CHAPTER ELEVEN
PROPERTIES OF SOLUTIONS
For Review
1.
Mass percent: the percent by mass of the solute in the solution.
Mole fraction: the ratio of the number of moles of a given component to the total number of
moles of solution.
Molarity: the number of moles of solute per liter of solution.
Molality: the number of moles of solute per kilogram of solvent.
Volume is temperature dependent, whereas mass and the number of moles are not. Only
molarity has a volume term so only molarity is temperature dependent.
2.
KF(s) → K+(aq) + F(aq) H = Hsoln;
K+(g) + Cl (g) → K+(aq) + F(aq) H = Hhyd
KF(s) → K+(g) + F(g)
H1 = HLE
K (g) + F(g) → K+(aq) + F(aq)
H2 = Hhyd
__________________________________________________________
KF(s) → K+(aq) + F(aq)
H = Hsoln = HLE + Hhyd
+
It is true that H1 and H2 have large magnitudes for their values; however, the signs are
opposite (H1 is large and positive because it is the reverse of the lattice energy and H2, the
hydration energy, is large and negative). These two H values basically cancel out each other
giving a Hsoln value close to zero.
3.
“Like dissolves like” refers to the nature of the intermolecular forces. Polar solutes and ionic
solutes dissolve in polar solvents because the types of intermolecular forces present in solute
and solvent are similar. When they dissolve, the strength of the intermolecular forces in
solution are about the same as in pure solute and pure solvent. The same is true for nonpolar
solutes in nonpolar solvents. The strength of the intermolecular forces (London dispersion
forces) are about the same in solution as in pure solute and pure solvent. In all cases of like
dissolves like, the magnitude of Hsoln is either a small positive number (endothermic) or a
small negative number (exothermic). For polar solutes in nonpolar solvents and vice versa,
Hsoln is a very large, unfavorable value (very endothermic). Because the energetics are so
unfavorable, polar solutes do not dissolve in nonpolar solvents and vice versa.
4.
Structure effects refer to solute and solvent having similar polarities in order for solution
formation to occur. Hydrophobic solutes are mostly nonpolar substances that are “waterfearing.” Hydrophilic solutes are mostly polar or ionic substances that are “water-loving.”
372
CHAPTER 11
PROPERTIES OF SOLUTIONS
373
Pressure has little effect on the solubilities of solids or liquids; it does significantly affect the
solubility of a gas. Henry’s law states that the amount of a gas dissolved in a solution is
directly proportional to the pressure of the gas above the solution (C = kP). The equation for
Henry’s law works best for dilute solutions of gases that do not dissociate in or react with the
solvent. HCl(g) does not follow Henry’s law because it dissociates into H+(aq) and Cl(aq) in
solution (HCl is a strong acid). For O2 and N2, Henry’s law works well since these gases do
not react with the water solvent.
An increase in temperature can either increase or decrease the solubility of a solid solute in
water. It is true that a solute dissolves more rapidly with an increase in temperature, but the
amount of solid solute that dissolves to form a saturated solution can either decrease or
increase with temperature. The temperature effect is difficult to predict for solid solutes.
However, the temperature effect for gas solutes is easier to predict as the solubility of a gas
typically decreases with increasing temperature.
5.
o
Raoult’s law: Psoln = χ solventPsolvent
; When a solute is added to a solvent, the vapor pressure of
a solution is lowered from that of the pure solvent. The quantity solvent, the mole fraction of
solvent, is the fraction that the solution vapor pressure is lowered.
For the experiment illustrated in Fig. 11.9, the beaker of water will stop having a net transfer
of water molecules out of the beaker when the equilibrium vapor pressure, PHo 2O , is reached.
This can never happen. The beaker with the solution wants an equilibrium vapor pressure of
PH 2O , which is less than PHo 2O . When the vapor pressure over the solution is above PH 2O , a
net transfer of water molecules into the solution will occur in order to try to reduce the vapor
pressure to PH 2O . The two beakers can never obtain the equilibrium vapor pressure they want.
The net transfer of water molecules from the beaker of water to the beaker of solution stops
after all of the water has evaporated.
If the solute is volatile, then we can get a transfer of both the solute and solvent back and
forth between the beakers. A state can be reached in this experiment where both beakers have
the same solute concentration and hence the same vapor pressure. When this state is reached,
no net transfer of solute or water molecules occurs between the beakers so the levels of
solution remain constant.
When both substances in a solution are volatile, then Raoult’s law applies to both. The total
vapor pressure above the solution is the equilibrium vapor pressure of the solvent plus the
equilibrium vapor pressure of the solute. Mathematically:
PTOT  PA  PB  χ A PAo  χ B PBo
where A is either the solute or solvent and B is the other one.
6.
An ideal liquid-liquid solution follows Raoult’s law:
PTOT 
χ A PAo  χ B PBo
A nonideal liquid-liquid solution does not follow Raoult’s law, either giving a total pressure
greater than predicted by Raoult’s law (positive deviation) or less than predicted (negative
deviation).
374
CHAPTER 11
PROPERTIES OF SOLUTIONS
In an ideal solution, the strength of the intermolecular forces in solution are equal to the
strength of the intermolecular forces in pure solute and pure solvent. When this is true, Hsoln
= 0 and Tsoln = 0. For positive deviations from Raoult’s law, the solution has weaker
intermolecular forces in solution than in pure solute and pure solvent. Positive deviations
have Hsoln > 0 (are endothermic) and Tsoln < 0. For negative deviations, the solution has
stronger intermolecular forces in solution than in pure solute or pure solvent. Negative
deviations have Hsoln < 0 (are exothermic) and Tsoln > 0. Examples of each type of solution
are:
ideal:
positive deviations:
negative deviations:
7.
benzene-toluene
ethanol-hexane
acetone-water
Colligative properties are properties of a solution that depend only on the number, not the
identity, of the solute particles. A solution of some concentration of glucose (C6H12O6) has
the same colligative properties as a solution of sucrose (C12H22O11) having the same concentration.
A substance freezes when the vapor pressure of the liquid and solid are identical to each
other. Adding a solute to a substance lowers the vapor pressure of the liquid. A lower
temperature is needed to reach the point where the vapor pressures of the solution and solid
are identical. Hence, the freezing point is depressed when a solution forms.
A substance boils when the vapor pressure of the liquid equals the external pressure. Because
a solute lowers the vapor pressure of the liquid, a higher temperature is needed to reach the
point where the vapor pressure of the liquid equals the external pressure. Hence, the boiling
point is elevated when a solution forms.
The equation to calculate the freezing point depression or boiling point elevation is:
T = Km
where K is the freezing point or boiling point constant for the solvent and m is the molality of
the solute. Table 11.5 lists the K values for several solvents. The solvent which shows the
largest change in freezing point for a certain concentration of solute is camphor; it has the
largest Kf value. Water, with the smallest Kb value, will show the smallest increase in boiling
point for a certain concentration of solute.
To calculate molar mass, you need to know the mass of the unknown solute, the mass and
identity of solvent used, and the change in temperature (T) of the freezing or boiling point
of the solution. Since the mass of unknown solute is known, one manipulates the freezing
point data to determine the number of moles of solute present. Once the mass and moles of
solute are known, one can determine the molar mass of the solute. To determine the moles of
solute present, one determines the molality of the solution from the freezing point data and
multiplies this by the kilograms of solvent present; this equals the moles of solute present.
8.
Osmotic pressure: the pressure that must be applied to a solution to stop osmosis; osmosis is
the flow of solvent into the solution through a semipermeable membrane. The equation to
calculate osmotic pressure, , is:
 = MRT
CHAPTER 11
PROPERTIES OF SOLUTIONS
375
where M is the molarity of the solution, R is the gas constant, and T is the Kelvin
temperature. The molarity of a solution approximately equals the molality of the solution
when 1 kg solvent  1 L solution. This occurs for dilute solutions of water since d H 2O = 1.00
g/cm3.
With addition of salt or sugar, the osmotic pressure inside the fruit cells (and bacteria) is less
than outside the cell. Water will leave the cells which will dehydrate bacteria present, causing
them to die.
Dialysis allows the transfer of solvent and small solute molecules through a membrane. To
purify blood, the blood is passed through a cellophane tube (the semipermeable membrane);
this cellophane tube is immersed in a dialyzing solution which contains the same concentrations of ions and small molecules as in blood, but has none of the waste products normally
removed by the kidney. As blood is passed through the dialysis machine, the unwanted waste
products pass through the cellophane membrane, cleansing the blood.
Desalination is the removal of dissolved salts from an aqueous solution. Here, a solution is
subjected to a pressure greater than the osmotic pressure and reverse osmosis occurs, i.e.,
water passes from the solution through the semipermeable membrane back into pure water.
Desalination plants can turn sea-water with its high salt content into drinkable water.
9.
A strong electrolyte completely dissociates into ions in solution, a weak electrolyte only
partially dissociates into ions in solution, and a nonelectrolyte does not dissociate into ions
when dissolved in solution. Colligative properties depend on the total number of solute
particles in solution. By measuring a property such as freezing point depression, boiling point
elevation, or osmotic pressure, we can determine the number of solute particles present from
the solute and thus characterize the solute as a strong, weak, or nonelectrolyte.
The van’t Hoff factor, i, is the number of moles of particles (ions) produced for every mol of
solute dissolved. For NaCl, i = 2 since Na+ and Cl are produced in water; for Al(NO3)3, i = 4
since Al3+ and 3 NO3 ions are produced when Al(NO3)3 dissolves in water. In real life, the
van’t Hoff factor is rarely the value predicted by the number of ions a salt dissolves into; i is
generally something less than the predicted number of ions. This is due to a phenomenon
called ion pairing where at any instant a small percentage of oppositely charged ions pair up
and act like a single solute particle. Ion pairing occurs most when the concentration of ions is
large. Therefore, dilute solutions behave most ideally; here i is close to that determined by the
number of ions in a salt.
10.
A colloidal dispersion is a suspension of particles in a dispersing medium. See Table 11.7 for
some examples of different types of colloids.
Both solutions and colloids have suspended particles in some medium. The major difference
between the two is the size of the particles. A colloid is a suspension of relatively large
particles as compared to a solution. Because of this, colloids will scatter light while solutions
will not. The scattering of light by a colloidal suspension is called the Tyndall effect.
Coagulation is the destruction of a colloid by the aggregation of many suspended particles to
form a large particle that settles out of solution.
376
CHAPTER 11
PROPERTIES OF SOLUTIONS
Solution Review
1 mol C3H 7 OH
60.09 g C3 H 7 OH
= 9.74 M
1.00 L
585 g C3H 7 OH 
9.
1 mol CaCl2
1L
1000 mL
= 19.9 mL


