CHM 1046. Chapter 12 Homework Solutions.
Problems: 2, 5, 8, 12, 20, 23, 32, 34, 36, 42, 52, 64, 72, 74, 78, 80, 84, 89, 98
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.
12)
The solubility of a solid in a liquid usually increases as temperature increases,
though as Fig 12.11 indicates there are a few exceptions to this general trend.
Recrystallization is a technique that makes use of the dependence of solubility on
temperature. In recrystallization the solid to be purified is dissolved in a minimum
amount of hot solvent to form a saturated or nearly saturated solution. As the solution
cools, and assuming the solubility of the solid decreases with decreasing temperature, the
solution becomes supersaturated, and a precipitate (solid) forms. Impurities in the
original sample of solid are generally present at low concentration and so tend to remain
in solution. The purified solid can be isolated by filtering the solution to remove the
solid.
NOTE: We can be more precise in identifying the factor that determines the
temperature dependence of the solubility of a solid. Based on the additional
thermodynamics we will learn in Chapter 17, we may say that if Hsoln < 0 (solution
formation is endothermic) then solubility will increase as temperature increases, while if
Hsoln > 0 (solution formation is exothermic) then solubility will decrease as temperature
increases. Since for most solids Hsoln < 0, most solids are more soluble at high
temperatures than at low temperatures.
20)
Just as an ideal gas is a gas whose components obey the ideal gas law, an ideal
solution of volatile liquids is a solution whose components obey Raoult’s law (p A =
XApA). Unlike gases, ideal solutions are relatively rare, but will form (to a good first
approximation) when mixing two similar liquids, such as benzene (C6H6) and toluene
(C6H5CH3).
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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 CCl4 and CH2Cl2 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 CCl4, and the molecule is
nonpolar. In CH2Cl2 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 CH2Cl2 should be more soluble in water than CCl4.
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.
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.
36)
a) Since the temperature of the water increases, heat must be released. Therefore
the process is exothermic.
b) Since the process is exothermic, Hsoln < 0. To a good first approximation
Hsoln  |Hlattice| - |Hhydration|. If this is negative, then |Hhydration| > |Hlattice|.
c) The diagram is given on the next page, along with a description
2
In the above diagram we begin with solid lithium iodide (LiI(s)). The energy
required to convert the solid into gas phase ions is |Hlattice|. At this point we have
formed Li+(g) and I-(g). The decrease in energy that occurs when these gas phase ions
dissolve in water and become hydrated (surrounded by water molecules) is |Hhydration|.
The difference between these two energies is (approximately) equal to the energy change
for solution formation
Hsoln  |Hlattice| - |Hhydration|
Since Hsoln < 0 it follows that |Hlattice| < |Hhydration|, as indicated in the diagram.
d) Solution formation becomes more likely when entropy increases (Ssoln > 0)
and energy decreases (Hsoln < 0). In this case solution formation both increases
randomness and decreases enthalpy, so both of these factors favor solution formation.
NOTE: In the above we equate the change in energy with the change in enthalpy.
This is not exact, but to a good first approximation (in most cases) H  E, and so the
enthalpy change and energy change are approximately the same.
42)
According to Figure 12.11, the solubility of potassium nitrate (KNO3) in water at
T = 25. C is about 37 g per 100 g water. So a solution containing 32 g KNO3 per 100 g
water is unsaturated.
52)
M(KNO3) = 101.1 g/mol
M(H2O) = 18.02 g/mol
g(KNO3) = 72.5 g
mol(KNO3) = 72.5 g 1 mol = 0.7171 mol
101.1 g
g solution = 2000 mL 1.05 g = 2100. g
1 mL
3
Since g solution = g water + g KNO3,
g water = g solution - g KNO3 = 2100. g – 72.5 g = 2027.5 g
So...
molarity (M) = 0.7171 mol KNO3 = 0.359 mol/L
2.00 L soln
molality(m) = 0.7171 mol KNO3 = 0.354 mol/kg
2.0275 kg water
mass percent KNO3 = 72.5 g KNO3 . 100% = 3.45 % KNO3 by mass
2100. g soln
NOTE: 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 KNO3 in the solution.
