CHAPTER 11 PROPERTIES OF SOLUTIONS by sdfgsg234

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```									378                                          CHAPTER 11          PROPERTIES OF SOLUTIONS

Questions
CHAPTER 11
17.   As the temperature increases, the gas molecules will have a greater average kinetic energy. A
greater fraction of the gas molecules in solution will have a kinetic energy greater than the
attractive forces between the gas molecules and the solvent molecules. More gas molecules           PROPERTIES OF SOLUTIONS
are able to escape to the vapor phase, and the solubility of the gas decreases.

18.   Henry’s law is obeyed most accurately for dilute solutions of gases that do not dissociate in       Solution Review
or react with the solvent. NH3 is a weak base and reacts with water by the following reaction:
1 mol C3H 7 OH
585 g C3 H 7 OH
NH3(aq) + H2O(l)       NH4+(aq) + OH (aq)                                                       11.
60.09 g C3H 7 OH
= 9.74 M
1.00 L
O2 will bind to hemoglobin in the blood. Due to these reactions in the solvent, NH3(g) in
water and O2(g) in blood do not follow Henry’s law.                                                                   0.100 mol      134.00 g
12.   0.250 L                             = 3.35 g Na2C2O4
19.   Because the solute is volatile, both the water and solute will transfer back and forth between                            L            mol
the two beakers. The volume in each beaker will become constant when the concentrations of
solute in the beakers are equal to each other. Because the solute is less volatile than water,                       0.040 mol HCl                                               1L
13.   1.00 L ×                 = 0.040 mol HCl; 0.040 mol HCl                        = 0.16 L
one would expect there to be a larger net transfer of water molecules into the right beaker                                L                                                0.25 mol HCl
than the net transfer of solute molecules into the left beaker. This results in a larger solution                                                                                            = 160 mL
volume in the right beaker when equilibrium is reached, i.e., when the solute concentration is
identical in each beaker.                                                                                                   1 mol CaCl 2            1L             1000 mL
14.   1.28 g CaCl2 ×                                                   = 19.9 mL
110.98 g CaCl 2    0.580 mol CaCl 2        L
20.   Solutions of A and B have vapor pressures less than ideal (see Figure 11.13 of the text), so
this plot shows negative deviations from Rault’s law. Negative deviations occur when the
intermolecular forces are stronger in solution than in pure solvent and solute. This results in                                        3.0 mol Na 2 CO 3
15.   Mol Na2CO3 = 0.0700 L ×                        = 0.21 mol Na2CO3
an exothermic enthalpy of solution. The only statement that is false is e. A substance boils                                                   L
when the vapor pressure equals the external pressure. Because B = 0.6 has a lower vapor
pressure at the temperature of the plot than either pure A or pure B, one would expect this               Na2CO3(s)      2 Na+(aq) + CO32 (aq); mol Na+ = 2(0.21) = 0.42 mol
solution to require the highest temperature in order for the vapor pressure to reach the
external pressure. Therefore, the solution with B = 0.6 will have a higher boiling point than                                           1.0 mol NaHCO 3
Mol NaHCO3 = 0.0300 L ×                       = 0.030 mol NaHCO3
either pure A or pure B. (Note that because P°B > P°A, B is more volatile than A, and B will                                                    L
have a lower boiling point temperature than A).
NaHCO3(s)        Na+(aq) + HCO3 (aq); mol Na+ = 0.030 mol
21.   No, the solution is not ideal. For an ideal solution, the strengths of intermolecular forces in
solution are the same as in pure solute and pure solvent. This results in Hsoln = 0 for an ideal
total mol Na      0.42 mol 0.030 mol        0.45 mol
solution. Hsoln for methanol-water is not zero. Because Hsoln < 0 (heat is released), this                M Na                                                            = 4.5 M Na+
solution shows a negative deviation from Raoult’s law.                                                                total volume      0.0700 L 0.030 L         0.1000 L

22.   The micelles form so that the ionic ends of the detergent molecules, the SO4 ends, are              16.   a. HNO3(l)        H+(aq) + NO3 (aq)               b. Na2SO4(s)       2 Na+(aq) + SO42 (aq)
exposed to the polar water molecules on the outside, whereas the nonpolar hydrocarbon
chains from the detergent molecules are hidden from the water by pointing toward the inside               c. Al(NO3)3(s)       Al3+(aq) + 3 NO3 (aq)        d. SrBr2(s)      Sr2+(aq) + 2 Br (aq)
of the micelle. Dirt, which is basically nonpolar, is stabilized in the nonpolar interior of the
micelle and is washed away. See the illustration on the following page.                                   e. KClO4(s)       K+(aq) + ClO4 (aq)              f.   NH4Br(s)      NH4+(aq) + Br (aq)

g. NH4NO3(s)        NH4+(aq) + NO3 (aq)           h. CuSO4(s)        Cu2+(aq) + SO42 (aq)

i.   NaOH(s)      Na+(aq) + OH (aq)

377
380                                          CHAPTER 11          PROPERTIES OF SOLUTIONS                  CHAPTER 11       PROPERTIES OF SOLUTIONS                                                            379

For statement a, the vapor pressure of a solution is directly related to the mole fraction of
solvent (not solute) by Raoult’s law. For statement c, colligative properties depend on the
number of solute particles present and not on the identity of the solute. For statement d, the                               = detergent
boiling point of water is increased because the sugar solute decreases the vapor pressure of                                   molecule
the water; a higher temperature is required for the vapor pressure of the solution to equal the
= SO4-
external pressure so boiling can occur.

26.   This is true if the solute will dissolve in camphor. Camphor has the largest Kb and Kf                                       = nonpolar
constants. This means that camphor shows the largest change in boiling point and melting                                       hydrocarbon
point as a solute is added. The larger the change in T, the more precise the measurement and
the more precise the calculated molar mass. However, if the solute won’t dissolve in
camphor, then camphor is no good and another solvent must be chosen which will dissolve
the solute.                                                                                                                  = dirt

27.   Isotonic solutions are those which have identical osmotic pressures. Crenation and hemolysis
refer to phenomena that occur when red blood cells are bathed in solutions having a mismatch        23.   Normality is the number of equivalents per liter of solution. For an acid or a base, an
in osmotic pressures inside and outside the cell. When red blood cells are in a solution having           equivalent is the mass of acid or base that can furnish 1 mole of protons (if an acid) or accept
a higher osmotic pressure than that of the cells, the cells shrivel as there is a net transfer of         1 mole of protons (if a base). A proton is an H+ ion. Molarity is defined as the moles of solute
water out of the cells. This is called crenation. Hemolysis occurs when the red blood cells are           per liter of solution. When the number of equivalents equals the number of moles of solute,
bathed in a solution having lower osmotic pressure than that inside the cell. Here, the cells             then normality = molarity. This is true for acids which only have one acidic proton in them
rupture as there is a net transfer of water to into the red blood cells.                                  and for bases that accept only one proton per formula unit. Examples of acids where
equivalents = moles solute are HCl, HNO3, HF, and HC2H3O2. Examples of bases where
28.   Ion pairing is a phenomenon that occurs in solution when oppositely charged ions aggregate
and behave as a single particle. For example, when NaCl is dissolved in water, one would                  equivalents = moles solute are NaOH, KOH, and NH3. When equivalents moles solute,
expect sodium chloride to exist as separate hydrated Na+ ions and Cl ions. A few ions,                    then normality molarity. This is true for acids that donate more than one proton (H2SO4,
however, stay together as NaCl and behave as just one particle. Ion pairing increases in a                H3PO4, H2CO3, etc.) and for bases that react with more than one proton per formula unit
solution as the ion concentration increases (as the molality increases).                                  [Ca(OH)2, Ba(OH)2, Sr(OH)2, etc.].

Exercises                                                                                                 24.   It is true that the sodium chloride lattice must be broken in order to dissolve in water, but a lot
of energy is released when the water molecules hydrate the Na+ and Cl ions. These two
Solution Composition                                                                                            processes have relatively large values for the amount of energy associated with them, but they
are opposite in sign. The end result is they basically cancel each other out resulting in a Hsoln
29.   Because the density of water is 1.00 g/mL, 100.0 mL of water has a mass of 100. g.                           0. So energy is not the reason why ionic solids like NaCl are so soluble in water. The
answer lies in nature’s tendency toward the higher probability of the mixed state. Processes,
mass     10.0 g H 3 PO 4 100. g H 2 O
Density =                                           = 1.06 g/mL = 1.06 g/cm3                              in general, are favored that result in an increase in disorder because the disordered state is the
volume                104 mL                                                                 easiest (most probable) state to achieve. The tendency of processes to increase disorder will
1 mol                                                                           be discussed in Chapter 16 when entropy, S, is introduced.
Mol H3PO4 = 10.0 g ×             = 0.102 mol H3PO4
97.99 g
25.   Only statement b is true. A substance freezes when the vapor pressure of the liquid and solid
1 mol                                                                              are the same. When a solute is added to water, the vapor pressure of the solution at 0 C is
Mol H2O = 100. g ×            = 5.55 mol H2O
18.02 g                                                                             less than the vapor pressure of the solid, and the net result is for any ice present to convert to
liquid in order to try to equalize the vapor pressures (which never can occur at 0 C). A lower
0.102 mol H 3 PO 4                                                            temperature is needed to equalize the vapor pressure of water and ice, hence, the freezing
Mole fraction of H3PO4 =                       = 0.0180
(0.102 5.55) mol                                                              point is depressed.
H 2O   = 1.000 – 0.0180 = 0.9820
382                                                 CHAPTER 11           PROPERTIES OF SOLUTIONS   CHAPTER 11           PROPERTIES OF SOLUTIONS                                                     381

Sulfuric acid (H2SO4):                                                                                            0.102 mol H 3 PO 4
Molarity =                        = 0.981 mol/L
95 g H 2SO 4        1.84 g soln    1000 cm 3          1 mol H 2SO 4                                                0.104 L
= 18 mol/L
100. g soln          cm 3 soln        L               98.1 g H 2SO 4                                          0.102 mol H 3 PO 4
Molality =                        = 1.02 mol/kg
0.100 kg
95 g H 2SO 4        1000 g      1 mol
= 194 mol/kg         200 mol/kg
5 g H 2O            kg        98.1 g                                                                         40.0 g EG        1000 g     1 mol EG
30.   Molality =                                           = 10.7 mol/kg
1 mol                              1 mol                                                         60.0 g H 2 O        kg        62.07 g
95 g H2SO4 ×        = 0.97 mol H2SO4; 5 g H2O ×        = 0.3 mol H2O
98.1 g                             18.0 g
40.0 g EG           1.05 g      1000 cm 3         1 mol
Molarity =                                                                = 6.77 mol/L
0.97                                                                                        100.0 g solution       cm 3          L             62.07 g
H 2SO 4                = 0.76
0.97 0.3
1 mol                                1 mol
40.0 g EG ×             = 0.644 mol EG; 60.0 g H2O ×         = 3.33 mol H2O
Acetic acid (CH3CO2H):                                                                                             62.07 g                              18.02 g

