Answers to selected
problems, Chapter 12
Review questions
2. Molarity, which is moles per liter of solution, depends on
temperature because the volume of a liquid depends on temperature (but in most
cases not very much). Molality, the number of moles per kilogram of solution,
does not depend on temperature because it is a mass/mass ratio, and mass
(obviously) does not depend on temperature.
16. These two statements may at first glance seem incompatible if one focuses only on vapor pressure of the solute, because it is the solvent, not the solute, that is diffusing during osmosis. The solvent is actually less concentrated in the solution with the solute because the solute dilutes the solvent. So the solvent diffuses from its pure liquid form into the solution (through the semipermeable membrane) because it is at lower concentration in the latter. The fact that this diffusion, or osmosis, increases the total pressure of the solution is almost incidental, and is certainly not the controlling mechanism. Osmosis is perfectly logical if you think about the solvent, which is diffusing into a zone of lower concentration just the way any gas does. Think partial pressures of gases, and osmosis will be easier to understand.
22. Which of each of these pairs of these solutions at 25 °C
has the higher osmotic pressure?
(a) 0.1 M NaHCO3 vs. 0.05 M NaHCO3?
Obviously the first one—since the two solutes
are the same, the osmotic pressure will vary directly with the molarity of the
solution. The greater molarities dilute the solvent in the upper solution more,
which increases the upward pressure from the pure solvent below.
(b) 1 M NaCl vs. 1 M glucose? Since glucose is
molecular and NaCl is ionic and the two have the same molarity, the solution of NaCl
will produce twice the concentration of particles (entities) that the glucose will. Since
the osmotic pressure depends on the concentration of particles irrespective of
their identity, the solution of NaCl will have twice the osmotic pressure that
the glucose will.
(c) 1 M NaCl vs. 1 M CaCl2? Since NaCl
produces two moles of particles per mole of formula unit and CaCl2
produces three, the latter will have the higher osmotic pressure.
(d) 1 M NaCl vs. 3 M glucose? The glucose, because it
creates 3 M particles, whereas the NaCl produces only 2 M particles.
Problems
25. What is the mass percent of solute in each of these
solutions?
(a) 4.12 g NaOH in 100.0 g water? (4.12 g NaOH)/(4.12
g + 100.00 g water) = 0.0396 = 3.96% by mass.
(b) 5.00 mL ethanol (d = 0.789 g/mL) in 50.0
g water (5.00 mL ethanol) x (0.789 g/mL) = 3.945 g ethanol. (3.945 g)/(3.945
g + 50.00 g water) = 0.0731 = 7.31% ethanol by mass.
(c) 1.50 mL glycerol (d = 1.324 g/mL) in
22.25 mL water (d = 0.998 g/mL) Mass of glycerol = (1.50 mL) (1.326
g/mL) = 1.986 g. Mass of water = (22.25 mL) (0.998 g/mL) = 22.21 g. Mass percent
glycerol = (1.986 g)/(1.986 g + 22.21 g water) x 100% = 8.21%.
29. To get the concentration in mg per dL, just find the concentration in mg per L and divide by 10. Concentration in mg/L = 0.10% x 106 mg/L = 10-3 x 106 = 103 mg/L. One tenth of this is 102 = 100 mg/dL.
31. Arrange these concentrations in increasing order of mass
percent: (a) 1% by mass; (b) 1 mg solute/dL solution; (c) 1 ppb; (d) 1 ppm; (e)
1 ppt.
(a) = 1% by mass (no manipulation needed)
(b) 1 mg/100 g solution = 10-3 g/102 g
= 10-3%
(c) 1 ppb = 10-9 = (102 x 10-9)/(102)
= 10-7/102 = 10-7%
(d) 1 ppm = 10-6 = (102 x 10-6)/(102)
= 10-4/102 = 10-4%
(e) 1 ppt = 10-12 = (102 x 10-12)/(102)
= 10-10/102 = 10-10%
Thus the increasing order is e, c, d, b, a.
That's the long way. The short way is to recognize that the
numerical values of each concentration is 1. That makes the order of
concentration inversely proportional to the unit in the denominator, i.e.,
1/1000 is less than 1/100 because 1000 is greater than 100. That means that ppt
< ppb < ppm < pph (%). Then to fit (b) into this sequence, you just
have to see that it is 10-3%. The full sequence then becomes ppt <
ppb < ppm < 10-3% < pph (%).
