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Silberberg7e solution manual ch 13

CHAPTER 13 THE PROPERTIES OF
MIXTURES: SOLUTIONS AND COLLOIDS
FOLLOW–UP PROBLEMS
13.1A

Plan: Compare the intermolecular forces in the solutes with the intermolecular forces in the solvent. The
intermolecular forces for the more soluble solute will be more similar to the intermolecular forces in the solvent
than the forces in the less soluble solute.
Solution:
a) 1,4–Butanediol is more soluble in water than butanol. Intermolecular forces in water are primarily hydrogen
bonding. The intermolecular forces in both solutes also involve hydrogen bonding with the hydroxyl groups.
Compared to butanol, each 1,4–butanediol molecule will form more hydrogen bonds with water because 1,4–
butanediol contains two hydroxyl groups in each molecule, whereas butanol contains only one –OH group. Since
1,4–butanediol has more hydrogen bonds to water than butanol, it will be more soluble than butanol.
b) Chloroform is more soluble in water than carbon tetrachloride because chloroform is a polar molecule and
carbon tetrachloride is nonpolar. Polar molecules are more soluble in water, a polar solvent.

13.1B

Plan: Compare the intermolecular forces in the solutes with the intermolecular forces in the solvent. The
intermolecular forces for the solvent that dissolves more solute will be more similar to the intermolecular forces in

the solute than the forces in the solvent that dissolves less solute.
Solution:
a) Both chloroform and chloromethane are polar molecules that experience dipole-dipole and dispersion
intermolecular forces. Methanol, on the other hand, has an O–H bond and, thus, can participate in hydrogen
bonding in addition to dipole-dipole and dispersion forces. Chloroform dissolves more chloromethane than
methanol because of similar dipole-dipole forces.
b) Pentanol has a polar O–H group that can participate in hydrogen bonding, dipole-dipole and dispersion forces.
However, it has a much larger non-polar section (CH3CH2CH2CH2-) that experiences only dispersion forces.
Because the nonpolar portion of the pentanol is much larger than the polar portion, the nonpolar portion
determines the overall solubility of the molecule. Thus, pentanol will be more soluble in nonpolar solvents like
hexane than polar solvents like water. Hexane dissolves more pentanol due to dispersion forces.

13.2A

Plan: Use the relationship Hsolution = Hlattice +Hhydration of the ions. Given Hsolution and Hlattice, Hhydration of the ions
can be calculated.
Solution:
The two ions in potassium nitrate are K+ and NO3–. Hhydration of the ions = Hhydration(K+) + Hhydration(NO3–)
Hsolution = Hlattice +Hhydration of the ions
Hhydration of the ions = Hsolution – Hlattice = 34.89 kJ/mol – 685 kJ/mol = –650.11 = –650. kJ/mol

13.2B

Plan: Use the relationship Hsolution = Hlattice +Hhydration of the ions. Given Hsolution and Hlattice, Hhydration of the ions
can be calculated. Hhydration of the ions and Hhydration (Na+) can then be used to solve for Hhydration(CN–).
Solution:
The two ions in sodium cyanide are Na+ and CN–.
Hsolution = Hlattice +Hhydration of the ions
Hhydration of the ions = Hsolution – Hlattice = 1.21 kJ/mol – 766 kJ/mol = –764.79 = –765 kJ/mol
Hhydration of the ions = Hhydration(Na+) + Hhydration(CN–)
Hhydration(CN–) = Hhydration of the ions – Hhydration(Na+)= –765 kJ/mol – (–410. kJ/mol) = –355 kJ/mol

13.3A

Plan: Solubility of a gas can be found from Henry’s law: Sgas = kH  Pgas. The problem gives kH for N2 but not its
partial pressure. To calculate the partial pressure, use the relationship from Chapter 5: Pgas = Xgas x Ptotal where X
represents the mole fraction of the gas.