99 g CH 3CO 2 H        1.05 g soln        1000 cm 3       1 mol                                            0.644
= 17 mol/L                     EG                 = 0.162 = mole fraction ethylene glycol
100. g soln           cm 3 soln            L           60.05 g                                        3.33 0.644

99 g CH 3CO 2 H        1000 g       1 mol                                                31.   Hydrochloric acid (HCl):
= 1600 mol/kg          2000 mol/kg
1 g H 2O             kg        60.05 g
38 g HCl       1.19 g soln        1000 cm 3     1 mol HCl
molarity =                                                              = 12 mol/L
1 mol                               1 mol                                                         100. g soln      cm 3 soln            L            36.5 g
99 g CH3CO2H ×         = 1.6 mol CH3CO2H; 1 g H2O ×        = 0.06 mol H2O
60.05 g                              18.0 g
38 g HCl         1000 g      1 mol HCl
molality =                                           = 17 mol/kg
1 .6                                                                                          62 g solvent        kg          36.5 g
CH 3CO 2 H               = 0.96
1.6 0.06
1 mol                            1 mol
38 g HCl ×           = 1.0 mol HCl; 62 g H2O ×        = 3.4 mol H2O
Ammonia (NH3):                                                                                                         36.5 g                           18.0 g
1 .0
28 g NH 3         0.90 g     1000 cm 3      1 mol                                                   mole fraction of HCl =       HCl             = 0.23
= 15 mol/L                                                                            3.4 1.0
100. g soln         cm 3         L           17.0 g
Nitric acid (HNO3):
28 g NH 3         1000 g      1 mol
= 23 mol/kg                                                     70. g HNO 3      1.42 g soln        1000 cm 3      1 mol HNO3
72 g H 2 O          kg        17.0 g                                                                                                                               = 16 mol/L
100. g soln       cm 3 soln            L              63.0 g
1 mol                                     1 mol
28 g NH3                  = 1.6 mol NH3; 72 g H2O                  = 4.0 mol H2O
17.0 g                                    18.0 g                                   70. g HNO 3       1000 g      1 mol HNO 3
= 37 mol/kg
1.6                                                                                 30. g solvent       kg           63.0 g
NH 3                    = 0.29
4.0 1.6
1 mol                              1 mol
70. g HNO3 ×           = 1.1 mol HNO3; 30. g H2O ×        = 1.7 mol H2O
32.   a. If we use 100. mL (100. g) of H2O, we need:                                                                           63.0 g                             18.0 g

2.0 mol KCl      74.55 g                                                               1 .1
0.100 kg H2O ×                                   = 14.9 g = 15 g KCl                              HNO 3             = 0.39
kg          mol KCl                                                            1.7 1.1
384                                                 CHAPTER 11        PROPERTIES OF SOLUTIONS           CHAPTER 11          PROPERTIES OF SOLUTIONS                                                                    383

mol pentane   0.22 mol                                                                    Dissolve 15 g KCl in 100. mL H2O to prepare a 2.0 m KCl solution. This will give us
Molality =                   =          = 7.3 mol/kg                                                       slightly more than 100 mL, but this will be the easiest way to make the solution. Because
kg hexane    0.030 kg
we don’t know the density of the solution, we can’t calculate the molarity and use a
volumetric flask to make exactly 100 mL of solution.
mol pentane      0.22 mol            1000 mL
Molarity =                   =                                = 3.1 mol/L
L solution   25 mL 45 mL              1L                                              b. If we took 15 g NaOH and 85 g H2O, the volume probably would be less than 100 mL.
To make sure we have enough solution, let’s use 100. mL H2O (100. g H2O). Let x =
0.867 g                                    0.874 g                                mass of NaCl.
34.   50.0 mL toluene ×                = 43.4 g toluene; 125 mL benzene ×         = 109 g benzene
mL                                         mL
x
Mass % = 15 =                 × 100, 1500 + 15x = (100.)x, x = 17.6 g           18 g
mass of toluene             43.4 g                                                                        100. x
Mass % toluene =                        × 100 =              × 100 = 28.5%
total mass            43.4 g 109 g
Dissolve 18 g NaOH in 100. mL H2O to make a 15% NaOH solution by mass.
43.4 g toluene            1000 mL      1 mol toluene
Molarity =                                                       = 2.69 mol/L                           c. In a fashion similar to part b, let’s use 100. mL CH3OH. Let x = mass of NaOH.
175 mL soln                 L        92.13 g toluene
0.79 g
43.4 g toluene      1000 g     1 mol toluene                                              100. mL CH3OH ×                       = 79 g CH3OH
Molality =                                                     = 4.32 mol/kg                                                               mL
109 g benzene         kg      92.13 g toluene
x
Mass % = 25 =                       × 100, 25(79) + 25x = (100.)x, x = 26.3 g   26 g
1 mol                                                                                                           79       x
43.4 g toluene ×         = 0.471 mol toluene
92.13 g                                                                                   Dissolve 26 g NaOH in 100. mL CH3OH.

1 mol benzene                                                  0.471                   d. To make sure we have enough solution, let’s use 100. mL (100. g) of H2O. Let x = mol
109 g benzene ×                 = 1.40 mol benzene;              toluene   =            = 0.252
78.11 g benzene                                              0.471 1.40                    C6H12O6.

35.   If we have 100.0 mL of wine:                                                                                                    1 mol H 2 O
100. g H2O ×                     = 5.55 mol H2O
18.02 g
0.789 g                                   1.00 g                                                             x
12.5 mL C2H5OH ×                  = 9.86 g C2H5OH and 87.5 mL H2O ×        = 87.5 g H2O                               = 0.10 =              , (0.10)x + 0.56 = x, x = 0.62 mol C6H12O6
mL                                       mL                                 C 6 H12 O 6
x     5.55

9.86 g                                                                                                         180.2 g
Mass % ethanol =                      × 100 = 10.1% by mass                                                0.62 mol C6H12O6 ×                      = 110 g C6H12O6
87.5 g 9.86 g                                                                                                        mol

9.86 g C 2 H 5 OH     1 mol                                                               Dissolve 110 g C6H12O6 in 100. mL of H2O to prepare a solution with               C 6 H12 O 6   = 0.10.
Molality =                                    = 2.45 mol/kg
0.0875 kg H 2 O      46.07 g
0.63 g                        0.63 g                  1 mol
33.   25 mL C5H12 ×        = 16 g C5H12; 25 mL ×                                  = 0.22 mol C5H12
1.00 mol acetone                                      1 mol                                                            mL                            mL                    72.15 g
36.                    = 1.00 molal; 1.00 × 103 g C2H5OH ×         = 21.7 mol C2H5OH
1.00 kg ethanol                                     46.07 g                                                          0.66 g                         0.66 g                  1 mol
45 mL C6H14 ×        = 30. g C6H14; 45 mL ×                                  = 0.34 mol C6H14
1.00                                                                                                 mL                             mL                    86.17 g
acetone   =             = 0.0441
1.00 21.7
mass pentane               16 g
Mass % pentane =                   × 100 =            × 100 = 35%
58.08 g CH 3COCH 3        1 mL                                                                     total mass           16 g 30. g
1 mol CH3COCH3                                              = 73.7 mL CH3COCH3                                     mol pentane           0.22 mol
mol CH 3COCH 3          0.788 g                                              pentane =              =                      = 0.39
total mol      0.22 mol 0.34 mol
386                                                   CHAPTER 11              PROPERTIES OF SOLUTIONS              CHAPTER 11         PROPERTIES OF SOLUTIONS                                                          385

5.3    10    2
mol H 3PO 4     3 equivalents         0.16 equivalents                                                 1 mL
c. Normality                                                                                                       1.00 × 103 g ethanol ×            = 1270 mL; total volume = 1270 + 73.7 = 1340 mL
L               mol H 3 PO 4                 L                                                        0.789 g

1.00 mol
Equivalent mass = 1/3(molar mass of H3PO4) = 1/3(97.09) = 32.66 g                                             Molarity =            = 0.746 M
1.34 L
0.134 mol NaOH            1 equivalent        0.134 equivalents
d. Normality                                                                                                 37.   If we have 1.00 L of solution:
L                  mol NaOH                    L
192.12 g
Equivalent mass = molar mass of NaOH = 40.00 g                                                                    1.37 mol citric acid ×            = 263 g citric acid (H3C6H5O7)
mol
0.00521 mol Ca (OH) 2            2 equivalents         0.0104 equivalents                                                     1.10 g
e. Normality                                                                                                           1.00 × 103 mL solution ×            = 1.10 × 103 g solution
L                       mol Ca (OH) 2                 L                                                               mL

263 g
Equivalent mass = 1/2[molar mass of Ca (OH) 2 ] = 1/2(74.10) = 37.05 g                                        Mass % of citric acid =                   × 100 = 23.9%
1.10    103 g
Energetics of Solutions and Solubility                                                                                   In 1.00 L of solution, we have 263 g citric acid and (1.10 × 103       263) = 840 g of H2O.

39.   Using Hess’s law:                                                                                                               1.37 mol citric acid
Molality =                        = 1.6 mol/kg
0.84 kg H 2 O
NaI(s)        Na+(g) + I (g)                           H=       HLE = ( 686 kJ/mol)
Na+(g) + I (g)       Na+(aq) + I (aq)                         H=       Hhyd = 694 kJ/mol
_______________________________________________________________________________________________________                            1 mol                                      1.37
840 g H2O ×            = 47 mol H2O;       citric acid           = 0.028
NaI(s)      Na+(aq) + I (aq)                         Hsoln =     8 kJ/mol                                               18.02 g                                   47 1.37