33. (a) 1 µg benzene/L of water is about 1 µg benzene/1000 g
water because the amount of benzene is too small to change the density of the
solution noticeably. So the concentration of benzene is then 1 x 10-6
g/103 g = 1 x 10-9 = 1/109 = 1 ppb.
(b) 0.0035% NaCl by mass is 35 x 10-4 x 10-2
= 35 x 10-6 = 35 ppm NaCl.
(c) 2.4 ppm F- is about 2.4 x 10-3 g/103
g water = 2.4 x 10-3 g/L water because there isn't enough F- to
change the density of the water noticeably. To convert the mass of F-
to moles, divide by 19.0 g mol-1 and get (2.4 x 10-3
g)/(19.0 g mol-1) = 1.3 x 10-4 M F-.
35. To calculate the molality of a solution of 18.0 g glucose in 80.0 g of water, you transform it to moles per 1000 g of water. The number of moles of glucose, C6H12O6, is 18.0 g/180.2 g mol-1. The way to convert 80.0 g water in the denominator to kg is just to divide it by 1000 kg/g. The final expression for molality is then (18.0 g/180.2 g mol-1)/(80 g/1000 g/kg) = 1.25 m.
39. It is straightforward to find how many moles of H2SO4 are in 375 mL of a 3.39 m solution of H2SO4 that has a density of 1.18 g/mL. First recognize that 3.39 m means 3.39 moles H2SO4 per 1000 g solution. Now just find the mass of the 375 mL of solution (375 mL x 1.18 g/mL) and multiply the previous concentration by that. This amounts to the expression (3.39 mol/1000 g solution)(375 mL solution x 1.18 g/mL solution) = 1.13 moles of H2SO4.
41. The mole fraction in the solution of naphthalene in benzene is just the number of moles of naphthalene divided by the number of moles of naphthalene and benzene. The number of moles of naphthalene, C10H8, is (23.5 g naphthalene/128.16 g mol-1 naphthalene). The number of moles of benzene is (315 g benzene/78.11 g mol-1 benzene). The mole fraction of naphthalene is then [(23.5 g/128.16 g mol-1)/(23.5 g/128.16 g mol-1) + (315 g/78.11 g mol-1)] = 0.0434.
45. Without detailed calculations, which of these
aqueous solutions has the greatest mole fraction of solute: (a) 1.00 m CH3OH,
(b) 5.0% CH3CH2OH by mass, or (c) 10.0% C12H22O11
by mass?
Rough calculation. Use 1000 g of solution for each
(a) has 32 g solute in 1000 - 32 = 968 g water, or 1 mole in
968 g water.
(b) has 50 g solute in 950 g water, or a bit more than 1 mole
solute in 950 g water. (MW of solute = 48)
(c) has 100 g solute (about 0.3 moles) in 900 g water. (MW of
solute = 342)
Conclusion, (b) has the greatest mole fraction of solute.
(More moles in less water than (a).)
47. Chloroform (CHCl3 = a) is insoluble in water because it does not have hydrogen bonds, the main intermolecular forces in water. Benzoic acid (b) is slightly soluble in water; the carboxylic acid group provides some hydrogen bonding, but the benzene ring is very unlike water. Propylene glycol (c) is soluble in water because its two hydroxyl groups provide abundant opportunities for hydrogen bonding with water.
49. Is a solution of 25 g NH4Cl and 55 g water at 60 °C saturated, unsaturated, or supersaturated? This combination, 25 g/55 g water, is equivalent to (25/55)(100 g/55 g) = 45 g solute per 100 g water. Figure 12.10 shows that the solubility of NH4Cl is about 47 g per 100 g water at 60 °C. Thus, the solution is unsaturated.
53. Is there a way to crystallize a solute from an unsaturated solution without changing the temperature? If you decreased the temperature you could make the solution saturated. The other obvious way to saturate the solution is to allow enough of the water to evaporate.