110.98 g CaCl2 0.580 mol CaCl2
L
10.
1.28 g CaCl2 ×
11.
mol Na2CO3 = 0.0700 L ×
3.0 mol Na 2 CO 3
= 0.21 mol Na2CO3
L
Na2CO3(s) → 2 Na+(aq) + CO32(aq); mol Na+ = 2(0.21) = 0.42 mol
mol NaHCO3 = 0.0300 L ×
1.0 mol NaHCO3
= 0.030 mol NaHCO3
L
NaHCO3(s) → Na+(aq) + HCO3(aq); mol Na+ = 0.030 mol
M Na  
12.
total mol Na 
0.42 mol  0.030 mol
0.45 mol


= 4.5 M Na+
total volume
0.0700 L  0.030 L
0.1000 L
a. HNO3(l) → H+(aq) + NO3(aq)
b. Na2SO4(s) → 2 Na+(aq) + SO42(aq)
c. Al(NO3)3(s) → Al3+(aq) + 3 NO3(aq)
d. SrBr2(s) → Sr2+(aq) + 2 Br(aq)
e. KClO4(s) → K+(aq) + ClO4(aq)
f.
g. NH4NO3(s) → NH4+(aq) + NO3(aq)
h. CuSO4(s) → Cu2+(aq) + SO42(aq)
i.
NH4Br(s) → NH4+(aq) + Br(aq)
NaOH(s) → Na+(aq) + OH(aq)
Questions
13.
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 kinetic energy greater than the
attractive forces between the gas molecules and the solvent molecules. More gas molecules
will escape to the vapor phase and the solubility of the gas will decrease.
14.
Henry’s law is obeyed most accurately for dilute solutions of gases that do not dissociate in
or react with the solvent. NH3 is a weak base and reacts with water by the following reaction:
NH3(aq) + H2O(l) → NH4+(aq) + OH(aq)
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.
CHAPTER 11
PROPERTIES OF SOLUTIONS
377
15
Because the solute is volatile, both the water and solute will transfer back and forth between
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,
one would expect there to be a larger net transfer of water molecules into the right beaker
than the net transfer of solute molecules into the left beaker. This results in a larger solution
volume in the right beaker when equilibrium is reached, i.e., when the solute concentration is
identical in each beaker.
16.
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
an exothermic enthalpy of solution. The only statement that is false is e. A substance boils
when the vapor pressure equals the external pressure. Since B = 0.6 has a lower vapor
pressure at the temperature of the plot than either pure A or pure B, then one would expect
this 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
either pure A or pure B. (Note that because P°B > P°A, B is more volatile than A, and B will
have a lower boiling point temperature than A).
17.
No, the solution is not ideal. For an ideal solution, the strength of intermolecular forces in
solution is the same as in pure solute and pure solvent. This results in ΔHsoln = 0 for an ideal
solution. ΔHsoln for methanol/water is not zero. Because ΔHsoln < 0 (heat is released), this
solution shows a negative deviation from Raoult’s law.
18.
The micelles form so the ionic ends of the detergent molecules, the SO4 ends, are exposed to
the polar water molecules on the outside, while the nonpolar hydrocarbon chains from the
detergent molecules are hidden from the water by pointing toward the inside of the micelle.
Dirt, which is basically nonpolar, is stabilized in the nonpolar interior of the micelle and is
washed away.
= detergent
molecule
= SO 4= nonpolar
hydrocarbon
= dirt
19.
Normality is the number of equivalents per liter of solution. For an acid or a base, an
equivalent is the mass of acid or base that can furnish 1 mol of protons (if an acid) or accept 1
378
CHAPTER 11
PROPERTIES OF SOLUTIONS
mol of protons (if a base). A proton is an H+ ion. Molarity is defined as the moles of solute
per liter of solution. When the number of equivalents equals the number of moles of solute,
then normality = molarity. This is true for acids which only have one acidic proton in them
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
equivalents = moles solute are NaOH, KOH, and NH3. When equivalents  moles solute,
then normality  molarity. This is true for acids that donate more than one proton (H2SO4,
H3PO4, H2CO3, etc.) and for bases that react with more than one proton per formula unit
[Ca(OH)2, Ba(OH)2, Sr(OH)2, etc.].
20.
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
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
 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,
in general, are favored that result in an increase in disorder because the disordered state is the
easiest (most probable) state to achieve. The tendency of processes to increase disorder will
be discussed in Chapter 16 when entropy, S, is introduced.
21.
Only statement b is true. A substance freezes when the vapor pressure of the liquid and solid
are the same. When a solute is added to water, the vapor pressure of the solution at 0C is
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
temperature is needed to equalize the vapor pressure of water and ice, hence the freezing
point is depressed.
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
boiling point of water is increased because the sugar solute decreases the vapor pressure of
the water; a higher temperature is required for the vapor pressure of the solution to equal the
external pressure so boiling can occur.
22.
This is true if the solute will dissolve in camphor. Camphor has the largest K b and Kf
constants. This means that camphor shows the largest change in boiling point and melting
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.
23.
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
in osmotic pressures inside and outside the cell. When red blood cells are in a solution having
a higher osmotic pressure than that of the cells, the cells shrivel as there is a net transfer of
CHAPTER 11
PROPERTIES OF SOLUTIONS
379
water out of the cells. This is called crenation. Hemolysis occurs when the red blood cells are
bathed in a solution having lower osmotic pressure than that inside the cell. Here, the cells
rupture as there is a net transfer of water to into the red blood cells.
24.
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
expect sodium chloride to exist as separate hydrated Na+ ions and Cl ions. A few ions,
however, stay together as NaCl and behave as just one particle. Ion pairing increases in a
solution as the ion concentration increases (as the molality increases).
Exercises
Concentration of Solutions
25.
Because the density of water is 1.00 g/mL, 100.0 mL of water has a mass of 100. g.
density =
10.0 g H3PO4  100. g H 2O
mass
= 1.06 g/mL = 1.06 g/cm3

volume
104 mL
mol H3PO4 = 10.0 g ×
mol H2O = 100. g ×
1 mol
= 0.102 mol H3PO4
97.99 g
1 mol
= 5.55 mol H2O
18.02 g
mole fraction of H3PO4 =
0.102 mol H3PO4
= 0.0180
(0.102  5.55) mol
χ H 2O = 1.000 – 0.0180 = 0.9820
26.
molarity =
0.102 mol H3PO4
= 0.981 mol/L
0.104 L
molality =
0.102 mol H3PO4
= 1.02 mol/kg
0.100 kg
molality =
molarity =
40.0 g EG 1000 g 1 mol EG
= 10.7 mol/kg


60.0 g H 2 O
kg
62.07 g
where EG = ethylene glycol (C2H6O2)
40.0 g EG
1.05 g 1000 cm3
1 mol



= 6.77 mol/L
3
100 .0 g solution
L
62.07 g
cm
380
CHAPTER 11
1 mol
1 mol
= 0.644 mol EG; 60.0 g H2O ×
= 3.33 mol H2O
18.02 g
62.07 g
40.0 g EG ×
χ EG 
27.
PROPERTIES OF SOLUTIONS
0.644
= 0.162 = mole fraction ethylene glycol
3.33  0.644
Hydrochloric acid:
molarity =
38 g HCl 1.19 g so ln 1000 cm3 1 mol HCl



= 12 mol/L
100 . g so ln
L
36.46 g
cm3 so ln
molality =
38 g HCl
1000 g 1 mol HCl
= 17 mol/kg


62 g solvent
kg
36.46 g
38 g HCl ×
1 mol
1 mol
= 1.0 mol HCl; 62 g H2O ×
= 3.4 mol H2O
36.46 g
18.02 g
mole fraction of HCl = χ HCl 
1.0
= 0.23
3.4  1.0
Nitric acid:
70. g HNO3 1.42 g so ln 1000 cm3 1 mol HNO3



= 16 mol/L
100 . g so ln
L
63.02 g
cm3 so ln
70. g HNO3 1000 g 1 mol HNO3
= 37 mol/kg