Hsoln refers to the heat released or gained when a solute dissolves in a solvent. Here, an                       Because citric acid is a triprotic acid, the number of protons citric acid can provide is three
ionic compound dissolves in water.                                                                                 times the molarity. Therefore, normality = 3 × molarity:
normality = 3 × 1.37 M = 4.11 N
40.   a.              CaCl2(s) Ca2+(g) + 2 Cl (g)                         H = HLE = ( 2247 kJ)
Ca2+(g) + 2 Cl (g) Ca2+(aq) + 2 Cl (aq)                        H = Hhyd
_____________________________________________________________________________________________________   38.   When expressing concentration in terms of normality, equivalents per liter are determined.
For acid-base reactions, equivalents are equal to the moles of H+ an acid can donate or the
CaCl2(s)         Ca2+(aq) + 2 Cl (aq)              Hsoln = 46 kJ
moles of H+ a base can accept. For monoprotic acids like HCl, the equivalents of H+
furnished equals the moles of acid present. Diprotic acids like H2SO4 furnish two equivalents
46 kJ = 2247 kJ + Hhyd,              Hhyd = 2293 kJ                                                         of H+ per mole of acid, whereas triprotic acids like H3PO4 furnish three equivalents of H+ per
2+                                                                          mole of acid. For the bases in this problem, the equivalents of H+ accepted equals the number
CaI2(s) Ca (g) + 2 I (g)                         H = HLE = ( 2059 kJ)                           of OH anions present in the formula (H+ + OH             H2O). Finally, the equivalent mass of
Ca2+(g) + 2 I (g) Ca2+(aq) + 2 I (aq)                        H = Hhyd
____________________________________________________________________________________________________         a substance is the mass of acid or base that can furnish or accept 1 mole of protons (H+ ions).
CaI2(s)      Ca2+(aq) + 2 I (aq)                 Hsoln = 104 kJ
0.250 mol HCl       1 equivalent       0.250 equivalents
a. Normality
104 kJ = 2059 kJ + Hhyd,              Hhyd = 2163 kJ                                                                                  L               mol HCl                  L

b. The enthalpy of hydration for CaCl2 is more exothermic than for CaI2. Any differences                               Equivalent mass = molar mass of HCl = 36.46 g
must be due to differences in hydration between Cl and I . Thus the chloride ion is more
strongly hydrated than the iodide ion.                                                                                             0.105 mol H 2SO 4       2 equivalents      0.210 equivalents
b. Normality
L                mol H 2SO 4                L
41.   Both Al(OH)3 and NaOH are ionic compounds. Since the lattice energy is proportional to the
charge of the ions, the lattice energy of aluminum hydroxide is greater than that of sodium
Equivalent mass = 1/2(molar mass of H2SO4) = 1/2(98.09) = 49.05 g
hydroxide. The attraction of water molecules for Al3+ and OH cannot overcome the larger
388                                             CHAPTER 11          PROPERTIES OF SOLUTIONS             CHAPTER 11         PROPERTIES OF SOLUTIONS                                                      387

44.   Water is a polar solvent and dissolves polar solutes and ionic solutes. Hexane (C6H14) is a             lattice energy, and Al(OH)3 is insoluble. For NaOH, the favorable hydration energy is large
nonpolar solvent and dissolves nonpolar solutes (like dissolves like).                                  enough to overcome the smaller lattice energy, and NaOH is soluble.

a. Water; Cu(NO3)2 is an ionic compound.                                                          42.   The dissolving of an ionic solute in water can be thought of as taking place in two steps. The
first step, called the lattice-energy term, refers to breaking apart the ionic compound into
b. C6H14; CS2 is a nonpolar molecule.                c. Water; CH3CO2H is polar.                        gaseous ions. This step, as indicated in the problem, requires a lot of energy and is
unfavorable. The second step, called the hydration-energy term, refers to the energy released
d. C6H14; the long nonpolar hydrocarbon chain favors a nonpolar solvent (the molecule                   when the separated gaseous ions are stabilized as water molecules surround the ions. Because
is mostly nonpolar).                                                                                 the interactions between water molecules and ions are strong, a lot of energy is released when
ions are hydrated. Thus the dissolution process for ionic compounds can be thought of as
e. Water; HCl is polar.                              f.   C6H14; C6H6 is nonpolar.                      consisting of an unfavorable and a favorable energy term. These two processes basically
cancel each other out, so when ionic solids dissolve in water, the heat released or gained is
45.   a. NH3; NH3 is capable of H-bonding, unlike PH3.                                                        minimal, and the temperature change is minimal.

b. CH3CN; CH3CN is polar, while CH3CH3 is nonpolar.                                               43.   Water is a polar solvent and dissolves polar solutes and ionic solutes. Carbon tetrachloride
(CCl4) is a nonpolar solvent and dissolves nonpolar solutes (like dissolves like). To predict
c. CH3CO2H; CH3CO2H is capable of H-bonding, unlike the other compound.                                 the polarity of the following molecules, draw the correct Lewis structure and then determine
if the individual bond dipoles cancel or not. If the bond dipoles are arranged in such a manner
46.   For ionic compounds, as the charge of the ions increases and/or the size of the ions decreases,         that they cancel each other out, then the molecule is nonpolar. If the bond dipoles do not
the attraction to water (hydration) increases.                                                          cancel each other out, then the molecule is polar.

a. Mg2+; smaller size, higher charge                 b. Be2+; smaller size                              a. KrF2, 8 + 2(7) = 22 e                     b. SF2, 6 + 2(7) = 20 e

c. Fe3+; smaller size, higher charge                 d. F ; smaller size                                                                                          S
F        Kr       F                        F              F
e. Cl ; smaller size                                 f.   SO42-; higher charge
nonpolar; soluble in CCl4                   polar; soluble in H2O
47.   As the length of the hydrocarbon chain increases, the solubility decreases. The OH end of
the alcohols can hydrogen-bond with water. The hydrocarbon chain, however, is basically                 c. SO2, 6 + 2(6) = 18 e                      d. CO2, 4 + 2(6) = 16 e
nonpolar and interacts poorly with water. As the hydrocarbon chain gets longer, a greater
portion of the molecule cannot interact with the water molecules, and the solubility decreases;
S            + 1 more
i.e., the effect of the OH group decreases as the alcohols get larger.                                                                                      O         C           O
O            O
48.   The main intermolecular forces are:
polar; soluble in H2O                      nonpolar; soluble in CCl4
hexane (C6H14): London dispersion; chloroform (CHCl3): dipole-dipole, London
dispersion; methanol (CH3OH): H-bonding; and H2O: H-bonding (two places)                            e. MgF2 is an ionic compound so it is soluble in water.

There is a gradual change in the nature of the intermolecular forces (weaker to stronger).              f.   CH2O, 4 + 2(1) + 6 = 12 e               g. C2H4, 2(4) + 4(1) = 12 e
Each preceding solvent is miscible in its predecessor because there is not a great change in
the strengths of the intermolecular forces from one solvent to the next.                                                 O
H                         H
4                                                                                                                                           C           C
8.21   10       mol                                                                                            C
49.   C = kP,                         = k × 0.790 atm, k = 1.04 × 10 3 mol/L atm                                                                                  H                         H
L                                                                                                   H        H

1.04     10 4 mol
C = kP, C =                     × 1.10 atm = 1.14 × 10 3 mol/L                                               polar; soluble in H2O                       nonpolar (like all compounds made up of
L atm                                                                                                                              only carbon and hydrogen); soluble in CCl4
390                                                           CHAPTER 11                  PROPERTIES OF SOLUTIONS                  CHAPTER 11            PROPERTIES OF SOLUTIONS                                                                             389

56.   19.6 torr =               (23.8 torr),           = 0.824;        solute   = 1.000        0.824 = 0.176                                              1 .3        10 3 mol               1 atm
50.   C = kP =                              × 120 torr ×          = 2.1 × 10 4 mol/L
L atm                  760 torr
0.176 is the mol fraction of all the solute particles present. Because NaCl dissolves to pro-
duce two ions in solution (Na+ and Cl ), 0.176 is the mole fraction of Na+ and Cl ions present                               Vapor Pressures of Solutions
(assuming complete dissociation of NaCl).
1 mol
51.   Mol C3H8O3 = 164 g ×                           = 1.78 mol C3H8O3
At 45°C, PH 2O = 0.824(71.9 torr) = 59.2 torr                                                                                                                             92.09 g

0.63 g       1 mol                                                                                                                        0.992 g        1 mol
57.   a. 25 mL C5H12 ×                                    = 0.22 mol C5H12                                                               Mol H2O = 338 mL ×                                           = 18.6 mol H2O
mL         72.15 g                                                                                                                         mL          18.02 g

0.66 g       1 mol                                                                                                    o                  18.6 mol
45 mL C6H14 ×                                  = 0.34 mol C6H14; total mol = 0.22 + 0.34 = 0.56 mol                           Psoln        H 2O PH 2O       =                   × 54.74 torr = 0.913 × 54.74 torr = 50.0 torr
mL         86.17 g                                                                                                                  (1.78 18.6) mol

L       mol pentane in solution                  0.22 mol                      L
pen                                                       = 0.39,              hex    = 1.00 - 0.39 = 0.61                                        o                                mol C 2 H 5 OH solution
total mol in solution                   0.56 mol                                                           52.   Psoln =     C 2 H 5 OH PC 2 H 5 OH ;     C 2 H 5OH   =
total mol in solution

Ppen            L    o
pen Ppen   = 0.39(511 torr) = 2.0 × 102 torr; Phex = 0.61(150. torr) = 92 torr                                                            1 mol C 3 H 8 O 3
53.6 g C3H8O3 ×                               = 0.582 mol C3H8O3
92.09 g
Ptotal      Ppen       Phex = 2.0 × 102 + 92 = 292 torr = 290 torr
1 mol C 2 H 5 OH
133.7 g C2H5OH ×                      = 2.90 mol C2H5OH; total mol = 0.582 + 2.90
b. From Chapter 5 on gases, the partial pressure of a gas is proportional to the number of                                                                46.07 g
moles of gas present. For the vapor phase:                                                                                                                                                                  = 3.48 mol
2.90 mol     o           o
113 torr =            PC 2H 5OH , PC2 H 5OH = 136 torr
V         mol pentane in vapor                Ppen         2.0 10 2 torr                                                       3.48 mol
pen                                                                      = 0.69
total mol vapor                   Ptotal          290 torr
53.   P = P°; 710.0 torr = (760.0 torr),                          = 0.9342 = mole fraction of methanol
Note: In the Solutions Guide, we added V or L to the mole fraction symbol to emphasize
which value we are solving. If the L or V is omitted, then the liquid phase is assumed.                                                         o                      o
54.   PB        B PB ,        B     PB / PB      = 0.900 atm/0.930 atm = 0.968
L o        L             0.0300 mol CH 2 Cl 2
58.   Ptotal = PCH 2Cl 2         PCH 2Br2 ; P             P ;      CH 2 Cl 2                          = 0.375                                     mol benzene                                   1 mol
0.0800 mol total                                      0.968 =                ; mol benzene = 78.11 g C6H6 ×          = 1.000 mol
total mol                                   78.11 g
Ptotal = 0.375(133 torr) + (1.000                 0.375)(11.4 torr) = 49.9 + 7.13 = 57.0 torr                                                                            1.000 mol
Let x = mol solute, then: B = 0.968 =            , 0.968 + (0.968)x = 1.000, x = 0.033 mol
V
PCH 2Cl 2     49.9 torr                        V
1.000 x
In the vapor:           CH 2 Cl 2                            = 0.875;               CH 2 Br2   = 1.000 – 0.875 = 0.125
Ptotal       57.0 torr                                                                                                 10.0 g
Molar mass =                      = 303 g/mol                 3.0 × 102 g/mol
0.033 mol
Note: In the Solutions Guide, we added V or L to the mole fraction symbol to emphasize
which value we are solving. If the L or V is omitted, then the liquid phase is assumed.
1 mol                            1 mol
55.   25.8 g CH4N2O ×                       = 0.430 mol; 275 g H2O ×         = 15.3 mol
59.   Ptotal = Pmeth + Pprop, 174 torr =               L                        L                         L                 L                                          60.06 g                          18.02 g
meth (303   torr )       prop ( 44.6   torr );     prop   1.000      meth

