
= 1, so X
P
=1 - X
H
, so p total
= (1 - X
H
) p
P

+ X
H
p
H

= p
P

- X
H
p
P

+ X
H
p
H

= p
P

+ X
H
(p
H

- p
P

)
Or, finally, X
H
= (p total
- p
P

) = (258 - 425) torr = 0.609
(p
H

- p
P

) (151 - 425) torr
And so X
P
= 1 - 0.609 = 0.391
4

78) For freezing point depression

T f
= K f
m , where m = molality of solute particles
So m =

T f
K f

K f
T f
= T f

- T f
= 0.0

C - (- 1.3

C) = 1.3

C
(Table 12.8) = 1.86 kg
.

C/mol
So m = 1.3

C = 0.699 mol/kg
(1.86 kg.

C/mol)
There are 150.0 g of solvent (water). The number of moles of solute is n solute
= 0.1500 kg solvent 0.699 mol = 0.105 mol solute
1 kg
So M = g solute = 35.9 g = 342. g/mol
mol solute 0.105 mol
82) The expression for osmotic pressure is

= [B] RT , where [B] = solute molarity
The moles of hemoglobin is n = 18.75 x 10 -3 g = 2.88 x 10 -7 mol
6.5 x 10
4
g/mol
So [B] = 2.88 x 10 -7 mol = 1.92 x 10
0.0150 L
-5 mol/L
And so

= (1.92 x 10
-5
mol/L) (0.08206 L
.
atm/mol
.
K) (298.2 K)
= 4.70 x 10
-4
atm (760 torr/1 atm) = 0.357 torr
87)

= [B] RT , so [B] =

/RT
So [B] = 8.3 atm = 0.339 mol/L
(0.08206 L
.
atm/mol
.
K) (298.2 K)
The above is the concentration of solute particles. Since the concentration of the ionic solute is 0.100 mol/L, then i = concentration of solute particles = 0.339 mol/L = 3.39
concentration of solute 0.100 mol/L
94) We first need to find the moles of gas. Assuming the ideal gas law applies pV = nRT , so n = pV/RT p = 725 torr (1 atm/760 torr) = 0.9539 atm n = (0.9539 atm) (1.65 L) = 0.06432 mol
(0.08206 L.atm/mol.K) (298.2 K)
5

Now, from Henry's law, [B] = k
H
p
B
, where B = solute
So [B] = (0.112 mol/L
, atm) (0.9539 atm) = 0.107 mol/L
Therefore, the volume of solution needed to completely dissolve the solute is
V = 0.06432 mol 1 L = 0.60 L
0.107 mol
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