64)
M(CH3OH) = 32.04 g/mol
M(H2O) = 18.02 g/mol
g(CH3OH) = 20.2 ml 0.782 g = 15.80 g
1 mL
g(H2O) = 100.0 mL 1.00 g = 100.0 g
1 mL
mol(CH3OH) = 15.80 g 1 mol = 0.493 mol mol(H2O) = 100.0 g 1 mol = 5.549 mol
32.04 g
18.02 g
a) molarity(M) = 0.493 mol CH3OH = 4.18 mol/L
0.118 L soln
b) molality(m) = 0.493 mol CH3OH = 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
4
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
72)
Since naphthalene is nonvolatile (no significant vapor pressure) we are only
concerned with the pressure due to the hexane. We will assume ideal solution behavior,
which means we can use Raoult’s law to find the vapor pressure of hexane above the
solution.
Let N = naphthalene and H = hexane
MN(C10H8) = 128.2 g/mol
MH(C6H14) = 86.17 g/mol
Assume 100.0 g of solution. Then there are 12.35 g naphthalene and (100.00 –
12.35) g = 87.65 g hexane.
To find the mole fraction of hexane (the solvent) we need to find the moles of
each substance
moles naphthalene = 12.35 g 1 mol = 0.0963 mol
128.2 g
moles hexane = 87.65 g
So XH =
1 mol
86.17 g
= 1.0172 mol
1.0172 mol
= 0.9135
(1.0172 + 0.0963) mol
Assuming Raoult's law is obeyed, then
pH = XH pH = (0.9135) (151 torr) = 138. torr
74)
Let P = pentane and H = hexane, and assume Raoult's law applies.
ptotal = pP + pH = XP pP + XH pH
We have one equation and two unknowns. But XP + XH = 1, so XP =1 - XH, so
ptotal = (1 - XH) pP + XH pH = pP - XH pP + XH pH
= pP + XH (pH - pP)
Or, finally, XH = (ptotal - pP) = (258 - 425) torr = 0.609
(pH - pP) (151 - 425) torr
And so XP = 1 - 0.609 = 0.391
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78)
To use the equations for boiling point elevation and freezing point depression we
need to find the molality of ethylene glycol in the solution. Let ET = ethylene glycol.
MET = 62.07 g/mol
So the moles of ethylene glycol is
mET =
21.2 g
= 0.3415 mol ET
62.07 g/mol
There are 85.4 mL of water, and since the density of water is 1.00 g/mL, there are 85.4 g
(or 0.0854 kg) of water. So the molality of the solution is
m = 0.3415 mol ET = 4.00 mol/kg
0.0854 kg water
The formulas for boiling point elevation and freezing point depression are
So
Tb = Kb m
Kb = 0.512 C.kg/mol
Tf = Kf m
Kf = 1.86 C.kg/mol
Tb = Kb m = (0.512 C.kg/mol) (4.00 mol/kg) = 2.05 C
Tf = Kf m = (1.86 C.kg/mol) (4.00 mol/kg) = 7.44 C
Since the normal boiling point and normal freezing point for water are 100.00 C and
0.00 C, respectively, the boiling point and freezing point for the solution are
Tb = 100.00 C + 2.05 C = 102.05 C
Tf = 0.00 C – 7.44 C = - 7.44 C
Note that the boiling point of the solution is higher than that for pure water and the
freezing point of the solution is lower than that for pure water.
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80)
For freezing point depression Tf = Kf m , where m = molality of solute particles
So m = Tf
Kf
So m =
Tf = Tf - Tf = 0.0C - (- 1.3 C) = 1.3 C
Kf (Table 12.8) = 1.86 C.kg/mol
1.3 C
= 0.699 mol/kg
(1.86 C.kg/mol)
There are 150.0 g of solvent (water). The number of moles of solute is
nsolute = 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
84)
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 104 g/mol
So [B] = 2.88 x 10-7 mol = 1.92 x 10-5 mol/L
0.0150 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
89)
 = [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
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98)
We first need to find the moles of gas. Assuming the ideal gas law applies
pV = nRT , so n = pV/RT
n=
p = 725 torr (1 atm/760 torr) = 0.9539 atm
(0.9539 atm) (1.65 L)
= 0.06432 mol
(0.08206 L.atm/mol.K) (298.2 K)
Now, from Henry's law, [B] = kH pB , 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
1L
0.107 mol
= 0.60 L
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