30. g solvent
kg
63.02 g
70. g HNO3 ×
χ HNO3 
1 mol
1 mol
= 1.1 mol HNO3; 30. g H2O ×
= 1.7 mol H2O
63.02 g
18.02 g
1.1
= 0.39
1.7  1.1
Sulfuric acid:
95 g H 2SO 4 1.84 g so ln 1000 cm3
1 mol H 2SO 4



= 18 mol/L
3
100 . g so ln
L
98.09 g H 2SO 4
cm so ln
95 g H 2SO 4 1000 g 1 mol
= 194 mol/kg ≈ 200 mol/kg


5 g H 2O
kg
98.09 g
95 g H2SO4 ×
χ H 2SO 4 
1 mol
1 mol
= 0.97 mol H2SO4; 5 g H2O ×
= 0.3 mol H2O
98.09 g
18.02 g
0.97
= 0.76
0.97  0.3
CHAPTER 11
PROPERTIES OF SOLUTIONS
381
Acetic Acid:
99 g HC2 H 3O 2 1.05 g so ln 1000 cm3
1 mol



= 17 mol/L
100 . g so ln
L
60.05 g H
cm3 so ln
99 g HC2 H3O2 1000 g 1 mol
= 1600 mol/kg ≈ 2000 mol/kg


1 g H 2O
kg
60.05 g
1 mol
1 mol
= 1.6 mol HC2H3O2; 1 g H2O ×
= 0.06 mol H2O
60.05 g
18.02 g
1.6
= 0.96

1.6  0.06
99 g HC2H3O2 ×
χ HC2H3O2
Ammonia:
28 g NH3
0.90 g 1000 cm3 1 mol



= 15 mol/L
100 . g so ln
L
17.03 g
cm3
28 g NH3 1000 g 1 mol
= 23 mol/kg


72 g H 2O
kg
17.03 g
28 g NH3 ×
χ NH3 
28.
1 mol
1 mol
= 1.6 mol NH3; 72 g H2O ×
= 4.0 mol H2O
17.03 g
18.02 g
1.6
= 0.29
4.0  1.6
a. If we use 100. mL (100. g) of H2O, we need:
0.100 kg H2O ×
2.0 mol KCl 74.55 g
= 14.9 g = 15 g KCl

kg
mol KCl
Dissolve 15 g KCl in 100. mL H2O to prepare a 2.0 m KCl solution. This will give us
slightly more than 100 mL, but this will be the easiest way to make the solution. Since 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.
b. If we took 15 g NaOH and 85 g H2O, the volume would probably be less than 100 mL.
To make sure we have enough solution, let’s use 100. mL H2O (100. g). Let x = mass of
NaCl.
mass % NaOH = 15 =
x
× 100, 1500 + 15 x = 100. x, x = 17.6 g ≈ 18 g
100.  x
Dissolve 18 g NaOH in 100. mL H2O to make a 15% NaOH solution by mass.
382
CHAPTER 11
PROPERTIES OF SOLUTIONS
c. In a fashion similar to part b, let’s use 100. mL CH3OH. Let x = mass of NaOH.
100. mL CH3OH ×
0.79 g
= 79 g CH3OH
mL
mass % NaOH = 25 =
x
× 100, 25(79) + 25 x = 100. x, x = 26.3 g ≈ 26 g
79  x
Dissolve 26 g NaOH in 100. mL CH3OH.
d. To make sure we have enough solution, let’s use 100. mL (100. g) of H2O. Let x = mol
C6H12O6.
1 mol H 2O
100. g H2O ×
= 5.55 mol H2O
18.02 g
χ C6H12O6  0.10 
x
; 0.10 x + 0.56 = x, x = 0.62 mol C6H12O6
x  5.55
0.62 mol C6H12O6 ×
180.16 g
= 110 g C6H12O6
mol
Dissolve 110 g C6H12O6 in 100. mL of H2O to prepare a solution with χ C6H12O6 = 0.10.
29.
25 mL C5H12 ×
0.63 g
0.63 g 1 mol
= 16 g C5H12; 25 mL ×
= 0.22 mol C5H12

mL
72.15 g
mL
45 mL C6H14 ×
0.66 g
0.66 g 1 mol
= 30. g C6H14; 45 mL ×
= 0.34 mol C6H14

mL
mL
86.17 g
mass % pentane =
χpentane =
molality =
molarity =
30.
mass pentane
16 g
× 100 =
= 35%
total mass
16 g  30. g
mol pentane
0.22 mol
=
= 0.39
total mol
0.22 mol  0.34 mol
mol pentane
0.22 mol
=
= = 7.3 mol/kg
kg hexane
0.030 kg
mol pentane
0.22 mol
1000 mL
=
= 3.1 mol/L

L solution
25 mL  45 mL
1L
If there are 100.0 mL of wine:
12.5 mL C2H5OH ×
0.789 g
1.00 g
= 9.86 g C2H5OH and 87.5 mL H2O ×
= 87.5 g H2O
mL
mL
CHAPTER 11
PROPERTIES OF SOLUTIONS
mass % ethanol =
molality =
31.
383
9.86
× 100 = 10.1% by mass
87.5  9.86
9.86 g C2 H5OH 1 mol
= 2.45 mol/kg

0.0875 kg H 2O 46.07 g
If we have 1.00 L of solution:
192.12 g
= 263 g citric acid (H3C6H5O7)
mol
1.10 g
1.00 × 103 mL solution ×
= 1.10 × 103 g solution
mL
1.37 mol citric acid ×
mass % of citric acid =
263 g
× 100 = 23.9%
1.10  10 3 g
In 1.00 L of solution, we have 263 g citric acid and (1.10 × 103  263) = 840 g of H2O.
molality =
1.37 mol citric acid
= 1.6 mol/kg
0.84 kg H 2 O
840 g H2O ×
1 mol
1.37
= 47 mol H2O; χ citric acid 
= 0.028
18.02 g
47  1.37
Since citric acid is a triprotic acid, the number of protons citric acid can provide is three times
the molarity. Therefore, normality = 3 × molarity:
normality = 3 × 1.37 M = 4.11 N
32.
1.00 mol acetone
1 mol
= 1.00 molal; 1.00 × 103 g C2H5OH ×
= 21.7 mol C2H5OH
1.00 kg ethanol
46.07 g
χacetone =
1.00
= 0.0441
1.00  21.7
1 mol CH3COCH3 ×
58.08 g CH3COCH3
1 mL
= 73.7 mL CH3COCH3

mol CH3COCH3
0.788 g
1.00 × 103 g ethanol ×
molarity =
1 mL
= 1270 mL; Total volume = 1270 + 73.7 = 1340 mL
0.789 g
1.00 mol
= 0.746 M
1.34 L
384
CHAPTER 11
PROPERTIES OF SOLUTIONS
Energetics of Solutions and Solubility
33.
Using Hess’s law:
NaI(s) → Na+(g) + I(g)
ΔH = ΔHLE = (686 kJ/mol)
Na+(g) + I(g) → Na+(aq) + I(aq)
ΔH = ΔHhyd = 694 kJ/mol
__________________________________________________________________
NaI(s) → Na+(aq) + I(aq)
ΔHsoln = 8 kJ/mol
ΔHsoln refers to the heat released or gained when a solute dissolves in a solvent. Here, an
ionic compound dissolves in water.
34.
a.
CaCl2(s) → Ca2+(g) + 2 Cl(g)
ΔH = ΔHLE =  (2247 kJ)
2+

2+

Ca (g) + 2 Cl (g) → Ca (aq) + 2 Cl (aq)
ΔH = ΔHhyd
_______________________________________________________________
CaCl2(s) → Ca2+(aq) + 2 Cl(aq)
ΔHsoln = 46 kJ
46 kJ = 2247 kJ + ΔHhyd, ΔHhyd = 2293 kJ
CaI2(s) → Ca2+(g) + 2 I(g)
ΔH = ΔHLE =  (2059 kJ)
2+