15.3                                                o
L                         L                     129          L                                                      H 2O   =                               0.973; Psoln =        H 2O PH 2O   = 0.973(23.8 torr) = 23.2 torr at 25°C
174 = 303       meth        (1.000        meth ) 44.6   torr,                meth       0.500                                                  15.3 0.430
258
L
prop    1.000 0.500                0.500                                                                                         Psoln = 0.973(71.9 torr) = 70.0 torr at 45°C
392                                               CHAPTER 11           PROPERTIES OF SOLUTIONS            CHAPTER 11              PROPERTIES OF SOLUTIONS                                                                                 391

b. From the data, the vapor pressures of the various solutions are greater than if the              60.   Ptol         L     o                  L    o                             V
tol Ptol ,   Ppen        ben Pben ;   for the vapor,        A   = PA/Ptotal. Because the mole fractions of
solutions behaved ideally (positive deviation from Raoult’s law). This occurs when the
intermolecular forces in solution are weaker than the intermolecular forces in pure                    benzene and toluene are equal in the vapor phase, Ptol                                  Pben .
solvent and pure solute. This gives rise to endothermic (positive) Hsoln values.
L     o          L    o                      L      o       L                                   L
tol Ptol         ben Pben        (1.00       tol ) Pben ,   tol   (28 torr )    (1.00           tol ) 95   torr
c. The interactions between propanol and water molecules are weaker than between the pure
substances because the solutions exhibit a positive deviation from Raoult’s law.                              L                  L                   L
123    tol       95,      tol     0.77;       ben   = 1.00 – 0.77 = 0.23
d. At H 2O = 0.54, the vapor pressure is highest as compared to the other solutions. Because
a solution boils when the vapor pressure of the solution equals the external pressure, the       61.   Compared to H2O, solution d (methanol-water) will have the highest vapor pressure since
o
methanol is more volatile than water (PH 2O = 23.8 torr at 25°C). Both solution b (glucose-
H 2 O = 0.54 solution should have the lowest normal boiling point; this solution will have
a vapor pressure equal to 1 atm at a lower temperature as compared to the other solutions.             water) and solution c (NaCl-water) will have a lower vapor pressure than water by Raoult's
law. NaCl dissolves to give Na+ ions and Cl ions; glucose is a nonelectrolyte. Because there
are more solute particles in solution c, the vapor pressure of solution c will be the lowest.
Colligative Properties
62.   Solution d (methanol-water); methanol is more volatile than water, which will increase the
mol solute     27.0 g N 2 H 4 CO    1000 g    1 mol N 2 H 4 CO                        total vapor pressure to a value greater than the vapor pressure of pure water at this
65.   Molality = m =                                                                       = 3.00 molal
kg solvent      150.0 g H 2 O         kg      60.06 g N 2 H 4 CO                      temperature.

0.51 C                                                                                                                           1 mol
Tb = Kbm =           × 3.00 molal = 1.5°C                                                          63.   50.0 g CH3COCH3 ×                            = 0.861 mol acetone
molal                                                                                                                          58.08 g
The boiling point is raised from 100.0 to 101.5°C (assuming P = 1 atm).
1 mol
50.0 g CH3OH ×                          = 1.56 mol methanol
Tb             1.35 C                                                                        32.04 g
66.    Tb = 77.85°C         76.50°C = 1.35°C; m =                             = 0.268 mol/kg
Kb          5.03 C kg / mol
L                  0.861                               L                          L
acetone     =              = 0.356;                    methanol    = 1.000        acetone   = 0.644
0.861 1.56
0.268 mol hydrocarbon
Mol biomolecule = 0.0150 kg solvent ×                       = 4.02 × 10 3 mol
kg solvent
Ptotal = Pmethanol + Pacetone = 0.644(143 torr) + 0.356(271 torr) = 92.1 torr + 96.5 torr
From the problem, 2.00 g biomolecule was used that must contain 4.02 × 10 3 mol                                                                                                                                              = 188.6 torr
biomolecule. The molar mass of the biomolecule is:
Because partial pressures are proportional to the moles of gas present, in the vapor phase:
2.00 g
3
= 498 g/mol                                                                                             Pacetone         96.5 torr
4.02     10       mol                                                                                         V                                                                    V
acetone                                      = 0.512;                methanol   = 1.000 - 0.512 = 0.488
1.86 C                                                                                                  Ptotal         188.6 torr
67.    Tf = Kfm,     Tf = 1.50°C =        × m, m = 0.806 mol/kg
molal                                                                       The actual vapor pressure of the solution (161 torr) is less than the calculated pressure
assuming ideal behavior (188.6 torr). Therefore, the solution exhibits negative deviations
0.806 mol C3H 8O 3  92.09 g C3H 8O3                                                 from Raoult’s law. This occurs when the solute-solvent interactions are stronger than in pure
0.200 kg H2O ×                                              = 14.8 g C3H8O3
kg H 2 O          mol C3H 8O 3                                                 solute and pure solvent.
0.91 C
68.    Tf = 25.50°C         24.59°C = 0.91°C = Kfm, m =                = 0.10 mol/kg                      64.   a. An ideal solution would have a vapor pressure at any mole fraction of H2O between that
9.1 C / molal                                             of pure propanol and pure water (between 74.0 and 71.9 torr). The vapor pressures of the
various solutions are not between these limits, so water and propanol do not form ideal
0.10 mol H 2 O   18.02 g H 2 O                                    solutions.
Mass H2O = 0.0100 kg t-butanol                                            = 0.018 g H2O
kg t - butanol     mol H 2 O
394                                                       CHAPTER 11          PROPERTIES OF SOLUTIONS           CHAPTER 11         PROPERTIES OF SOLUTIONS                                                               393

1.0 g protein            1 mol                                                                                              50.0 g C 2 H 6 O 2       1000 g       1 mol
73.   a. M =                                               = 1.1 × 10 5 mol/L;      = MRT                       69.   Molality = m =                                                     = 16.1 mol/kg
L                9.0       10 4 g                                                                                       50.0 g H 2 O             kg        62.07 g

5
Tf = Kfm = 1.86°C/molal × 16.1 molal = 29.9°C; Tf = 0.0°C                  29.9°C = 29.9°C
1.1       10        mol       0.08206 L atm           760 torr
At 298 K:     =                                                × 298 K ×          ,     = 0.20 torr
L                       K mol                 atm                               Tb = Kbm = 0.51°C/molal × 16.1 molal = 8.2°C; Tb = 100.0°C + 8.2°C = 108.2°C
3
Because d = 1.0 g/cm , 1.0 L solution has a mass of 1.0 kg. Because only 1.0 g of protein                          Tf             25.0 C
is present per liter of solution, 1.0 kg of H2O is present to the correct number of signifi-          70.   m=                                = 13.4 mol C2H6O2/kg
Kf          1.86 C kg / mol
cant figures, and molality equals molarity.
Because the density of water is 1.00 g/cm3, the moles of C2H6O2 needed are:
1.86 C
Tf = Kfm =          × 1.1 × 10 5 molal = 2.0 × 10 5 °C
molal                                                                                                                1.00 kg H 2 O          13.4 mol C 2 H 6 O 2
15.0 L H2O ×                                                   = 201 mol C2H6O2
b. Osmotic pressure is better for determining the molar mass of large molecules. A tem-                                                    L H 2O                   kg H 2 O
perature change of 10 5 °C is very difficult to measure. A change in height of a column
of mercury by 0.2 mm (0.2 torr) is not as hard to measure precisely.                                                                                            62.07 g          1 cm 3
Volume C2H6O2 = 201 mol C2H6O2 ×                                           = 11,200 cm3 = 11.2 L
mol C 2 H 6 O 2   1.11 g
1 atm       0.08206 L atm
74.    = MRT,      = 18.6 torr ×                 =M×               × 298 K, M = 1.00 × 10 3 mol/L                                       0.51 C
760 torr         K mol                                                         Tb = Kbm =              × 13.4 molal = 6.8°C; Tb = 100.0°C + 6.8°C = 106.8°C
molal
3
1.00       10       mol protein
Mol protein = 0.0020 L ×                                         = 2.0 × 10 6 mol protein                                                Tf             2.63o C         6 .6   10 2 mol reserpine
L                                                     71.    Tf = Kfm, m
Kf              o                       kg solvent
40. C kg/mol
0.15 g
Molar mass =                 6
= 7.5 × 104 g/mol
2 .0    10       mol                                                                            The mol of reserpine present is:
15 atm
75.    = MRT, M                                                  0.62 M                                                                                6 .6       10 2 mol reserpine
RT         0.08206 295 K                                                                           0.0250 kg solvent ×                                       = 1.7 × 10 3 mol reserpine
kg solvent
0.62 mol       342.30 g
= 212 g/L                 210 g/L                                                 From the problem, 1.00 g reserpine was used, which must contain 1.7 × 10 3 mol reserpine.
L         mol C12 H 22 O11
The molar mass of reserpine is:
Dissolve 210 g of sucrose in some water and dilute to 1.0 L in a volumetric flask. To get 0.62
±0.01 mol/L, we need 212 ±3 g sucrose.                                                                                      1.00 g
3
= 590 g/mol (610 g/mol if no rounding of numbers)
1.7     10       mol
15 atm
76.   M=                                                  = 0.62 M solute particles
RT      0.08206 L atm
295 K                                                                           Tb              0.55o C
K mol                                                                                    72.   m                                            0.32 mol/kg
Kb          1.71 o C kg/mol
This represents the total molarity of the solute particles. NaCl is a soluble ionic compound
0.32 mol hydrocarbon
that breaks up into two ions, Na+ and Cl . Therefore, the concentration of NaCl needed is                       Mol hydrocarbon = 0.095 kg solvent ×                                    = 0.030 mol hydrocarbon
0.62/2 = 0.31 M; this NaCl concentration will produce a 0.62 M solute particle solution                                                                                 kg solvent
assuming complete dissociation.                                                                                 From the problem, 3.75 g hydrocarbon was used, which must contain 0.030 mol hydrocarbon.
The molar mass of the hydrocarbon is:
0.31 mol NaCl          58.44 g NaCl
1.0 L ×                                       = 18.1              18 g NaCl
L                 mol NaCl                                                                             3.75 g
= 130 g/mol (120 g/mol if no rounding of numbers)
0.030 mol
Dissolve 18 g of NaCl in some water and dilute to 1.0 L in a volumetric flask. To get 0.31
±0.01 mol/L, we need 18.1 g ±0.6 g NaCl in 1.00 L solution.
396                                             CHAPTER 11          PROPERTIES OF SOLUTIONS                CHAPTER 11        PROPERTIES OF SOLUTIONS                                                          395

80.   NaCl(s)     Na+(aq) + Cl (aq), i = 2.0                                                               Properties of Electrolyte Solutions
0.10 mol      0.08206 L atm                                                      77.   Na3PO4(s)      3 Na+(aq) + PO43 (aq), i = 4.0; CaBr2(s)      Ca2+(aq) + 2 Br (aq), i = 3.0
= iMRT = 2.0 ×                                 × 293 K = 4.8 atm
L             K mol
KCl(s)     K+(aq) + Cl (aq), i = 2.0
A pressure greater than 4.8 atm should be applied to ensure purification by reverse osmosis.
The effective particle concentrations of the solutions are (assuming complete dissociation):