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13-1


Solution:
To find partial pressure use the 78% N2 given for the mole fraction:
Pgas = Xgas x Ptotal
PN 2 = 0.78 x 1 atm = 0.78 atm
Use Henry’s law to find solubility at this partial pressure:
Sgas = kH  Pgas
S N 2 = (7x10–4 mol/L•atm)(0.78 atm) = 5.46x10–4 mol/L = 5x10–4 mol/L
13.3B

Plan: Solubility of a gas can be found from Henry’s law: Sgas = kH  Pgas. The problem gives kH for N2O but not its
partial pressure. To calculate the partial pressure, use the relationship from Chapter 5: Pgas = Xgas x Ptotal where X
represents the mole fraction of the gas.
Solution:
To find partial pressure use the 40.% N2O given for the mole fraction:
Pgas = Xgas x Ptotal
PN 2 = 0.40 x 1.2 atm = 0.48 atm
Use Henry’s law to find solubility at this partial pressure:
Sgas = kH  Pgas
S (N2O) = (2.5x10–2 mol/L•atm)(0.48 atm) = 0.012 mol/L

13.4A

Plan: Molality (m) is defined as amount (mol) of solute per kg of solvent. Use the molality and the mass of solvent
given to calculate the amount of glucose in moles. Then convert amount (mol) of glucose to mass (g) of glucose
by multiplying by its molar mass.
Solution:
Mass of solvent must be converted to kg.
 1 kg 
Mass of solvent (kg) =  563 g   3  = 0.563 kg
 10 g 


 2.40 x 102 mol C6 H12 O 6
Mass (g) of glucose =  0.563 kg solvent  

1 kg solvent

= 2.4343 = 2.43 g C6H12O6

13.4B

Plan: Molality (m) is defined as amount (mol) of solute per kg of solvent. Calculate the moles of solute, I2, from
the mass. Find the mass (kg) of the solvent, diethyl ether, from the given number of moles. Then divide the moles
of solute by the kg of solvent to find the molality.
Solution:
Amount (mol) of solute = (15.20 g I2)

1 mol I2

Molality (m) =

mol solute
kg solvent

=

0.05989 mol
0.0986 kg

= 0.05989 mol I2

253.8 g I2
74.12 g (CH3 CH2 )2 O

Mass of solvent (kg) = (1.33 mol (CH3CH2)2O

13.5A

  180.16 C6 H12 O 6 
 

  1 mol C6 H12 O 6 

1 mol (CH3 CH2 )2 O

1 kg
1000 g

= 0.0986 kg (CH3CH2)2O

= 0.607 m

Plan: Mass percent is the mass (g) of solute per 100 g of solution. For each alcohol, divide the mass of the alcohol
by the total mass. Multiply this number by 100 to obtain mass percent. To find mole fraction, first find the amount
(mol) of each alcohol, then divide by the total moles.
Solution:
Mass percent:
35.0 g
mass of propanol
x 100%
Mass % propanol =
x 100% =
mass of propanol + mass of ethanol
 35.0  150. g
= 18.9189 = 18.9% propanol

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13-2


Mass % of ethanol =

150. g
mass of ethanol
x 100%
x 100% =
mass of propanol + mass of ethanol
 35.0  150. g

= 81.08108 = 81.1% ethanol
Mole fraction:

 1 mol C3 H7 OH 
Moles of propanol =  35.0 g C3H7 OH  
 = 0.5824596 mol propanol
 60.09 g C3 H7 OH 

 1 mol C2 H5 OH 
Moles of ethanol = 150. g C2 H5 OH  
 = 3.2559149 mol ethanol
 46.07 g C2 H5OH 
0.5824596 mol propanol
Xpropanol =
= 0.151746 = 0.152
0.5824596 mol propanol + 3.2559149 mol ethanol
Xethanol =
13.5B

3.2559149 mol ethanol
= 0.84825 = 0.848
0.5824596 mol propanol + 3.2559149 mol ethanol