2+

Ca (g) + 2 I (g) → Ca (aq) + 2 I (aq)
ΔH = ΔHhyd
______________________________________________________________
CaI2(s) → Ca2+(aq) + 2 I(aq)
ΔHsoln = 104 kJ
104 kJ = 2059 kJ + ΔHhyd, ΔHhyd = 2163 kJ
b. The enthalpy of hydration for CaCl2 is more exothermic than for CaI2. Any differences
must be due to differences in hydration between Cl and I. Thus, the chloride ion is more
strongly hydrated as compared to the iodide ion.
35.
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
hydroxide. The attraction of water molecules for Al3+ and OH cannot overcome the larger
lattice energy and Al(OH)3 is insoluble. For NaOH, the favorable hydration energy is large
enough to overcome the smaller lattice energy and NaOH is soluble.
36.
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
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
when the separated gaseous ions are stabilized as water molecules surround the ions. Since
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
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
minimal, and the temperature change is minimal.
CHAPTER 11
37.
PROPERTIES OF SOLUTIONS
385
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
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
that they cancel each other out, then the molecule is nonpolar. If the bond dipoles do not
cancel each other out, then the molecule is polar.
a. KrF2, 8 + 2(7) = 22 e
b. SF2, 6 + 2(7) = 20 e
S
F
Kr
F
F
nonpolar; soluble in CCl4
c. SO2, 6 + 2(6) = 18 e
polar; soluble in H2O
d. CO2, 4 + 2(6) = 16 e
+ 1 more
S
O
F
O
O
polar; soluble in H2O
C
O
nonpolar; soluble in CCl4
e. MgF2 is an ionic compound so it is soluble in water.
f.
CH2O, 4 + 2(1) + 6 = 12 e
g. C2H4, 2(4) + 4(1) = 12 e
O
H
C
C
H
H
C
H
H
H
polar; soluble in H2O
b. water
nonpolar (like all compounds made up of
only carbon and hydrogen); soluble in CCl4
38.
a. water
c. hexane
d. water
39.
Water is a polar molecule capable of hydrogen bonding. Polar molecules, especially
molecules capable of hydrogen bonding, and ions are all attracted to water. For covalent
compounds, as polarity increases, the attraction to water increases. For ionic compounds, as
the charge of the ions increases and/or the size of the ions decreases, the attraction to water
increases.
a. CH3CH2OH; CH3CH2OH is polar while CH3CH2CH3 is nonpolar.
b. CHCl3; CHCl3 is polar while CCl4 is nonpolar.
c. CH3CH2OH; CH3CH2OH is much more polar than CH3(CH2)14CH2OH.
386
CHAPTER 11
PROPERTIES OF SOLUTIONS
40.
For ionic compounds, as the charge of the ions increases and/or the size of the ions decreases,
the attraction to water (hydration) increases.
a. Mg2+; smaller size, higher charge
b. Be2+; smaller
c. Fe3+; smaller size, higher charge
d. F; smaller
e. Cl; smaller
f.
SO42-; higher charge
41.
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
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,
i.e., the effect of the ‒OH group decreases as the alcohols get larger.
42.
Benzoic acid is capable of hydrogen bonding, but a significant part of benzoic acid is the nonpolar benzene ring which is composed of only carbon and hydrogen. In benzene, a hydrogen
bonded dimer forms:
O
H
O
C
C
O
H
O
The dimer is relatively nonpolar since the polar part of benzoic acid is hidden in the dimer
formation. Thus, benzoic acid is more soluble in benzene than in water due to the dimer
formation.
Benzoic acid would be more soluble in 0.1 M NaOH because of the reaction:
C6H5CO2H + OH → C6H5CO2 + H2O
By removing the proton from benzoic acid, an anion forms, and like all anions, the species
becomes more soluble in water.
8.21  10 4 mol
= k × 0.790 atm, k = 1.04 × 10 3 mol/Latm
L
1.04  10 4 mol
C = kP, C =
× 1.10 atm = 1.14 × 10 3 mol/L
L atm
43.
C = kP,
44.
750. mL grape juice ×
12 mL C2 H5OH 0.79 g C2 H5OH 1 mol C2 H5OH


100. mL juice
mL
46.07 g

2 mol CO2
= 1.54 mol CO2 (carry extra significant figure)
2 mol C2 H5OH
1.54 mol CO2 = total mol CO2 = mol CO2(g) + mol CO2(aq) = ng + naq
CHAPTER 11
PCO 2 =
PCO 2 =
PROPERTIES OF SOLUTIONS
n g RT
V
 0.08206 L atm 
 298 K 
n g 
mol K


=
= 326 ng
3
75  10 L
n aq
C
0.750 L
=
= 43.0 naq = 43.0 naq
k
3.1  10  2 mol
L atm
PCO 2 = 326 ng = 43.0 naq and from above naq = 1.54  ng; Solving:
326 ng = 43.0(1.54  ng), 369 ng = 66.2, ng = 0.18 mol
PCO 2 = 326(0.18) = 59 atm in gas phase
C = kPCO 2 =
3.1  10 2 mol
× 59 atm, C = 1.8 mol CO2/L in wine
L atm
Vapor Pressures of Solutions
45.
mol C3H8O3 = 164 g ×
mol H2O = 338 mL ×
PH2O  χ H2O PHo 2O =
46.
1 mol
= 1.78 mol C3H8O3
92.09 g
0.992 g 1 mol
= 18.6 mol H2O

mL
18.02 g
18.6 mol
× 54.74 torr = 0.913 × 54.74 torr = 50.0 torr
(1.78  18.6) mol
PC2 H5OH = χ C2H5OHPCo2H5OH ; χ C2H5OH =
53.6 g C3H8O3 ×
mol C 2 H 5 OH solution
total mol in solution
1 mol C3H8O3
= 0.582 mol C3H8O3
92.09 g
1 mol C 2 H 5OH
= 2.90 mol C2H5OH; total mol = 0.582 + 2.90
46.07 g
= 3.48 mol
2.90 mol
113 torr =
 PCo2H5OH , PCo2H5OH = 136 torr
3.48 mol
133.7 g C2H5OH ×
387
388
47.
CHAPTER 11
PROPERTIES OF SOLUTIONS
PB  χ B PBo , χ B  PB / PBo = 0.900 atm/0.930 atm = 0.968
mol benzene
1 mol
; mol benzene = 78.11 g C6H6 ×
= 1.000 mol
total mol
78.11 g
1.000 mol
Let x = mol solute, then: χB = 0.968 =
, 0.968 + 0.968 x = 1.000, x = 0.033 mol
1.000  x
0.968 =
molar mass =
48.
10.0 g
= 303 g/mol ≈ 3.0 × 102 g/mol
0.033 mol
19.6 torr = χ Η Ο (23.8 torr), χ Η Ο = 0.824; χ solute = 1.000  0.824 = 0.176
0.176 is the mol fraction of all the solute particles present. Since NaCl dissolves to produce
two ions in solution (Na+ and Cl), 0.176 is the mole fraction of Na+ and Cl ions present
(assuming complete dissociation of NaCl).
At 45°C, PH 2O = 0.824 (71.9 torr) = 59.2 torr
49.
a. 25 mL C5H12 ×
0.63 g 1 mol
= 0.22 mol C5H12

mL
72.15 g
0.66 g 1 mol
= 0.34 mol C6H14; total mol = 0.22 + 0.34 = 0.56 mol

mL
86.17 g
mol pentane in solution 0.22 mol
= 0.39, χ Lhex = 1.00 - 0.39 = 0.61


total mol in solution
0.56 mol
45 mL C6H14 ×
χ Lpen
o
= 0.39(511 torr) = 2.0 × 102 torr; Phex = 0.61(150. torr) = 92 torr
Ppen  χ Lpen Ppen
Ptotal  Ppen  Phex = 2.0 × 102 + 92 = 292 torr = 290 torr
b. From Chapter 5 on gases, the partial pressure of a gas is proportional to the number of
moles of gas present. For the vapor phase:
χ Vpen 
mol pentane in vapor Ppen 2.0  10 2 torr


= 0.69
total mol vapor
Ptotal
290 torr
Note: In the Solutions Guide, we have added V or L to the mole fraction symbol to emphasize the value for which we are solving. If the L or V is omitted, then the liquid phase is
assumed.
50.
L
=
Ptotal  PCH 2Cl 2  PCH 2Br2 ; P  χ L Po ; χ CH
2Cl 2
0.0300 mol CH 2 Cl 2
= 0.375
0.0800 mol total
Ptotal = 0.375 (133 torr) + (1.000  0.375) (11.4 torr) = 49.9 + 7.13 = 57.0 torr
CHAPTER 11
PROPERTIES OF SOLUTIONS
V

In the vapor: χ CH
2 Cl 2
51.
PCH 2Cl 2

Ptotal
49.9 torr
V
= 0.875; χ CH
= 1.000  0.875 = 0.125
2 Br2
57.0 torr
Ptotal  Pmeth  Pprop , 174 torr = χ Lmeth (303 torr) + χ Lprop (44.6 torr); χ Lprop = 1.000  χ Lmeth
174 = 303 χ Lmeth + (1.000  χ Lmeth ) 44.6,
52.
389
129
 χ Lmeth = 0.500;
258
χ Lprop = 1.000  0.500 = 0.500
o
o
Ptol  χ LtolPtol
; Pben  χ Lben Pben
; For the vapor, χ VA  PA / Ptotal. Since the mole fractions of
benzene and toluene are equal in the vapor phase, then Ptol  Pben .
o
o
o
= (1.00  χ Ltol ) Pben
χ LtolPtol
 χ Lben Pben
, χ Ltol (28 torr) = (1.00  χ Ltol ) 95 torr
123 χ tol = 95, χ tol = 0.77; χ ben = 1.00  0.77 = 0.23
L
L
L
53.
Compared to H2O, solution d (methanol/water) will have the highest vapor pressure because
methanol is more volatile than water. Both solution b (glucose/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. Since there are more solute particles
in solution c, the vapor pressure of solution c will be the lowest.
54.
Solution d (methanol/water); Methanol is more volatile than water, which will increase the
total vapor pressure to a value greater than the vapor pressure of pure water at this
temperature.
55.
50.0 g CH3COCH3 ×
50.0 g CH3OH ×
L
=
χ acetone
1 mol
= 0.861 mol acetone
58.08 g
1 mol
= 1.56 mol methanol
32.04 g
0.861
L
= 0.356; χ Lmethanol = 1.000  χ acetone
= 0.644
0.861  1.56
Ptotal = Pmethanol + Pacetone = 0.644(143 torr) + 0.356(271 torr) = 92.1 torr + 96.5 torr
= 188.6 torr
Because partial pressures are proportional to the moles of gas present, then in the vapor
phase:
V
χ acetone