4.0(0.010 molal) = 0.040 molal for the Na3PO4 solution; 3.0(0.020 molal) = 0.060 molal
81.   a. MgCl2(s)       Mg2+(aq) + 2 Cl (aq), i = 3.0 mol ions/mol solute
for the CaBr2 solution; 2.0(0.020 molal) = 0.040 molal for the KCl solution; slightly
greater than 0.020 molal for the HF solution because HF only partially dissociates in
Tf = iKfm = 3.0 × 1.86 °C/molal × 0.050 molal = 0.28°C; Tf = -0.28°C (Assuming water
water (it is a weak acid).
freezes at 0.00°C.)
a. The 0.010 m Na3PO4 solution and the 0.020 m KCl solution both have effective particle
Tb = iKbm = 3.0 × 0.51 °C/molal × 0.050 molal = 0.077°C; Tb = 100.077°C (Assuming
concentrations of 0.040 m (assuming complete dissociation), so both of these solutions
water boils at 100.000°C.)
should have the same boiling point as the 0.040 m C6H12O6 solution (a nonelectrolyte).
b. FeCl3(s)      Fe3+(aq) + 3 Cl (aq), i = 4.0 mol ions/mol solute
b. P = P°; as the solute concentration decreases, the solvent’s vapor pressure increases
Tf = iKfm = 4.0 × 1.86 °C/molal × 0.050 molal = 0.37°C; Tf = 0.37°C                                     because increases. Therefore, the 0.020 m HF solution will have the highest vapor
Tb = iKbm = 4.0 × 0.51 °C/molal × 0.050 molal = 0.10°C; Tb = 100.10°C                                   pressure because it has the smallest effective particle concentration.

82.   a. MgCl2, i (observed) = 2.7                                                                               c.     T = Kfm; the 0.020 m CaBr2 solution has the largest effective particle concentration, so
it will have the largest freezing point depression (largest T).
Tf = iKfm = 2.7 × 1.86 °C/molal × 0.050 molal = 0.25°C; Tf = 0.25°C
Tb = iKbm = 2.7 × 0.51 °C/molal × 0.050 molal = 0.069°C; Tb = 100.069°C                        78.   The solutions of C12H22O11, NaCl, and CaCl2 will all have lower freezing points, higher
boiling points, and higher osmotic pressures than pure water. The solution with the largest
b. FeCl3, i (observed) = 3.4                                                                               particle concentration will have the lowest freezing point, the highest boiling point, and the
highest osmotic pressure. The CaCl2 solution will have the largest effective particle
Tf = iKfm = 3.4 × 1.86 °C/molal × 0.050 molal = 0.32°C; Tf = -0.32°C                                 concentration because it produces three ions per mole of compound.
Tb = iKbm = 3.4 × 0.51°C/molal × 0.050 molal = 0.087°C; Tb = 100.087°C
a. pure water             b. CaCl2 solution             c. CaCl2 solution
Tf                     0.110 C                                                            d. pure water             e. CaCl2 solution
83.    Tf = iKfm, i =                                   = 2.63 for 0.0225 m CaCl2
Kf m                 0.0225 molal
1.86 C / molal
0.440                                     1.330                                                               5.0 g NaCl      1 mol
i=             = 2.60 for 0.0910 m CaCl2; i =              = 2.57 for 0.278 m CaCl2                  79.   a. m =                           = 3.4 molal; NaCl(aq)       Na+(aq) + Cl (aq), i = 2.0
1.86 0.0910                                1.86 0.278                                                              0.025 kg      58.44 g

iave = (2.63 + 2.60 + 2.57)/3 = 2.60                                                                             Tf = iKfm = 2.0 × 1.86°C/molal × 3.4 molal = 13°C; Tf = 13°C

Note that i is less than the ideal value of 3.0 for CaCl2. This is due to ion pairing in solution.               Tb = iKbm = 2.0 × 0.51°C/molal × 3.4 molal = 3.5°C; Tb = 103.5°C
Also note that as molality increases, i decreases. More ion pairing appears to occur as the
solute concentration increases.                                                                                      2.0 g Al( NO3 ) 3    1 mol
b.   m=                                = 0.63 mol/kg
0.015 kg        213.01 g
Tf                   0.440 o C
84.   For CaCl2: i                                                = 2.6
Kf m       1.86 o C/molal    0.091 molal                                                       Al(NO3)3(aq)      Al3+(aq) + 3 NO3 (aq), i = 4.0

Tf = iKfm = 4.0 × 1.86°C/molal × 0.63 molal = 4.7°C; Tf = 4.7°C
2.6 1.0
Percent CaCl2 ionized =             × 100 = 80.%; 20.% ion association occurs.
3.0 1.0                                                                          Tb = iKbm = 4.0 × 0.51°C/molal × 0.63 molal = 1.3°C; Tb = 101.3°C
398                                             CHAPTER 11              PROPERTIES OF SOLUTIONS       CHAPTER 11       PROPERTIES OF SOLUTIONS                                                                 397

Tf                      0.320 o C
O      H       O                                                                    For CsCl: i                                                        = 1.9
Kf m              o
1.86 C/molal         0.091 molal
C                      C
O      H       O                                                                                                            1.9 1.0
Percent CsCl ionized =                    × 100 = 90.%; 10% ion association occurs.
2.0 1.0
The dimer is relatively nonpolar and thus more soluble in benzene than in water.
Benzoic acid would be more soluble in a basic solution because of the reaction C6H5CO2H +             The ion association is greater in the CaCl2 solution.
OH       C6H5CO2 + H2O. By removing the acidic proton from benzoic acid, an anion
forms, and like all anions, the species becomes more soluble in water.                          85.   a. TC = 5(TF      32)/9 = 5( 29          32)/9 = 34°C
o
Tf          1.32 C                                                                            Assuming the solubility of CaCl2 is temperature independent, the molality of a saturated
88.   a. m =                               = 0.258 mol/kg
Kf     5.12 o C kg/mol                                                                         CaCl2 solution is:

0.258 mol unknown                                                               74.5 g CaCl 2       1000 g         1 mol CaCl 2            6.71 mol CaCl 2
Mol unknown = 0.01560 kg ×                     = 4.02 × 10 3 mol
kg                                                                      100.0 g H 2 O         kg          110.98 g CaCl 2              kg H 2 O
1.22 g
Molar mass of unknown =                 = 303 g/mol
4.02 10 3 mol                                                                  Tf = iKfm = 3.00 × 1.86 °C kg/mol × 6.71 mol/kg = 37.4°C

Assuming i = 3.00, a saturated solution of CaCl2 can lower the freezing point of water to
0.04
Uncertainty in temperature =          × 100 = 3%                                                       37.4°C. Assuming these conditions, a saturated CaCl2 solution should melt ice at 34°C
1.32                                                                 ( 29°F).
A 3% uncertainty in 303 g/mol = 9 g/mol.
b. From Exercise 83, i        2.6;        Tf = iKfm = 2.6 × 1.86 × 6.71 = 32°C; Tf = 32°C.
So molar mass = 303 ±9 g/mol.
Assuming i = 2.6, a saturated CaCl2 solution will not melt ice at 34°C ( 29°F).
b. No, codeine could not be eliminated since its molar mass is in the possible range
including the uncertainty.                                                                                                                      2.50 atm                                  2
86.     = iMRT, M =            =                                               = 5.11 × 10       mol/L
iRT                    0.08206 L atm
2.00                              298 K
c. We would like the uncertainty to be ±1 g/mol. We need the freezing-point depression to                                                          K mol
be about 10 times what it was in this problem. Two possibilities are:
0.500 g
Molar mass of compound =                                       2
= 97.8 g/mol
(1) make the solution 10 times more concentrated (may be solubility problem)                                                                  5.11     10       mol
0.1000 L
L
(2) use a solvent with a larger Kf value, e.g., camphor
Connecting to Biochemistry
2
Tf           0.300 C           5.86   10       mol thyroxine
89.    Tf = Kfm, m =
Kf         5.12 C kg/mol               kg benzene                             87.

The moles of thyroxine present are:                                                                                 O

2
C    O     H
5.86   10       mol thyroxine                 4
0.0100 kg benzene ×                                       = 5.86 × 10       mol thyroxine
kg benzene
Benzoic acid is capable of hydrogen-bonding, but a significant part of benzoic acid is the
From the problem, 0.455 g thyroxine was used; this must contain 5.86 × 10 4 mol thyroxine.
nonpolar benzene ring. In benzene, a hydrogen-bonded dimer forms.
The molar mass of the thyroxine is:
400                                          CHAPTER 11          PROPERTIES OF SOLUTIONS         CHAPTER 11              PROPERTIES OF SOLUTIONS                                                 399

94.   a. The average values for each ion are:                                                                                               0.455 g
molar mass =                                     = 776 g/mol
+          +               2+
5.86      10 4 mol
300. mg Na , 15.7 mg K , 5.45 mg Ca , 388 mg Cl , and 246 mg lactate (C3H5O3 )