Plan: Mass percent is the mass (g) of solute per 100 g of solution. For each component of the mixture, divide the
mass of the component by the total mass. Multiply this number by 100 to obtain mass percent. To find mole
percent, first find the amount (mol) of each component, then divide by the total moles and multiply by 100%.
Solution:
Mass percent:
Mass % ethanol =

mass of ethanol
mass of ethanol+mass of iso-octane + mass of heptane
1.87 g
1.87 g+27.4 g + 4.10 g

Mass % iso-octane =

x 100% = 5.6038 = 5.60% ethanol

mass of iso-octane
mass of ethanol+mass of iso-octane + mass of heptane
27.4 g
1.87 g+27.4 g + 4.10 g

Mass % heptane =

x 100% =

x 100% = 82.1097 = 82.1% iso-octane

mass of heptane
mass of ethanol+mass of iso-octane + mass of heptane
4.10 g
1.87 g+27.4 g + 4.10 g

x 100% =

x 100% =

x 100% = 12.2865 = 12.3% heptane

Mole percent:
1 mole ethanol

Moles of ethanol = (1.87 g ethanol)

46.07 g ethanol

Moles of iso-octane = (27.4 g iso-octane)
Moles of heptane = (4.10 g heptane)
Mole percent ethanol =

= 0.0406 mol ethanol

1 mole iso-octane
114.22 g iso-octane

1 mole heptane
100.20gheptane

= 0.240 mol iso-octane

= 0.0409 mol heptane

moles of ethanol
moles of ethanol+moles of iso-octane + moles of heptane
0.0406 mol
0.0406mol+0.240 mol + 0.0409 mol

Mole percent iso-octane =

x 100% =

x 100% = 12.6283 = 12.6% ethanol

moles of iso-octane
moles of ethanol+moles of iso-octane + moles of heptane

x 100% =

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13-3


0.240 mol
0.0406mol+0.240 mol + 0.0409 mol

Mole percent heptane =

x 100% = 74.6501 = 74.6% iso-octane

moles of heptane
moles of ethanol+moles of iso-octane + moles of heptane
0.0409 mol
0.0406mol+0.240 mol + 0.0409 mol

x 100% =

x 100% = 12.7216 = 12.7% heptane

(Slight differences from 100% are due to rounding.)
13.6A

Plan: To find the mass percent, molality and mole fraction of HCl, the following is needed:
1) Moles of HCl in 1 L solution (from molarity)
2) Mass of HCl in 1L solution (from molarity times molar mass of HCl)
3) Mass of 1L solution (from volume times density)
4) Mass of solvent in 1L solution (by subtracting mass of solute from mass of solution)
5) Moles of solvent (by dividing the mass of solvent by molar mass of water)
Mass percent is calculated by dividing mass of HCl by mass of solution and multiplying by 100.
Molality is calculated by dividing moles of HCl by mass of solvent in kg.
Mole fraction is calculated by dividing mol of HCl by the sum of mol of HCl plus mol of solvent.
Solution:
Assume the volume is exactly 1 L.
 11.8 mol HCl 
1) Mole of HCl in 1 L solution = 1.0 L  
 = 11.8 mol HCl
1.0 L



 36.46 g HCl 
2) Mass (g) of HCl in 1 L solution = 11.8 mol HCl  
 = 430.228 g HCl
 1 mol HCl 
 1 mL  1.190 g 
3) Mass (g) of 1 L solution = 1.0 L   3 
 = 1190. g solution
 10 L  1 mL 
4) Mass (g) of solvent in 1 L solution = 1190. g – 430.228 g = 759.772 g solvent (H2O)
 1 kg 
Mass (kg) of solvent in 1 L solution =  759.772 g   3  = 0.759772 kg solvent
 10 g 