Pacetone 96.5 torr
= 0.512; χ Vmethanol = 1.000 - 0.512 = 0.488

Ptotal
188.6 torr
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
390
CHAPTER 11
PROPERTIES OF SOLUTIONS
from Raoult’s law. This occurs when the solute-solvent interactions are stronger than in pure
solute and pure solvent.
56.
a. An ideal solution would have a vapor pressure at any mole fraction of H2O between that
of pure propanol and pure water (between 74.0 torr and 71.9 torr). The vapor pressures
of the solutions are not between these limits, so water and propanol do not make ideal
solutions.
b. From the data, the vapor pressures of the various solutions are greater than in the ideal
solution (positive deviation from Raoult’s law). This occurs when the intermolecular
forces in solution are weaker than the intermolecular forces in pure solvent and pure
solute. This gives rise to endothermic (positive) ΔHsoln values.
c. The interactions between propanol and water molecules are weaker than between the pure
substances since this solution exhibits a positive deviation from Raoult’s law.
d. At χ H 2O = 0.54, the vapor pressure is highest as compared to the other solutions. Since a
solution boils when the vapor pressure of the solution equals the external pressure, the
χ H 2O = 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 than the other solutions.
Colligative Properties
57.
molality = m =
ΔTb = Kbm =
mol solute 27.0 g N 2 H 4CO 1000 g 1 mol N 2 H 4CO
= 3.00 molal



kg solvent
150.0 g H 2O
kg
60.06 g N 2 H 4CO
0.51 C
× 3.00 molal = 1.5°C
molal
The boiling point is raised from 100.0°C to 101.5°C (assuming P = 1 atm).
58.
ΔTb = 77.85°C  76.50°C = 1.35°C; m =
mol biomolecule = 0.0150 kg solvent ×
ΔTb
1.35C
= 0.268 mol/kg

K b 5.03 C kg / mol
0.268 mol hydrocarbon
= 4.02 × 10 3 mol
kg solvent
From the problem, 2.00 g biomolecule was used that must contain 4.02 × 10 3 mol
biomolecule. The molar mass of the biomolecule is:
2.00 g
= 498 g/mol
4.02  10 3 mol
59.
ΔTf = Kfm, ΔTf = 1.50°C =
0.200 kg H2O ×
1.86 C
× m, m = 0.806 mol/kg
molal
0.806 mol C3H8O3 92.09 g C3H8O3
= 14.8 g C3H8O3

kg H 2O
mol C3H8O3
CHAPTER 11
60.
PROPERTIES OF SOLUTIONS
ΔTf = 25.50°C  24.59°C = 0.91°C = Kfm, m =
391
0.91 C
= 0.10 mol/kg
9.1 C / molal
 0.10 mol H 2 O   18.02 g H 2 O
 
mass H2O = 0.0100 kg t-butanol 
 kg t  butanol   mol H 2 O
61.
molality = m =

 = 0.018 g H2O

50.0 g C 2 H 6O 2 1000 g 1 mol
= 16.1 mol/kg


50.0 g H 2O
kg
62.07 g
ΔTf = Kfm = 1.86°C/molal × 16.1 molal = 29.9°C; Tf = 0.0°C  29.9°C = 29.9°C
ΔTb = Kbm = 0.51°C/molal × 16.1 molal = 8.2°C; Tb = 100.0°C + 8.2°C = 108.2°C
62.
m=
ΔTf
25.0 C
= 13.4 mol C2H6O2/kg

K f 1.86 C kg / mol
Since the density of water is 1.00 g/cm3, the moles of C2H6O2 needed are:
15.0 L H2O ×
1.00 kg H 2O 13.4 mol C2 H 6O 2
= 201 mol C2H6O2

L H 2O
kg H 2O
Volume C2H6O2 = 201 mol C2H6O2 ×
ΔTb = Kbm =
63.
62.07 g
1 cm3

= 11,200 cm3 = 11.2 L
mol C 2 H 6 O 2
1.11 g
0.51 C
× 13.4 molal = 6.8°C , Tb = 100.0°C + 6.8°C = 106.8°C
molal
ΔTf = Kfm, m =
5.86  10 2 mol thyroxine
ΔTf
0.300 C


Kf
5.12 C kg / mol
kg benzene
The mol of thyroxine present is:
0.0100 kg benzene ×
5.86  10 2 mol thyroxine
kg benzene
= 5.86 × 10 4 mol thyroxine
From the problem, 0.455 g thyroxine were used; this must contain 5.86 × 10 4 mol thyroxine.
The molar mass of the thyroxine is:
molar mass =
64.
0.455 g
5.86  10 4 mol
= 776 g/mol
empirical formula mass ≈ 7(12) + 4(1) + 16 = 104 g/mol
ΔTf = Kfm, m =
ΔTf
22.3 C
= 0.56 molal

Kf
40. C / molal
392
CHAPTER 11
mol anthraquinone = 0.0114 kg solvent ×
1.32 g
molar mass =
6.4  10 3 mol
molar mass
empirical formula mass
65.
a. M =

PROPERTIES OF SOLUTIONS
0.56 mol anthraquin one
kg solvent
= 6.4 × 10 3 mol
= 210 g/mol
210
= 2.0; molecular formula = C14H8O2
104
1.0 g protein
1 mol
= 1.1 × 10 5 mol/L; π = MRT

4
L
9.0  10 g
At 298 K: π =
1.1  10 5 mol
L

760 torr
0.08206 L atm
× 298 K ×
, π = 0.20 torr
mol K
atm
Because d = 1.0 g/cm3, 1.0 L solution has a mass of 1.0 kg. Because only 1.0 g of protein
is present per liter of solution, 1.0 kg of H2O is present and molality equals molarity.
ΔTf = Kfm =
1.86 C
× 1.1 × 10 5 molal = 2.0 × 10 5 °C
molal
b. Osmotic pressure is better for determining the molar mass of large molecules. A
temperature 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.
1 atm
π
760 torr
M=
= 4.0 × 10 5 mol/L

0
.
08206
L
atm
RT
 300 . K
mol K
0.74 torr 
66.
1.00 L ×
4.0  10 5 mol
= 4.0 × 10 5 mol catalase
L
molar mass =
10.00 g
= 2.5 × 105 g/mol
4.0  10 5 mol
67.
π = MRT, M =
68.
M=
π
8.00 atm
= 0.327 mol/L

RT 0.08206 L atm  298 K
mol K
π
15 atm

= 0.62 M solute particles
RT 0.08206 L atm  295 K
mol K
This represents the total molarity of the solute particles. NaCl is a soluble ionic compound
that breaks up into two ions, Na+ and Cl. Therefore, the concentration of NaCl needed is
0.62/2 = 0.31 M; this NaCl concentration will produce a 0.62 M solute particle solution
assuming complete dissociation.
CHAPTER 11
1.0 L ×
PROPERTIES OF SOLUTIONS
393
0.31 mol NaCl 58.44 g NaCl
= 18.1 ≈ 18 g NaCl

L
mol NaCl
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.
Properties of Electrolyte Solutions
69.
Na3PO4(s) → 3 Na+(aq) + PO34 (aq), i = 4.0; CaBr2(s) → Ca2+(aq) + 2 Br(aq), i = 3.0
KCl(s) → K+(aq) + Cl(aq), i = 2.0.
The effective particle concentrations of the solutions are:
4.0(0.010 molal) = 0.040 molal for Na3PO4 solution; 3.0(0.020 molal) = 0.060 molal for
CaBr2 solution; 2.0(0.020 molal) = 0.040 molal for KCl solution; slightly greater than
0.020 molal for HF solution since HF only partially dissociates in water (it is a weak
acid).
a. The 0.010 m Na3PO4 solution and the 0.020 m KCl solution both have effective particle
concentrations of 0.040 m (assuming complete dissociation), so both of these solutions
should have the same boiling point as the 0.040 m C6H12O6 solution (a nonelectrolyte).
b. P = χP°; As the solute concentration decreases, the solvent’s vapor pressure increases
since χ increases. Therefore, the 0.020 m HF solution will have the highest vapor
pressure since it has the smallest effective particle concentration.
c. ΔT = Kfm; The 0.020 m CaBr2 solution has the largest effective particle concentration so
it will have the largest freezing point depression (largest ΔT).
70.
71.
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
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
concentration because it produces three ions per mol of compound.
a. pure water
b. CaCl2 solution
d. pure water
e. CaCl2 solution
c. CaCl2 solution
a. MgCl2(s) → Mg2+(aq) + 2 Cl(aq), i = 3.0 mol ions/mol solute
ΔTf = iKfm = 3.0 × 1.86°C/molal × 0.050 molal = 0.28°C; Tf = -0.28°C (Assuming water
freezes at 0.00°C.)
ΔTb = iKbm = 3.0 × 0.51°C/molal × 0.050 molal = 0.077°C; Tb = 100.077°C (Assuming
water boils at 100.000°C.)
394
CHAPTER 11
PROPERTIES OF SOLUTIONS
b. FeCl3(s) → Fe3+(aq) + 3 Cl(aq), i = 4.0 mol ions/mol solute
ΔTf = iKfm = 4.0 × 1.86°C/molal × 0.050 molal = 0.37°C; Tf = 0.37°C
ΔTb = iKbm = 4.0 × 0.51°C/molal × 0.050 molal = 0.10°C; Tb = 100.10°C
72.
NaCl(s) → Na+(aq) + Cl(aq), i = 2.0
π = iMRT = 2.0 ×
0.10 mol
L

0.08206 L atm
× 293 K = 4.8 atm
mol K
A pressure greater than 4.8 atm should be applied to insure purification by reverse osmosis.
73.
ΔTf = iKfm, i =
i=
ΔTf
0.110  C