Note: Because we can precisely weigh to ±0.1 mg on an analytical balance, we'll carry                                                     1 atm
0.745 torr
extra significant figures and calculate results to ±0.1 mg.                                                                              760 torr
90.   M=                                                     = 3.98 × 10 5 mol/L
RT          0.08206 L atm
300. K
The only source of lactate is NaC3H5O3.                                                                                K mol
5
112.06 mg NaC3 H 5 O 3                                                              3.98       10       mol
246 mg C3H5O3 ×                                = 309.5 mg sodium lactate                  1.00 L ×                                  = 3.98 × 10 5 mol catalase
89.07 mg C 3 H 5 O 3                                                                          L
10.00 g
The only source of Ca2+ is CaCl2 2H2O.                                                        Molar mass =                          = 2.51 × 105 g/mol
3.98 10 5 mol
147.01 mg CaCl 2 2H 2 O
5.45 mg Ca2+ ×                                 = 19.99 or 20.0 mg CaCl2 2H2O                                                              8.00 atm
40.08 mg Ca 2                                                  91.     = MRT, M =                                                            = 0.327 mol/L
RT         0.08206 L atm/K mol        298 K
The only source of K+ is KCl.
T           0.406 o C
92.   m=                                = 0.218 mol/kg
+74.55 mg KCl                                                                           Kf        1.86 o C/molal
15.7 mg K ×              = 29.9 mg KCl
39.10 mg K
= MRT, where M = mol/L; we must assume that molarity = molality so that we can
calculate the osmotic pressure. This is a reasonable assumption for dilute solutions when 1.00
From what we have used already, let's calculate the mass of Na+ added.                        kg of water 1.00 L of solution. Assuming complete dissociation of NaCl, a 0.218 m solution
corresponds to 6.37 g NaCl dissolved in 1.00 kg of water. The volume of solution may be a
309.5 mg sodium lactate       246.0 mg lactate = 63.5 mg Na+                              little larger than 1.00 L but not by much (to three sig. figs.). The assumption that molarity =
molality will be good here.
Thus we need to add an additional 236.5 mg Na+ to get the desired 300. mg.
= (0.218 M)(0.08206 L atm/K mol)(298 K) = 5.33 atm
58.44 mg NaCl
236.5 mg Na+ ×                   = 601.2 mg NaCl
22.99 mg Na                                                                                                       0.995 g
93.   Mass of H2O = 160. mL                              = 159 g = 0.159 kg
mL
Now let's check the mass of Cl added:
0.378 mol
Mol NaDTZ = 0.159 kg                                 = 0.0601 mol
kg
70.90 mg Cl
20.0 mg CaCl2 2H2O ×                              = 9.6 mg Cl
147.01 mg CaCl 2 2H 2 O                                                                                     38.4 g
Molar mass of NaDTZ =                                = 639 g/mol
0.0601 mol
20.0 mg CaCl2 2H2O = 9.6 mg Cl
29.9 mg KCl 15.7 mg K+ = 14.2 mg Cl                                                            o                                  1 mol
Psoln =       H 2 O PH 2 O ;   mol H2O = 159 g                 = 8.82 mol
601.2 mg NaCl 236.5 mg Na+ = 364.7 mg Cl                                                                                              18.02 g
_______________________________________
Sodium diatrizoate is a salt because there is a metal (sodium) in the compound. From the
Total Cl = 388.5 mg Cl
short-hand notation for sodium diatrizoate, NaDTZ, we can assume this salt breaks up into
Na+ and DTZ ions. So the moles of solute particles are 2(0.0601) = 0.120 mol solute
This is the quantity of Cl we want (the average amount of Cl ).
particles.
An analytical balance can weigh to the nearest 0.1 mg. We would use 309.5 mg sodium                               8.82 mol
lactate, 20.0 mg CaCl2 2H2O, 29.9 mg KCl, and 601.2 mg NaCl.                                    H 2O   =                       = 0.987; Psoln = 0.987 × 34.1 torr = 33.7 torr
0.120 mol 8.82 mol
402                                                          CHAPTER 11           PROPERTIES OF SOLUTIONS                  CHAPTER 11        PROPERTIES OF SOLUTIONS                                                          401

NH4+(g) + NO3 (g)                        NH4NO3(s)        H = lattice energy                                 b. To get the range of osmotic pressure, we need to calculate the molar concentration of
each ion at its minimum and maximum values. At minimum concentrations, we have:
NH4+(g) + NO3 (g)                      NH4+(aq) + NO3 (aq)              H=       Hhyd = 630. kJ/mol
NH4+(aq) + NO3 (aq)                    NH4NO3(s)                    H = Hsoln = 26.6 kJ/mol                                  285 mg Na        1 mmol             14.1 mg K         1 mmol
_________________________________________________________________________________________________________                                             = 0.124 M;                           = 0.00361 M
100. mL       22.99 mg             100. mL         39.10 mg
+
NH4 (g) + NO3 (g)                    NH4NO3(s)                        H = Hhyd          Hsoln
= 657 kJ/mol                      4.9 mg Ca 2       1 mmol              368 mg Cl         1 mmol
= 0.0012 M;                           = 0.104 M
100. mL        40.08 mg              100. mL         35.45 mg
12 mL C 2 H 5 OH            0.79 g C 2 H 5 OH   1 mol C 2 H 5 OH
96.   750. mL grape juice ×
100. mL juice                    mL               46.07 g                                           231 mg C 3 H 5 O 3    1 mmol
= 0.0259 M (Note: Molarity = mol/L = mmol/mL.)
100. mL          89.07 mg
2 mol CO 2
= 1.54 mol CO2 (carry extra significant figure)                     Total = 0.124 + 0.00361 + 0.0012 + 0.104 + 0.0259 = 0.259 M
2 mol C 2 H 5 OH
0.259 mol       0.08206 L atm
1.54 mol CO2 = total mol CO2 = mol CO2(g) + mol CO2(aq) = ng + naq                                                               = MRT =                                   × 310. K = 6.59 atm
L               K mol

0.08206 L atm                                                                                  Similarly, at maximum concentrations, the concentration for each ion is:
ng                   (298 K )
n g RT                    mol K                                                                                            Na+: 0.137 M; K+: 0.00442 M; Ca2+: 0.0015 M; Cl : 0.115 M; C3H5O3 : 0.0293 M
PCO 2                                             3
= 326ng
V                            75   10       L

n aq                                                                                             The total concentration of all ions is 0.287 M.
C              0.750 L
PCO 2                                        = (43.0)naq                                                                            0.287 mol      0.08206 L atm
k        3.1    10 2 mol                                                                                               =                                × 310. K = 7.30 atm
L              K mol
L atm
Osmotic pressure ranges from 6.59 atm to 7.30 atm.
PCO 2 = 326ng = (43.0)naq, and from above, naq = 1.54                      ng; solving:

326ng = 43.0(1.54                ng), 369ng = 66.2, ng = 0.18 mol
PCO 2 = 326(0.18) = 59 atm in gas phase;
95.   a. NH4NO3(s)          NH4+(aq) + NO3 (aq)      Hsoln = ?
2                                                                                         Heat gain by dissolution process = heat loss by solution; we will keep all quantities posi-
3.1    10 mol            1.8 mol CO 2
C = kPCO 2 =                     × 59 atm =              (in wine)                                                             tive in order to avoid sign errors. Because the temperature of the water decreased, the
L atm                   L
dissolution of NH4NO3 is endothermic ( H is positive). Mass of solution = 1.60 + 75.0
= 76.6 g.
97.   a. Water boils when the vapor pressure equals the pressure above the water. In an open pan,
Patm 1.0 atm. In a pressure cooker, Pinside > 1.0 atm, and water boils at a higher temper-                                                             4.18 J
ature. The higher the cooking temperature, the faster is the cooking time.                                                  Heat loss by solution =           × 76.6 g × (25.00°C    23.34°C) = 532 J
Cg
b. Salt dissolves in water, forming a solution with a melting point lower than that of pure                                                                    80.05 g NH 4 NO3
532 J
water ( Tf = Kfm). This happens in water on the surface of ice. If it is not too cold, the                                   Hsoln =                                         = 2.66 × 104 J/mol = 26.6 kJ/mol
ice melts. This won't work if the ambient temperature is lower than the depressed                                                      1.60 g NH 4 NO 3       mol NH 4 NO3
freezing point of the salt solution.
b. We will use Hess’s law to solve for the lattice energy. The lattice-energy equation is:
c. When water freezes from a solution, it freezes as pure water, leaving behind a more
concentrated salt solution. Therefore, the melt of frozen sea ice is pure water.
404                                           CHAPTER 11          PROPERTIES OF SOLUTIONS                  CHAPTER 11        PROPERTIES OF SOLUTIONS                                                         403

Tf      5.20 C                                                                                      d. On the CO2 phase diagram in Chapter 10, the triple point is above 1 atm, so CO2(g) is the
m=                       = 2.80 molal                                                                         stable phase at 1 atm and room temperature. CO2(l) can't exist at normal atmospheric
Kf    1.86 C/molal
pressures. Therefore, dry ice sublimes instead of boils. In a fire extinguisher, P > 1 atm,
2.80 mol solute                                                                   and CO2(l) can exist. When CO2 is released from the fire extinguisher, CO2(g) forms, as
Mol solute = 0.0250 kg ×                    = 0.0700 mol solute                                               predicted from the phase diagram.
kg

10.56 g                                                                                   e. Adding a solute to a solvent increases the boiling point and decreases the freezing point
Molar mass =              = 151 g/mol                                                                         of the solvent. Thus the solvent is a liquid over a wider range of temperatures when a
0.0700 mol
solute is dissolved.
The empirical formula mass of C2H4O3 = 76.05 g/mol. Because the molar mass is about twice
the empirical mass, the molecular formula is C4H8O6, which has a molar mass of 152.10               98.    A 92 proof ethanol solution is 46% C2H5OH by volume. Assuming 100.0 mL of solution:
g/mol.                                                                                                                                                          1 mol C 2 H 5 OH
0.79 g
mol ethanol = 46 mL C2H5OH ×                                      = 0.79 mol C2H5OH
Note: We use the experimental molar mass to determine the molecular formula. Knowing                                                                   mL          46.07 g
this, we calculate the molar mass precisely from the molecular formula using the atomic                                    0.79 mol
molarity =            = 7.9 M ethanol
masses in the periodic table.                                                                                              0.1000 L
99.    Because partial pressures are proportional to the moles of gas present, then
102.   a. As discussed in Figure 11.18 of the text, the water would migrate from right to left.                     V
Initially, the level of liquid in the right arm would go down, and the level in the left arm              CS2   PCS2 /Ptotal .
would go up. At some point the rate of solvent transfer will be the same in both
V
directions, and the levels of the liquids in the two arms will stabilize. The height                    PCS2     CS2 Ptotal   = 0.855(263 torr) = 225 torr
difference between the two arms is a measure of the osmotic pressure of the NaCl
solution.                                                                                                        L    o          L
PCS2     225 torr
PCS2     CS2 PCS2 ,      CS2    o
= 0.600
PCS2     375 torr
b. Initially, H2O molecules will have a net migration into the NaCl side. However, Na+ and
Cl ions can now migrate into the H2O side. Because solute and solvent transfer are both                              0.1 mol       0.08206 L atm
possible, the levels of the liquids will be equal once the rate of solute and solvent transfer   100.    = MRT =                                      298 K = 2.45 atm   2 atm
L              K mol
is equal in both directions. At this point the concentration of Na+ and Cl ions will be
equal in both chambers, and the levels of liquid will be equal.
760 mm Hg
= 2 atm ×                      2000 mm      2m
103.   If ideal, NaCl dissociates completely, and i = 2.00. Tf = iKfm; assuming water freezes at                                    atm
0.00°C:                                                                                                    The osmotic pressure would support a mercury column of approximately 2 m. The height of
1.28°C = 2 × 1.86°C kg/mol × m, m = 0.344 mol NaCl/kg H2O                                             a fluid column in a tree will be higher because Hg is more dense than the fluid in a tree. If we
assume the fluid in a tree is mostly H2O, then the fluid has a density of 1.0 g/cm3. The
Assume an amount of solution that contains 1.00 kg of water (solvent).                                     density of Hg is 13.6 g/cm3.