55. (a) The molarity of a saturated solution of oxygen in
water at 20 °C is just the number of moles of O2 in one liter, or
(4.43 x 10-3 g O2/32 g mol-1 O2)/(100
mL x (1 L/1000 mL)) = 1.38 x 10-3 M.
(b) A saturated solution with a concentration of 0.010 M O2
is about 7.2 times greater than normal saturation (0.010/0.00138 = 7.25). Thus,
about 7.2 atmospheres of oxygen would be required to produce this concentration
at saturation.
59. The vapor pressure of water over the solution of glucose will be the saturation vapor pressure for pure water (Table 11.2) multiplied by the mole fraction of the water in the glucose solution. The vapor pressure of pure water at 20 °C is 17.5 mmHg. The mole fraction of water is (moles glucose)/(moles glucose + moles water). In the 0.2 m solution of glucose, there will be 0.2 moles glucose per 1000 g water, or per 55.6 moles of water (because 1000 g water/18 g mol-1 water = 55.6 moles water). Thus the mole fraction of water will be 55.6/(55.6 + 0.2) = 55.6/55.8 = 0.996. This makes the vapor pressure of water 17.5 mmHg x 0.996 = 17.4 mmHg.
61. (a) The freezing point will be depressed relative to
that of pure water by 1.86 °C per molal unit of solute (Table 12.2). The 0.25 m
concentration of the urea here will therefore depress the freezing point by
(0.25 m)(1.86 °C/m) = 0.46 °C. Thus the freezing point of the
solution will be -0.46 °C because it started at 0 °C.
(b) The freezing point of benzene will be depressed by 5.12 °C
m-1 of solute. To find the molality of the 5.0% para-dichlorobenzene
in benzene, note that the solute makes up 5% of the mass and the solvent the
other 95%. Thus in 1000 g of solution there will be 50 g of solute ( = 50 g/147
g mol-1 = 50/147 mol) in 950 g solvent. This corresponds to a
molality of [(50/147 mol)/(950 g)](1000 g/950 g) = 0.358 m. The freezing
point will then be depressed by (5.12 °C m-1)(0.358 m)
= 1.83 °C. Since the FP started at 5.53 °C, its final value will be 5.53 -
1.83 = 3.70 °C.
67. This problem requires two steps to solve. The first is
the find the empirical formula of the compound. The second is to use the FP
depression by a certain concentration of the compound to find the molality and
from that the molecular weight.
(a) To get the empirical formula, we determine the molar
ratios of C, H, O, and N by dividing the mass percent of each by its atomic
weight and reducing these results to a series of small whole numbers.
C: 33.81/12.011 = 2.81
H: 1.42/1.00794 = 1.41
O: 45.05/15.9994 = 2.82
N: 19.72/14.0067 = 1.41
The empirical formula is obviously C2HO2N,
whose molecular weight is 71.
(b) To get the molecular weight of the compound, we write the
general formula for freezing-point depression, solve algebraically for the
molecular weight, and fill in the numbers. Using the symbols from Chapter 12 and
earlier, the freezing-point depression will be:
ΔTf = -mKf = -Kf[(m/W)/(kg solvent)]
This can be rearranged to W = -(m/kg solvent)(Kf/ΔTf) =
-{1.505 g/[(50.00 mL x 0.879 g mL-1)/(1000 g)]}{5.12 °C m-1/[-0.83 °C]} = 211 g mol-1
The actual molecular weight is three times the empirical molecular weight. That makes the actual formula C6H3O6N3.
71. What is the osmotic pressure at 37 °C of an aqueous solution that is 1.80% (mass/vol) of CH3CH2OH? Just use the formula π = MRT (page 539), being careful to convert the 1.80% (M/V) properly.
π = [1.80 g/46 g mol-1)/0.1 L](0.08206 L atm mol-1 K-1)(37 °C + 273 °C) = 9.95 atm
Is this solution isotonic, hypotonic, or hypertonic? To answer this part, we compare this osmotic pressure to that of a 5.5% (M/V) solution of glucose, C6H12O6.
πglucose = [5.5 g/180 g mol-1)/0.1 L](0.08206 L atm mol-1 K-1)(37 °C + 273 °C) = 7.77 atm
Therefore the ethanol is hypertonic because its osmotic pressure is greater than that of standard glucose in the bloodstream.