= 2.63 for 0.0225 m CaCl2
K f m 1.86  C / molal  0.0225 molal
0.440
1.330
= 2.60 for 0.0910 m CaCl2; i =
= 2.57 for 0.278 m CaCl2
1.86  0.0910
1.86  0.278
iave = (2.63 + 2.60 + 2.57)/3 = 2.60
Note that i is less than the ideal value of 3.0 for CaCl2. This is due to ion pairing in solution.
Also note that as molality increases, i decreases. More ion pairing occurs as the solute concentration increases.
74.
a. MgCl2, i(observed) = 2.7
ΔTf = iKfm = 2.7 × 1.86°C/molal × 0.050 molal = 0.25°C; Tf = 0.25°C
ΔTb = iKbm = 2.7 × 0.51°C/molal × 0.050 molal = 0.069°C; Tb = 100.069°C
b. FeCl3, i(observed) = 3.4
ΔTf = iKfm = 3.4 × 1.86 °C/molal × 0.050 molal = 0.32°C; Tf = 0.32°C
ΔTb = iKbm = 3.4 × 0.51°C/molal × 0.050 molal = 0.087°C; Tb = 100.087°C
75.
a. TC = 5(TF  32)/9 = 5(29  32)/9 = 34°C; Assuming the solubility of CaCl2 is
temperature independent, the molality of a saturated CaCl2 solution is:
74.5 g CaCl2 1000 g
1 mol CaCl2
6.71 mol CaCl2



100.0 g
kg
110.98 g CaCl2
kg H 2O
ΔTf = 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
37.4°C. Assuming these conditions, a saturated CaCl2 solution should melt ice at 34°C
(29°F).
CHAPTER 11
PROPERTIES OF SOLUTIONS
395
b. From Exercise 11.73, iave = 2.60; ΔTf = iKfm = 2.60 × 1.86 × 6.71 = 32.4°C
Tf = 32.4°C
Assuming i = 2.60, a saturated CaCl2 solution will not melt ice at 34°C(29°F).
76.
 = iMRT, M =
2.50 atm
π
=
= 5.11 × 10 2 mol/L
0.08206 L atm
iRT
2.00 
 298 K
K mol
molar mass of compound =
0.500 g
5.11  10  2 mol
0.1000 L 
L
= 97.8 g/mol
Additional Exercises
77
a. NH4NO3(s) → NH4+(aq) + NO3(aq) ΔHsoln = ?
Heat gain by dissolution process = heat loss by solution; We will keep all quantities positive in order to avoid sign errors. Since the temperature of the water decreased, the dissolution of NH4NO3 is endothermic (ΔH is positive). Mass of solution = 1.60 + 75.0 =
76.6 g.
heat loss by solution =
ΔHsoln =
4.18 J
× 76.6 g × (25.00°C  23.34°C) = 532 J
g C
80.05 g NH 4 NO3
532 J
= 2.66 × 104 J/mol = 26.6 kJ/mol

1.60 g NH 4 NO3
mol NH 4 NO3
b. We will use Hess’s law to solve for the lattice energy. The lattice energy equation is:
NH4+(g) + NO3(g) → NH4NO3(s)
NH4+(g) + NO3(g) → NH4+(aq) + NO3(aq)
ΔH = lattice energy
ΔH = ΔHhyd = 630. kJ/mol
NH4 (aq) + NO3 (aq) → NH4NO3(s)
ΔH = ΔHsoln = 26.6 kJ/mol
____________________________________________________________________
NH4+(g) + NO3(g) → NH4NO3(s)
ΔH = ΔHhyd  ΔHsoln
= 657 kJ/mol
+
78.

The main intermolecular forces are: hexane (C6H14): London dispersion; chloroform
(CHCl3): dipole-dipole; London dispersion; methanol (CH3OH): H bonding; H2O:
H bonding (two places)
There is a gradual change in the nature of the intermolecular forces (weaker to stronger).
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.
79.
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 temperature. The higher the cooking temperature, the faster the cooking time.
396
CHAPTER 11
PROPERTIES OF SOLUTIONS
b. Salt dissolves in water forming a solution with a melting point lower than that of pure
water (ΔTf = Kfm). This happens in water on the surface of ice. If it is not too cold, the
ice melts. This won't work if the ambient temperature is lower than the depressed
freezing point of the salt solution.
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.
d. On the CO2 phase diagram in chapter 10, the triple point is above 1 atm, so CO2(g) is the
stable phase at 1 atm and room temperature. CO2(l) can't exist at normal atmospheric
pressures. Therefore, dry ice sublimes instead of boils. In a fire extinguisher, P > 1 atm
and CO2(l) can exist. When CO2 is released from the fire extinguisher, CO2(g) forms as
predicted from the phase diagram.
e. Adding a solute to a solvent increases the boiling point and decreases the freezing point
of the solvent. Thus, the solvent is a liquid over a wider range of temperatures when a
solute is dissolved.
80.
A 92 proof ethanol solution is 46% C2H5OH by volume. Assuming 100.0 mL of solution:
mol ethanol = 46 mL C2H5OH ×
molarity =
81.
0.79 g 1 mol C 2 H 5 OH
= 0.79 mol C2H5OH

mL
46.07 g
0.79 mol
= 7.9 M ethanol
0.1000 L
V
Because partial pressures are proportional to the moles of gas present, then χ CS
 PCS2 / Ptot .
2
V
PCS2  χ CS
Ptot = 0.855 (263 torr) = 225 torr
2
L
L
PCS2  χ CS
P o , χ CS

2 CS 2
2
82.
0.1mol
PCS2
o
PCS
2

225 torr
= 0.600
375 torr
0.08206 L atm
× 298 K = 2.45 atm ≈ 2 atm
mol K
L
760 mm Hg
π = 2 atm ×
≈ 2000 mm ≈ 2 m
atm
π = MRT =

The osmotic pressure would support a mercury column of ≈ 2 m. The height of 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
density of Hg is 13.6 g/cm3.
Height of fluid ≈ 2 m × 13.6 ≈ 30 m
83.
Out of 100.00 g, there are:
31.57 g C ×
2.629
1 mol C
= 2.629 mol C;
= 1.000
12.01 g
2.629
CHAPTER 11
PROPERTIES OF SOLUTIONS
5.30 g H ×
5.26
1 mol H
= 5.26 mol H;
= 2.00
2.629
1.008 g
63.13 g O ×
3.946
1 mol O
= 3.946 mol O;
= 1.501
2.629
16.00 g
397
empirical formula: C2H4O3; Use the freezing point data to determine the molar mass.
m=
ΔTf
5.20 C
= 2.80 molal

K f 1.86 C / molal
mol solute = 0.0250 kg ×
molar mass =
2.80 mol solute
= 0.0700 mol solute
kg
10.56 g
= 151 g/mol
0.0700 mol
The empirical formula mass of C2H4O3 = 76.05 g/mol. Since the molar mass is about twice
the empirical mass, the molecular formula is C4H8O6, which has a molar mass of 152.10
g/mol.
Note: We use the experimental molar mass to determine the molecular formula. Knowing
this, we calculate the molar mass precisely from the molecular formula using the atomic
masses in the periodic table.
84.
a. As discussed in Figure 11.18 of the text, the water would migrate from right to left.
Initially, the level of liquid in the right arm would go down and the level in the left arm
would go up. At some point, the rate of solvent transfer would be the same in both
directions and the levels of the liquids in the two arms would stabilize. The height
difference between the two arms is a measure of the osmotic pressure of the NaCl
solution.
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
possible, the levels of the liquids will be equal once the rate of solute and solvent transfer
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.
85.
If ideal, NaCl dissociates completely and i = 2.00. ΔTf = iKfm; Assuming water freezes at
0.00°C:
1.28°C = 2 × 1.86°C kg/mol × m, m = 0.344 mol NaCl/kg H2O
Assume an amount of solution which contains 1.00 kg of water (solvent).
0.344 mol NaCl ×
58.44 g
20.1 g
= 20.1 g NaCl; mass % NaCl =
× 100
mol
1.00  10 3 g  20.1 g
= 1.97%
398
86.
CHAPTER 11
PROPERTIES OF SOLUTIONS
The main factor for stabilization seems to be electrostatic repulsion. The center of a colloid
particle is surrounded by a layer of same charged ions, with opposite charged ions forming
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
outer layers are the same charge, the particles repel each other and do not easily aggregate for
precipitation to occur.
Heating increases the velocities of the colloidal particles. This causes the particles to collide
with enough energy to break the ion barriers, allowing the colloids to aggregate and
eventually precipitate out. Adding an electrolyte neutralizes the adsorbed ion layers which
allows colloidal particles to aggregate and then precipitate out.
87.
T = Kfm, m =
ΔT
2.79 o C

= 1.50 molal
Kf
1.86 o C / molal
a. T = Kbm, T = (0.51C/molal)(1.50 molal) = 0.77C, Tb = 100.77C
b.
o
Pwater  χ water Pwater
, χ water =
mol H 2 O
mol H 2 O  mol solute
Assuming 1.00 kg of water, we have 1.50 mol solute and:
mol H2O = 1.00 × 103 g H2O ×
water =
1 mol H 2 O
= 55.5 mol H2O
18.02 g H 2 O
55.5 mol
= 0.974; Pwater = (0.974)(23.76 mm Hg) = 23.1 mm Hg
1.50  55.5
c. We assumed ideal behavior in solution formation and assumed i = 1 (no ions form).
Challenge Problems
88.
For the second vapor collected, χ VB, 2 = 0.714 and χ TV, 2 = 0.286, where χ LB, 2 = mole fraction
of benzene in the second solution and χ TL, 2 = mole fraction of toluene in the second solution.
χ LB, 2  χ TL, 2 = 1.000
χ VB, 2 = 0.714 =
χ LB, 2 (750 .0 torr )
PB
PB
= L