Height of fluid       2 m × 13.6   30 m
58.44 g
0.344 mol NaCl ×           = 20.1 g NaCl
mol                                                                             101.   Out of 100.00 g, there are:
20.1 g                                                                                               1 mol C                2.629
Mass % NaCl =                              × 100 = 1.97%                                                       31.57 g C ×            = 2.629 mol C;       = 1.000
1.00   103 g     20.1 g                                                                                      12.01 g                2.629

104.   The main factor for stabilization seems to be electrostatic repulsion. The center of a colloid                               1 mol H               5.26
5.30 g H ×            = 5.26 mol H;       = 2.00
particle is surrounded by a layer of same charged ions, with oppositely charged ions forming                                 1.008 g               2.629
another charged layer on the outside. Overall, there are equal numbers of charged and
oppositely charged ions, so the colloidal particles are electrically neutral. However, since the                              1 mol O                3.946
63.13 g O ×            = 3.946 mol O;       = 1.501
outer layers are the same charge, the particles repel each other and do not easily aggregate for                              16.00 g                2.629
precipitation to occur.
Empirical formula: C2H4O3; use the freezing-point data to determine the molar mass.
406                                                              CHAPTER 11                            PROPERTIES OF SOLUTIONS               CHAPTER 11                      PROPERTIES OF SOLUTIONS                                                                                405

Ax                                    Ax
Heating increases the velocities of the colloidal particles. This causes the particles to collide
0.30 =                       , 0.30                                                                                                     with enough energy to break the ion barriers, allowing the colloids to aggregate and
Ax         By                     Ax    (1.00                 A)y
eventually precipitate out. Adding an electrolyte neutralizes the adsorbed ion layers, which
allows colloidal particles to aggregate and then precipitate out.
A   x = 0.30(    A   x) + 0.30 y           0.30(   A   y),       A   x       (0.30)      A   x + (0.30)       A   y = 0.30 y

0.30 y                                                                                                    T            2.79 o C
105.    T = Kfm, m =                                               = 1.50 molal
A(x      0.30 x + 0.30 y) = 0.30 y,                A   =                 ;                    B   = 1.00          A                                                     Kf         1.86 o C/molal
0.70 x 0.30 y

y                                        y                       a.          T = Kbm,                T = (0.51 C/molal)(1.50 molal) = 0.77 C, Tb = 100.77 C
Similarly, if vapor above is 50.% A:                      A                         ;       B       1.00
x       y                                x       y
o                               mol H 2 O
0.80 y                                                                                 b.          Psoln           water Pwater ,    water     =
If vapor above is 80.% A:                 A   =                 ;                   B   = 1.00           A                                                                                                mol H 2 O mol solute
0.20 x 0.80 y
Assuming 1.00 kg of water, we have 1.50 mol solute, and:
If the liquid solution is 30.% A by moles,                          A   = 0.30.                                                                                                                                   1 mol H 2 O
mol H2O = 1.00 × 103 g H2O ×                                      = 55.5 mol H2O
18.02 g H 2 O
V           PA                 0.30 x                            V                            0.30 x
Thus       A                      =                 and                     B           1.00
PA        PB        0.30 x 0.70 y                                                  0.30 x 0.70 y                                                     55.5 mol
water   =             = 0.974; Psoln = (0.974)(23.76 mm Hg) = 23.1 mm Hg
1.50 55.5
V           x                     V                       V
If solution is 50.% A:                A                   and           B        1.00           A                                            c. We assumed ideal behavior in solution formation, we assumed the solute was nonvolatile,
x       y
and we assumed i = 1 (no ions formed).
V           0.80 x                                V                    V
If solution is 80.% A:                A                     and                         B       1.00         A
0.80 x 0.20 y                                                                           Challenge Problems
108.   a. Freezing-point depression is determined using molality for the concentration units,                                                106.                                                      V
For the second vapor collected, V, 2 = 0.714 and T , 2 = 0.286. Let L , 2 = mole fraction of
B                                      B
whereas molarity units are used to determine osmotic pressure. We need to assume that                                                                                          L
the molality of the solution equals the molarity of the solution.                                                                         benzene in the second solution and T , 2 = mole fraction of toluene in the second solution.
L      L
B, 2   T , 2 = 1.000
moles solvent                moles solvent
b. Molarity =                          ; molality =                                                                                                                                                                               L
liters solution               kg solvent                                                                                                                PB               PB                                B, 2 (750.0    torr )
V
B, 2    = 0.714 =                                       =   L                                      L
When the liters of solution equal the kilograms of solvent present for a solution, then                                                                                  Ptotal       PB        PT       B, 2 (750.0    torr ) (1.000           B, 2 )(300.0   torr )
molarity equals molality. This occurs for an aqueous solution when the density of the
L                         L
solution is equal to the density of water, 1.00 g/cm3. The density of a solution is close to                                            Solving:                 B, 2   = 0.500 =          T, 2
1.00 g/cm3 when not a lot of solute is dissolved in solution. Therefore, molarity and
molality values are close to each other only for dilute solutions.                                                                      This second solution came from the vapor collected from the first (initial) solution, so, V, 1 =
B
V                    L                                                      L
T , 1 = 0.500. Let   B, 1 = mole fraction benzene in the first solution and T , 1 = mole fraction
T      0.621o C                                                                                            of toluene in first solution. L , 1
B
L
T, 1 = 1.000.
c.     T = Kf m, m =                =                 = 0.334 mol/kg
Kf   1.86 o C kg/mol                                                                                                                                                                             L
V                            PB              PB                                 B, 1 (750.0   torr )
B, 1   = 0.500 =                                        =   L                                     L
Assuming 0.334 mol/kg = 0.334 mol/L:                                                                                                                                     Ptotal      PB        PT        B, 1 (750.0   torr ) (1.000           B, 1 )(300.0   torr )
L
0.334 mol          0.08206 L atm                                                                            Solving:                 B, 1   = 0.286
= MRT =                                                                  298 K = 8.17 atm
L                  K mol                                                                                The original solution had                             = 0.286 and          = 0.714.
B                    T
T       2.0o C
d. m =            =                 = 3.92 mol/kg                                                                                                                                                                       PA
Kb   0.51 o C kg/mol                                                                                                   107.   For 30.% A by moles in the vapor, 30. =                                              × 100:
PA        PB
408                                                                 CHAPTER 11             PROPERTIES OF SOLUTIONS                        CHAPTER 11                     PROPERTIES OF SOLUTIONS                                               407

Mass of 1.000 L solution = 1071 g; mass of NaCl = 10.7 g                                                                                       This solution is much more concentrated than the isotonic solution in part c. Here, water
will leave the plant cells in order to try to equilibrate the ion concentration both inside
Mass of solvent in 1.000 L solution = 1071 g                               10.7 g = 1060. g                                                    and outside the cell. Because there is such a large concentration discrepancy, all the
water will leave the plant cells, causing them to shrivel and die.
0.320 mol
T = 1.86 °C kg/mol ×                          = 0.562°C
1.060 kg                                                                                                   Tf              0.426 o C
109.   m=                                        = 0.229 molal
Kf            1.86 o C / molal
Assuming water freezes at 0.000°C, then Tf = 0.562°C.
Assuming a solution density = 1.00 g/mL, then 1.00 L contains 0.229 mol solute.
V
Ppen                       L    o                                         L              L
111.    pen    0.15                  ; Ppen            pen Ppen ;    Ptotal   Ppen     Phex           pen (511)      hex (150.)                  NaCl                Na+ + Cl      i = 2; so: 2(mol NaCl) + mol C12H22O11 = 0.229 mol
Ptotal

L                         L                       L                               L                            L
Mass NaCl + mass C12H22O11 = 20.0 g
Because       hex      1.000            pen    : Ptotal         pen (511)      (1.000           pen )(150.)   150.     361   pen
2nNaCl + nC12 H 22O11 = 0.229 and 58.44(nNaCl) + 342.3(nC12 H 22O11 ) = 20.0
L
V
Ppen                           pen   (511)                                   L               L
pen               , 0.15                              L
, 0.15(150.      361      pen )     511   pen                                Solving: nC12 H 22O11 = 0.0425 mol = 14.5 g and nNaCl = 0.0932 mol = 5.45 g
Ptotal                  150.         361     pen

14.5 g
L                 L          L          23                                                                                    Mass % C12H22O11 =                       × 100 = 72.5 % and 27.5% NaCl by mass
23 + 54     pen    = 511      pen   ,    pen   =       = 0.050                                                                                                              20.0 g
457
0.0425 mol
Tf             5.40o C                                                                                            C12 H 22 O11   =                  = 0.313
112.    Tf = Kfm, m =                                           = 2.90 molal                                                                                      0.0425 mol 0.0932 mol
Kf          1.86 o C / molal                                                                                                                            7.83 atm
110.   a.          = iMRT, iM                                                         0.320 mol/L
2.90 mol solute                  n                                                                                                                                  RT 0.08206 L atm K 1 mol               1
298 K
, n = 0.145 mol of ions in solution
kg solvent                 0.0500 kg                                                                                                           Assuming 1.000 L of solution:

Because NaNO3 and Mg(NO3)2 are strong electrolytes:                                                                                                      total mol solute particles = mol Na+ + mol Cl + mol NaCl = 0.320 mol
1.071 g
n = 2(x mol of NaNO3) + 3[y mol Mg(NO3)2] = 0.145 mol ions                                                                                            mass solution = 1000. mL ×                = 1071 g solution
mL
85.00 g                  148.3 g                                                               mass NaCl in solution = 0.0100 × 1071 g = 10.7 g NaCl
In addition: 6.50 g = x mol NaNO3                                 + y mol Mg(NO3)2
mol                       mol
1 mol
mol NaCl added to solution = 10.7 g ×                 = 0.183 mol NaCl
We have two equations: 2x + 3y = 0.145 and (85.00)x + (148.3)y = 6.50                                                                                                                                  58.44 g

Solving by simultaneous equations:                                                                                                                Some of this NaCl dissociates into Na+ and Cl (two moles of ions per mole of NaCl),
and some remains undissociated. Let x = mol undissociated NaCl = mol ion pairs.
(85.00)x (127.5)y = 6.16
(85.00)x + (148.3)y = 6.50                                                                                                                 Mol solute particles = 0.320 mol = 2(0.183            x) + x
______________________________________
0.320 = 0.366           x, x = 0.046 mol ion pairs
(20.8)y = 0.34,                 y = 0.016 mol Mg(NO3)2
0.046
Fraction of ion pairs =              = 0.25, or 25%
Mass of Mg(NO3)2 = 0.016 mol × 148.3 g/mol = 2.4 g Mg(NO3)2, or 37% Mg(NO3)2 by mass                                                                                             0.183
Mass of NaNO3 = 6.50 g - 2.4 g = 4.1 g NaNO3, or 63% NaNO3 by mass                                                                        b.       T = Kfm, where Kf = 1.86 °C kg/mol; from part a, 1.000 L of solution contains 0.320
mol of solute particles. To calculate the molality of the solution, we need the kilograms
of solvent present in 1.000 L of solution.
410                                                      CHAPTER 11                   PROPERTIES OF SOLUTIONS                    CHAPTER 11          PROPERTIES OF SOLUTIONS                                                                     409