75. (Restating the problem to make it easier to understand) When HCl is dissolved in water and benzene to give 0.01 m solutions, the freezing-point depressions are about 0.04 °C and 0.05 °C, respectively. How can this be, if the molal freezing-point depressions for the solvents are 1.88 and 5.12 °C m-1, respectively? The key is to note that the molality here refers to the total for all entities, not just for formula units. Because HCl dissociates in water but not in benzene, there are twice as many entities in water, effectively 0.02 m for purposes of the freezing point. Thus the FP is depressed by (0.02)(1.88 °C), or nearly 0.04 °C, but the FP of benzene is depressed by only (0.01)(5.12 °C), or about 0.05 °C.
81. To solve this problem, just write out the algebraic expression for boiling-point elevation, solve algebraically for the variable of interest (m = mass of NaCl), and insert the numbers. Remember that i, the van't Hoff factor, is 2 for NaCl. (Recall that m and m are not the same.)
ΔTb = imKb = iKb[(m/W)/(kg solvent)]
This can be rearranged to m = [WΔTb (kg solvent)]/iKb = [(58 g mol-1)(0.6 °C)(3.50 kg solvent)]/[(2)(0.512 °C m-1)]
= 120 g NaCl
83. AlCl3(aq) would be the best because the high charge on Al3+ would more effectively negate the negative charge on the silica and allow it to coagulate.
85. Consider an aqueous solution with density 0.980 g/mL
at 20 °C, which was prepared by dissolving 11.3 mL of CH3OH (d =
0.793 g/mL) in enough water to produce 75.0 mL of solution. What is the percent
CH3OH, expressed as:
(a) volume percent: (11.3 mL CH3OH)/(75 mL
solution) = 0.151 = 15.1% by volume
(b) mass percent: (11.3 mL CH3OH x 0.793 g/mL)/(75.0
mL solution x 0.980 g/mL) = 0.122 = 12.2% by mass
(c) mass/volume percent: (11.3 mL CH3OH x
0.793 g/mL)/(75.0 mL solution) = 0.119 = 11.9% mass/volume
(d) mole percent: This one has several steps. First
calculate the mass of CH3OH from its volume and density. Then
calculate the mass of the solution in the same way and get the mass of water by
subtracting. Then calculate the moles of each and calculate the mole fraction
and percent for CH3OH as its moles divided by the total number of
moles in the solution.
(1) Mass of CH3OH = 11.3 mL CH3OH x
0.793 g/mL = 8.961 g
(2) Mass of solution = 75.0 mL solution x 0.980 g/mL = 73.5 g
(3) Mass of water = 73.5 g - 9.0 g = 64.5 g
(4) Moles of CH3OH = (8.961 g)/(32.04 g mol-1)
= 0.2797 mol
(5) Moles of water = (64.5 g)/(18.02 g mol-1) =
3.579 mol
(6) Mole fraction of CH3OH = (0.2797 mol)/(0.2797
mol + 3.579 mol) = 0.0725 = 7.25%
93. The solution to this problem requires two conditions to be met: the total vapor pressure of the solution at 100.0 °C must be 760 mmHg (1 atmosphere) and the mole fractions of benzene and toluene must add to 1. As usual, set up the appropriate general equations:
Psolution = 750 mmHg = xbenzeneP°benzene + xtolueneP°toluene
xbenzene + xtoluene = 1
Psolution = 750 mmHg = xbenzeneP°benzene + (1-xbenzene)P°toluene
Solving for xbenzene gives:
xbenzene = (Psolution - P°toluene)/(P°benzene - P°toluene)
xbenzene = (760 - 556.3)/(1351 - 556.3) = 0.256 = 25.6%
That makes xtoluene = 1 - xbenzene = 0.744 = 74.4%
95. To see how much a 0.92% (mass/vol) solution of NaCl depresses the freezing point of water, we must convert it to molality in the formula:
ΔTf = -imKf = -iKf[(m/W)/(kg solvent)]
= -2(1.86 °C/m)[(0.92 g/58.44 g mol-1)/(0.1 kg)] = -0.59 °C
(quite close to the given change of -0.52 °C)
99.