χ B, 2 (750 .0 torr )  (1.000  χ LB, 2 )(300 .0 torr )
PTOT
PB  PT
Solving: χ LB, 2 = 0.500 = χ TL, 2
This second solution came from the vapor collected from the first (initial) solution. So, χ VB, 1
= χ TV, 1 = 0.500. Let χ LB, 1 = mole fraction benzene in the first solution and χ TL, 1 = mole fraction of toluene in first solution. χ LB, 1  χ TL, 1 = 1.000
CHAPTER 11
χ VB, 1
PROPERTIES OF SOLUTIONS
399
χ LB, 1 (750 .0 torr )
PB
PB
= 0.500 =
= L

χ B, 1 (750 .0 torr )  (1.000  χ LB, 1 )(300 .0 torr )
PTOT
PB  PT
Solving: χ LB, 1 = 0.286
The original solution had B = 0.286 and T = 0.714.
89.
PA
× 100:
PA  PB
χ Ax
χ Ax
0.30 =
, 0.30 
χ Ax  χ By
χ A x  (1.00  χ A ) y
For 30.% A by moles in the vapor, 30. =
χA x = 0.30(χA x) + 0.30 y  0.30 χA y, χA x  0.30 χA x + 0.30 χA y = 0.30 y
χA(x  0.30 x + 0.30 y) = 0.30 y, χA =
0.30 y
; χB = 1.00  χA
0.70 x  0.30 y
Similarly, if vapor above is 50.% A: χ A 
If vapor above is 80%A: χA =
y
y
; χ B  1.00 
x y
x y
0.80 y
; χB = 1.00  χA
0.20 x  0.80 y
If the liquid solution is 30.%A by moles, χA = 0.30.
PA
0.30 x
0.30 x
=
and χ VB  1.00 
PA  PB 0.30 x  0.70 y
0.30 x  0.70 y
x
If solution is 50.%A: χ VA 
and χ VB  1.00  χ VA
x y
0.80 x
If solution is 80.%A: χ VA 
and χ VB  1.00  χ VA
0.80 x  0.20 y
Thus, χ VA 
90.
m=
ΔT
0.406 o C

= 0.218 mol/kg
Kf
1.86 o C / molal
 = MRT where M = mol/L; We must assume that molarity = molality so we can calculate
the osmotic pressure. This is a reasonable assumption for dilute solutions when 1.00 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
little larger than 1.00 L, but not by much (to three sig figs). The assumption that molarity =
molality will be good here.
 = (0.218 M)(0.08206 L atm/Kmol)(298 K) = 5.33 atm
91.
m=
ΔT
0.426 o C

= 0.229 molal
Kf
1.86 o C / molal
400
CHAPTER 11
PROPERTIES OF SOLUTIONS
Let x = mol NaCl in mixture and y = mol C12H22O11.
NaCl(aq) → Na+(aq) + Cl(aq); In solution, NaCl exists as separate Na+ and Cl ions.
0.229 mol = mol Na+ + mol Cl + mol C12H22O11 = x + x + y = 2x + y
The molar mass of NaCl is 58.44 g/mol and the molar mass of C12H22O11 = 342.3 g/mol.
Setting up another equation for the mass of the mixture:
20.0 g = x(58.44) + y(342.3)
Substituting: 20.0 = 58.44 x + (0.229 – 2x) 342.3
Solving:
x = mol NaCl = 0.0932 mol and
y = mol C12H22O11 = 0.229 – 2(0.0932) = 0.043 mol
mass NaCl = 0.0932 mol × 58.44 g/mol = 5.45 g NaCl
mass% NaCl =
sucrose =
92.
5.45 g
× 100 = 27.3%, mass % C12H22O11 = 100.0 – 27.3 = 72.7%
20.0 g
0.043 mol
= 0.32
(0.043  0.0932 ) mol
a. π = iMRT, iM =
π
7.83 atm
= 0.320 mol/L

RT 0.08206 L atm  298 K
mol K
Assuming 1.000 L of solution:
total mol solute particles = mol Na+ + mol Cl + mol NaCl = 0.320 mol
mass solution = 1000. mL ×
1.071 g
= 1071 g solution
mL
mass NaCl in solution = 0.0100 × 1071 g = 10.7 g NaCl
mol NaCl added to solution = 10.7 g ×
1 mol
= 0.183 mol NaCl
58.44 g
Some of this NaCl dissociates into Na+ and Cl (two mol ions per mol NaCl) and some
remains undissociated. Let x = mol undissociated NaCl = mol ion pairs.
mol solute particles = 0.320 mol = 2(0.183  x) + x
0.320 = 0.366  x, x = 0.046 mol ion pairs
CHAPTER 11
PROPERTIES OF SOLUTIONS
fraction of ion pairs =
401
0.046
= 0.25, or 25%
0.183
b. ΔT = 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 kg of
solvent present in 1.000 L solution.
mass of 1.000 L solution = 1071 g; mass of NaCl = 10.7 g
mass of solvent in 1.000 L solution = 1071 g  10.7 g = 1060. g
ΔT = 1.86 °C kg/mol ×
0.320 mol
= 0.562°C
1.060 kg
Assuming water freezes at 0.000°C, then Tf = 0.562°C.
93.
χ Vpen = 0.15 =
Ppen
o
; Ppen = χ Lpen Ppen
= χ Lpen (511 torr); Ptotal = Ppen + Phex = χ Lpen (511) +
Ptotal
χ Lhex (150.)
Since χ Lhex = 1.000  χ Lpen , then: Ptotal = χ Lpen (511 torr) + (1.000 χ Lpen )(150.) = 150. +
361 χ Lpen
χ Vpen
χ Lpen (511)
Ppen
=
, 0.15 =
, 0.15 (150. + 361 χ Lpen ) = 511 χ Lpen
L
150 .  361 χ pen
Ptotal
23 + 54 χ Lpen = 511 χ Lpen , χ Lpen =
94.
a. m =
23
= 0.050
457
ΔTf
1.32 C
= 0.258 mol/kg

Kf
5.12 C kg / mol
0.258 mol unknown
= 4.02 × 10 3 mol
kg
1.22 g
molar mass of unknown =
= 303 g/mol
4.02  10 3 mol
mol unknown = 0.01560 kg ×
Uncertainty in temperature =
0.04
× 100 = 3%; A 3% uncertainty in 303 g/mol
1.32
= 9 g/mol.
So, molar mass = 303 ± 9 g/mol.
b. No, codeine could not be eliminated since its molar mass is in the possible range
including the uncertainty.
402
CHAPTER 11
PROPERTIES OF SOLUTIONS
c. We would really like the uncertainty to be ± 1 g/mol. We need the freezing point
depression to be about 10 times what it was in this problem. Two possibilities are:
1. make the solution ten times more concentrated (may be a solubility problem) or
2. use a solvent with a larger Kf value, e.g., camphor
95.
ΔTf = 5.51C  2.81C = 2.70°C; m =
ΔTf
2.70 C
= 0.527 molal

Kf
5.12 C / molal
Let x = mass of naphthalene (molar mass = 128.2 g/mol). Then, 1.60  x = mass of anthracene
(molar mass = 178.2 g/mol).
x
1.60  x
= moles napthalene and
= moles anthracene
128.2
178.2
x
1.60  x

0.527 mol solute
178.2 , 1.05 × 10 2 = 178.2 x  1.60(128.2)  128.2 x
= 128.2
0.0200 kg solvent
kg solvent
128.2 (178.2)
50.0 x + 205 = 240., 50.0 x = 35, x = 0.70 g naphthalene
So mixture is:
96.
iM =
0.70 g
× 100 = 44% naphthalene by mass and 56% anthracene by mass
1.60 g
π
0.3950 atm
= 0.01614 mol/L = total ion concentration

0
.
08206
L atm
RT
 298 .2 K
mol K
0.01614 mol/L = M Mg 2 + M Na + M Cl  ; M Cl  = 2 M Mg 2 + M Na (charge balance)
Combining: 0.01614 = 3 M Mg 2 + 2 M Na 
Let x = mass MgCl2 and y = mass NaCl, then x + y = 0.5000 g.
M Mg 2  =
x
y
and M Na =
(Because V = 1.000 L.)
95.21
58.44
3x
2y
= 0.01614 mol/L; Rearranging: 3 x + 3.258 y