115.   HCO2H H+ + HCO2 ; only 4.2% of HCO2H ionizes. The amount of H+ or HCO2                                                                                                      Tf         2.70o C
produced is 0.042 × 0.10 M = 0.0042 M.                                                                                    113.    Tf = 5.51 - 2.81 = 2.70°C; m                                     = 0.527 molal
Kf       5.12 o C/molal
The amount of HCO2H remaining in solution after ionization is 0.10 M                            0.0042 M = 0.10 M.               Let x = mass of naphthalene (molar mass = 128.2 g/mol). Then 1.60                       x = mass of
anthracene (molar mass = 178.2 g/mol).
The total molarity of species present = M HCO 2 H                  MH          M HCO
2

= 0.10 + 0.0042 + 0.0042 = 0.11 M                   x                           1.60 x
= moles naphthalene and         = moles anthracene
128.2                           178.2
Assuming 0.11 M = 0.11 molal, and assuming ample significant figures in the freezing point and
boiling point of water at P = 1 atm:                                                                                                                        x     1.60 x
0.527 mol solute         128.2     178.2 , 1.05                 2       (178.2) x    1.60(128.2) (128.2) x
10
T = Kfm = 1.86°C/molal × 0.11 molal = 0.20°C; freezing point = 0.20°C                                                     kg solvent            0.0200 kg solvent                                          128.2(178.2)

T = Kbm = 0.51°C/molal × 0.11 molal = 0.056°C; boiling point = 100.056°C                                               (50.0)x + 205 = 240., (50.0)x = 240.            205, (50.0)x = 35, x = 0.70 g naphthalene
L                                                            L       L                         V        L
116.   Let       A   = mole fraction A in solution, so 1.000                  A   =   B   . From the problem,   A   =2   A   .          So the mixture is:

L                                                                                           0.70 g
V         PA                            A (350.0 torr )
A    =          =                                                                                                                          × 100 = 44% naphthalene by mass and 56% anthracene by mass
L                             L                                                                             1.60 g
Ptotal         A   (350.0 torr ) (1.000      A )(100.0 torr )

0.3950 atm
(350.0) L                                                                                      114.   iM =                                    = 0.01614 mol/L = total ion concentration
V
=2       L
=               A
, ( 250.0)         L
= 75.0,          L
= 0.300                                              RT      0.08206 L atm
A             A
(250.0) L 100.0
A                    A                                                                                 298.2 K
A                                                                                                                    K mol

The mole fraction of A in solution is 0.300.                                                                                     0.01614 mol/L = M Mg 2 + M Na + M Cl ; M Cl = 2 M Mg 2 + M Na (charge balance)

117.   a. Assuming MgCO3(s) does not dissociate, the solute concentration in water is:                                                  Combining: 0.01614 = 3M Mg 2 + 2M Na
3
560 µg MgCO 3 (s)               560 mg      560    10        g       1 mol MgCO 3
Let x = mass MgCl2 and y = mass NaCl; then x + y = 0.5000 g.
mL                         L               L                     84.32 g
= 6.6 × 10 3 mol MgCO3/L                                  x                y
M Mg 2 =           and M Na =       (Because V = 1.000 L.)
95.21            58.44
An applied pressure of 8.0 atm will purify water up to a solute concentration of:
3x         2y
8.0 atm                            0.32 mol                                          Total ion concentration =                        = 0.01614 mol/L
M                                                                                                                                                  95.21      58.44
RT        0.08206 L atm/K mol          300. K              L
Rearranging: 3x + (3.258)y = 1.537
When the concentration of MgCO3(s) reaches 0.32 mol/L, the reverse osmosis unit can no
longer purify the water. Let V = volume (L) of water remaining after purifying 45 L of H2O.                                Solving by simultaneous equations:
When V + 45 L of water has been processed, the moles of solute particles will equal:
3x     + (3.258)y = 1.537
6.6 × 10 3 mol/L × (45 L + V) = 0.32 mol/L × V                                                                            3(x    +       y) = 3(0.5000)
_________________________________________________________________

Solving: 0.30 = (0.32             0.0066) × V, V = 0.96 L                                                                                    (0.258)y =        0.037, y = 0.14 g NaCl

The minimum total volume of water that must be processed is 45 L + 0.96 L = 46 L.                                                                                                                         0.36 g
Mass MgCl2 = 0.5000 g           0.14 g = 0.36 g; mass % MgCl2 =                        × 100 = 72%
0.5000 g
412                                                   CHAPTER 11        PROPERTIES OF SOLUTIONS          CHAPTER 11          PROPERTIES OF SOLUTIONS                                                              411

2.016 mg H                        0.185 mg                                                Note: If MgCO3 does dissociate into Mg2+ and CO32 ions, then the solute concentration
1.65 mg H2O ×                    = 0.185 mg H; % H =          × 100 = 3.85% H                                increases to 1.3 × 10 2 M, and at least 47 L of water must be processed.
18.02 mg H 2 O                     4.80 mg
b. No; a reverse osmosis system that applies 8.0 atm can only purify water with a solute
Mass % O = 100.00          (80.8 + 3.85) = 15.4% O                                                          concentration of less than 0.32 mol/L. Salt water has a solute concentration of 2(0.60 M) =
1.2 mol/L ions. The solute concentration of salt water is much too high for this reverse
Out of 100.00 g:                                                                                            osmosis unit to work.

1 mol                 6.73
80.8 g C ×           = 6.73 mol C;       = 6.99        7                                      Integrative Problems
12.01 g               0.963

1L         10 dL      1.0 mg       1g           1 mol
1 mol                3.82                                                      118.   10.0 mL                                                                = 8.8 × 10 7 mol C4H7N3O
3.85 g H ×            = 3.82 mol H;       = 3.97       4                                                          1000 mL       1L         1 dL      1000 mg       113.14 g
1.008 g               0.963

1.025 g
1 mol                 0.963                                                           Mass of blood = 10.0 mL               = 10.3 g
15.4 g O ×            = 0.963 mol O;       = 1.00                                                                                    mL
16.00 g                0.963
Therefore, the empirical formula is C7H4O.                                                                            8.8 10 7 mol                 5
Molality =                = 8.5 × 10          mol/kg
Tf        22.3 o C                                                                                   0.0103 kg
Tf = Kfm, m =                            = 0.56 molal
Kf      40. o C / molal
8.8 10 7 mol                  5
0.56 mol anthraquinone                                   = MRT, M =                      = 8.8 × 10           mol/L
Mol anthraquinone = 0.0114 kg camphor ×                                 = 6.4 × 10 3 mol                                        0.0100 L
kg camphor

1.32 g                                                                                         5           0.08206 L atm                           3
Molar mass =                       = 210 g/mol                                                            = 8.8 × 10       mol/L                 × 298 K = 2.2 × 10            atm
6 .4     10 3 mol                                                                                                     K mol

The empirical mass of C7H4O is 7(12) + 4(1) + 16 104 g/mol. Because the molar mass is twice
the empirical mass, the molecular formula is C14H8O2.                                                                        T             2.79 o C
119.    T = imKf, i =          =                       = 3.00
mK f   0.250 mol 1.86 o C kg
0.500 kg        mol
Marathon Problem
121.   a. From part a information we can calculate the molar mass of NanA and deduce the formula.               We have three ions in solutions, and we have twice as many anions as cations. Therefore, the
formula of Q is MCl2. Assuming 100.00 g of compound:
0.02313 mol
Mol NanA = mol reducing agent = 0.01526 L ×                     = 3.530 × 10 4 mol NanA                                  1 mol Cl
L                                               38.68 g Cl               = 1.091 mol Cl
3                                                                                 35.45 g
30.0 10 g
Molar mass of NanA =                      = 85.0 g/mol
3.530 10 4 mol                                                                mol M = 1.091 mol Cl
1 mol M
= 0.5455 mol M
2 mol Cl
To deduce the formula, we will assume various charges and numbers of oxygens present in
the oxyanion and then use the periodic table to see if an element fits the molar mass data.                                  61.32 g M
Molar mass of M =                  = 112.4 g/mol; M is Cd, so Q = CdCl2.
Assuming n = 1, the formula is NaA. The molar mass of the oxyanion A- is 85.0 23.0 =                                       0.5455 mol M
62.0 g/mol. The oxyanion part of the formula could be EO or EO2 or EO3 , where E is some
element. If EO , then the molar mass of E is 62.0 16.0 = 46.0 g/mol; no element has this
12.01 mg C                      3.88 mg
molar mass. If EO2 , molar mass of E = 62.0 32.0 = 30.0 g/mol. Phosphorus is close, but       120.   14.2 mg CO2 ×                   = 3.88 mg C; % C =         × 100 = 80.8% C
PO2 anions are not common. If EO3 , molar mass of E = 62.0 48.0 = 14.0. Nitrogen has                                   44.01 mg CO 2                    4.80 mg
this molar mass, and NO3 anions are very common. Therefore, NO3 is a possible formula
for A .
CHAPTER 11         PROPERTIES OF SOLUTIONS                                                             413

Next, we assume Na2A and Na3A formulas and go through the same procedure as above. In
all cases, no element in the periodic table fits the data. Therefore, we assume the oxyanion is
NO3 = A .

b. The crystal data in part b allow determination of the metal M in the formula. See Exercise
10.55 for a review of relationships in body-centered cubic cells. In a body-centered cubic
unit cell, there are two atoms per unit cell, and the body diagonal of the cubic cell is related to
the radius of the metal by the equation 4r = l 3 , where l = cubic edge length.

8
4r     4(1.984    10       cm)
l=                                    = 4.582 × 10 8 cm
3                3

Volume of unit cell = l3 = (4.582 × 10 8)3 = 9.620 × 10         23
cm3

23           5.243 g
Mass of M in a unit cell = 9.620 × 10            cm3 ×           = 5.044 × 10       22
gM
cm 3
1 mol                           24
Mol M in a unit cell = 2 atoms ×                    = 3.321 × 10                mol M
6.022 10 23

5.044 10 22 g M
Molar mass of M =                        = 151.9 g/mol
3.321 10 24 mol M

From the periodic table, M is europium (Eu). Given that the charge of Eu is +3, then the
formula of the salt is Eu(NO3)3 zH2O.

c. Part c data allow determination of the molar mass of Eu(NO3)3 zH2O, from which we can
determine z, the number of waters of hydration.

1 atm
558 torr
760 torr
= iMRT, iM =                                            = 0.0300 mol/L
RT        0.08206 L atm/K mol 298 K

The total molarity of solute particles present is 0.0300 M. The solute particles are Eu3+ and
NO3 ions (the waters of hydration are not solute particles). Because each mole of
Eu(NO3)3 zH2O dissolves to form four ions (Eu3+ + 3 NO3 ), the molarity of
Eu(NO3)3 zH2O is 0.0300/4 = 0.00750 M.

0.00750 mol
Mol Eu(NO3)3 zH2O = 0.01000 L ×                        = 7.50 × 10 5 mol
L
33.45 10 3 g
Molar mass of Eu(NO3)3 zH2O =                         = 446 g/mol
7.50 10 5 mol

446 g/mol = 152.0 + 3(62.0) + z(18.0), z(18.0) = 108, z = 6.00

The formula for the strong electrolyte is Eu(NO3)3 6H2O.

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