95.21 58.44
= 1.537
Solving by simultaneous equations:
Total ion concentration =
3 x
+ 3.258 y
= 1.537
3 (x
+
y) = 3(0.5000)
__________________________________________
0.258 y = 0.037, y = 0.14 g NaCl
CHAPTER 11
PROPERTIES OF SOLUTIONS
403
mass MgCl2 = 0.5000 g  0.14 g = 0.36 g; mass % MgCl2 =
97.
0.36 g
× 100 = 72%
0.5000 g
HCO2H → H+ + HCO2; Only 4.2% of HCO2H ionizes. The amount of H+ or HCO2
produced is:
0.042 × 0.10 M = 0.0042 M
The amount of HCO2H remaining in solution after ionization is:
0.10 M  0.0042 M = 0.10 M
The total molarity of species present = M HCO2H  M H  M HCO 
2
= 0.10 + 0.0042 + 0.0042 = 0.11 M
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:
ΔT = Kfm = 1.86°C/molal × 0.11 molal = 0.20°C; freezing point = 0.20°C
ΔT = Kbm = 0.51°C/molal × 0.11 molal = 0.056°C; boiling point = 100.056°C
98.
a. The average values for each ion are:
300. mg Na+; 15.7 mg K+; 5.45 mg Ca2+; 388 mg Cl; 246 mg lactate, C3H5O3
Note:
Since we can precisely weigh to ± 0.1 mg on an analytical balance, we'll carry extra
significant figures and calculate results to ± 0.1 mg.
The only source of lactate is NaC3H5O3.
246 mg C3H5O3 ×
112 .06 mg NaC3 H 5O3
89.07 mg C3 H 5O3

= 309.5 mg sodium lactate
The only source of Ca2+ is CaCl22H2O.
5.45 mg Ca2+ ×
147.0 mg CaCl2  2H 2O
= 19.99 or 20.0 mg CaCl22H2O
40.08 mg Ca 2
The only source of K+ is KCl.
15.7 mg K+ ×
74.55 mg KCl
= 29.9 mg KCl
39.10 mg K 
From what we have used already, let's calculate the mass of Na+ added.
309.5 mg sodium lactate = 246.0 mg lactate + 63.5 mg Na+
Thus, we need to add an additional 236.5 mg Na+ to get the desired 300. mg.
404
CHAPTER 11
236.5 mg Na+ ×
PROPERTIES OF SOLUTIONS
58.44 mg NaCl
= 601.2 mg NaCl
22.99 mg Na 
Now, let's check the mass of Cl added:
20.0 mg CaCl22H2O ×
70.90 mg Cl 
= 9.6 mg Cl
147 .0 mg CaCl2  2H 2 O
20.0 mg CaCl22H2O = 9.6 mg Cl
29.9 mg KCl  15.7 mg K+ = 14.2 mg Cl
601.2 mg NaCl  236.5 mg Na+ = 364.7 mg Cl
_______________________________________
Total Cl = 388.5 mg Cl
This is the quantity of Cl we want (the average amount of Cl).
An analytical balance can weigh to the nearest 0.1 mg. We would use 309.5 mg sodium
lactate, 20.0 mg CaCl22H2O, 29.9 mg KCl and 601.2 mg NaCl.
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:
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
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

231 mg C 3 H 5 O 3
1 mmol

= 0.0259 M (Note: molarity = mol/L = mmol/mL.)
100 . mL
89.07 mg
Total = 0.124 + 0.00361 + 0.0012 + 0.104 + 0.0259 = 0.259 M
π = MRT =
0.259 mol 0.08206 L atm
× 310. K = 6.59 atm

L
mol K
Similarly at maximum concentrations, the concentration of each ion is:
Na+: 0.137 M; K+: 0.00442 M; Ca2+: 0.0015 M; Cl: 0.115 M; C3H5O3: 0.0293 M
The total concentration of all ions is the sum, 0.287 M.
π=
0.287 mol 0.08206 L atm
× 310. K = 7.30 atm

L
mol K
Osmotic pressure ranges from 6.59 atm to 7.30 atm.
99.
a. Assuming MgCO3(s) does not dissociate, the solute concentration in water is:
CHAPTER 11
PROPERTIES OF SOLUTIONS
560 μg MgCO 3 (s) 560 mg 560  10 3 g 1 mol MgCO 3



= 6.6 × 10 3 mol MgCO3/L
mL
L
L
84.32 g
An applied pressure of 8.0 atm will purify water up to a solute concentration of:
π
8.0 atm
0.32 mol
M=


RT 0.08206 L atm  300 . K
L
mol K
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.
When V + 45 L of water have been processed, the moles of solute particles will equal:
6.6 × 10 3 mol/L × (45 L + V) = 0.32 mol/L × V
Solving: 0.30 = (0.32  0.0066) × V, V = 0.96 L
The minimum total volume of water that must be processed is 45 L + 0.96 L = 46 L.
Note: If MgCO3 does dissociate into Mg2+ and CO32 ions, the solute concentration will
increase to 1.3 × 10 2 M and at least 47 L of water must be processed.
b. No; A reverse osmosis system that applies 8.0 atm can only purify water with a solute
concentration less than 0.32 mol/L. Salt water has a solute concentration of 2(0.60 M) =
1.20 M ions. The solute concentration of salt water is much too high for this reverse osmosis
unit to work.
Integrative Problems
100.
10.0 mL 
1L
10 dL
1.0 mg
1g
1 mol
= 8.8 × 10 7 mol C4H7N3O




1000 mL
1L
1 dL
1000 mg 113.14 g
mass of blood = 10.0 mL 
molality =
1.025 g
= 10.3 g
mL
8.8  10 7 mol
= 8.5 × 10 5 mol/kg
0.0103 kg
 = MRT, M =
8.8  10 7 mol
= 8.8 × 10 5 mol/L
0.0100 L
 = 8.8 × 10 5 mol/L 
0.08206 L atm
× 298 K = 2.2 × 10 3 atm
K mol
405
406
101.
CHAPTER 11
T = imKf, i =
PROPERTIES OF SOLUTIONS
2.79 o C
ΔT
=
= 3.00
mK f
0.250 mol 1.86 o C kg

0.500 kg
mol
We have 3 ions in solutions and we have twice as many anions as cations. Therefore, the formula
of Q is MCl2. Assuming 100.00 g compound:
38.68 g Cl 
1 mol Cl
= 1.091 mol Cl
35.45 g
mol M = 1.091 mol Cl 
molar mass of M =
102.
1 mol M
= 0.5455 mol M
2 mol Cl
61.32 g M
= 112.4 g/mol; M is Cd so Q = CdCl2.
0.5455 mol M
mass of H2O = 160. mL 
0.995 g
= 159 g = 0.159 kg
mL
mol NaDTZ = 0.159 kg 
0.378 mol
= 0.0601 mol
kg
molar mass of NaDTZ =
38.4 g
= 639 g/mol
0.0601 mol
Psoln = χ H2O PHo 2O ; mol H2O = 159 g 
1 mol
= 8.82 mol
18.02 g
Sodium diatrizoate is a salt because there is a metal (sodium) in the compound. From the shorthand 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 particles.
χ H 2O =
8.82 mol
= 0.987; Psoln = 0.987 × 34.1 torr = 33.7 torr
0.120 mol  8.82 mol
Marathon Problem
103.
a. From part a information, we can calculate the molar mass of NanA and deduce the formula.
mol NanA = mol reducing agent = 0.01526 L 
molar mass of NanA =
0.02313 mol
= 3.530 × 10 4 mol NanA
L
30.0  10 3 g
= 85.0 g/mol
3.530  10  4 mol
To deduce the formula, we will assume various charges and numbers of oxygens present in
the oxyanion, then use the periodic table to see if an element fits the molar mass data.
Assuming n = 1 so the formula is NaA. The molar mass of the oxyanion, A, is 85.0  23.0 =
62.0 g/mol. The oxyanion part of the formula could be EO or EO2 or EO3 where E is some
CHAPTER 11
PROPERTIES OF SOLUTIONS
407
element. If EO, then the molar mass of E is 62.0  16.0 = 46.0 g/mol; no element has this
molar mass. If EO2, molar mass of E = 62.0  32.0 = 30.0 g/mol. Phosphorus is close, but
PO2 anions are not common. If EO3, molar mass of E = 62.0  48.0 = 14.0. Nitrogen has this
molar mass and NO3 anions are very common. Therefore, NO3 is a possible formula for A.
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 allows determination of the metal, M, in the formula. See Exercise
10.49 for a review of relationships in body-centered cubic cells. In a bcc unit cell, there are 2
atoms per unit cell and the body diagonal of the cubic cell is related to the radius of the metal
by the equation 4r = 3 ℓ where ℓ = cubic edge length.
ℓ=
4r
3

4(1.984  10 8 cm)
3
= 4.582 × 10 8 cm
volume of unit cell = ℓ 3 = (4.582 × 10 8 )3 = 9.620 × 10 23 cm3
mass of M in a unit cell = 9.620 × 10 23 cm3 ×
mol M in a unit cell = 2 atoms ×
molar mass of M =
5.243 g
cm
3
= 5.044 × 10 22 g M
1 mol
= 3.321 × 10 24 mol M
23
6.022  10 atoms
5.044  10 22 g M
3.321  10  24 mol M
= 151.9 g/mol
From the periodic table, M is europium, Eu. Given the charge of Eu is +3, then the formula of
the salt is Eu(NO3)3zH2O.
c. Part c data allows 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 
π

π = iMRT, iM =
=
= 0.0300 mol/L
0.08206 L atm
RT
 298 K
mol 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 mol of
Eu(NO3)3zH2O dissolves to form 4 ions (Eu3+ + 3 NO3), the molarity of Eu(NO3)3zH2O
is 0.0300/4 = 0.00750 M.
mol Eu(NO3)3zH2O = 0.01000 L ×
0.00750 mol
= 7.50 × 10 5 mol
L
408
CHAPTER 11
molar mass of Eu(NO3)3zH2O =
PROPERTIES OF SOLUTIONS
33.45  10 3 g
= 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.