 1 mol H 2 O 
5) Moles of solvent in 1 L solution =  759.772 g H 2 O  
 = 42.1627 mol solvent
 18.02 g H 2 O 
mass HCl
430.228 g HCl
Mass percent of HCl =
100  =
100  = 36.1536 = 36.2%
mass solution
1190 g solution
Molality of HCl =

mole HCl
11.8 mol HCl
=
= 15.530975 = 15.5 m
kg solvent
0.759772 kg

Mole fraction of HCl =
13.6B

11.8 mol
mol HCl
=
= 0.21866956 = 0.219
11.8 mol + 42.1627 mol
mol HCl + mol H 2 O

Plan: To find the molality, molarity and mole percent of CaBr2, the following is needed:
1) Mass of CaBr2 and mass of H2O in 100 g of CaBr2
2) Moles of CaBr2 and mass of H2O in 100 g of CaBr2
3) Volume of 100 g of solution
Molality is calculated by dividing moles of CaBr2 by mass of solvent in kg.
Molarity is calculated by dividing the moles of CaBr2 by the volume of the solution (L).
Mole percent is calculated by dividing moles of CaBr2 by the total moles in the solution and multiplying by 100.
Solution:
Assume you have 100 g of solution.
1) Mass of CaBr2 in 100 g solution = (100 g solution)

52.1 g CaBr2
100 g solution

= 52.1 g CaBr2

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13-4


Mass of H2O in 100 g solution = 100 g – 52.1 g = 47.9 g H2O
1 mol CaBr2

2) Moles of CaBr2 in 100 g solution = (52.1 g CaBr2)
Moles of H2O in 100 g solution = (47.9 g H2O)

199.88 g CaBr2
1 mol H2 O

Molality =
Molarity =

moles solute
kg solvent
moles solute
L solution

Mole percent =
13.7A

0.261 mol CaBr2

1000 g

47.9 g H2 O

1 kg

=

0.261 mol CaBr2

molessolute moles solvent

1L

1.70 g

=

0.0588 L
moles solute

= 2.66 mol H2O

18.02 g H2 O
1 mL

3) Volume (L) of 100 g solution = (100 g solution)

= 0.261 mol CaBr2

1000 mL

= 0.0588 L solution

= 5.45 m

= 4.44 M
x 100%

=

0.261 mol
0.261 mol + 2.66 mol

x 100% = 8.94%

Plan: Raoult’s law states that the vapor pressure of a solvent is proportional to the mole fraction of the solvent:
Psolvent = Xsolvent  P°solvent. To calculate the drop in vapor pressure, a similar relationship is used with the mole
fraction of the solute substituted for that of the solvent.
Solution:
Mole fraction of aspirin in methanol:
 1 mol aspirin 
 2.00 g  

 180.15 g aspirin 

 1 mol aspirin 
 1 mol methanol 
   50.0 g  

 180.15 g aspirin 
 32.04 g methanol 

 2.00 g  

Xaspirin = 7.06381x10–3
P = Xaspirin P°methanol = (7.06381x10–3) (101 torr) = 0.71344 = 0.713 torr
13.7B

Plan: The drop in vapor pressure of a solvent is calculated by the following equation derived from Raoult’s law:
ΔP = Xsolute  P°solvent.
Solution:
Amount (mol) menthol = (6.49 g menthol)
Amount (mol) ethanol = (25.0 g ethanol)
ΔP =

13.8A

0.0415 mol
0.0415 mol+0.543 mol

(6.87 kPa)

1 mol menthol
156.26 g menthol
1 mol ethanol
46.07 g ethanol

= 0.0415 mol menthol

= 0.543 mol ethanol

1 atm

760 torr

101.325 kPa

1 atm

= 3.66 torr

Plan: Find the molality of the solution by dividing the moles of P4 by the mass of the CS2 in kg. The change in
freezing point is calculated from Tf = iKfm, where Kf is 3.83°C/m when CS2 is the solvent, i is the van’t Hoff
factor, and m is the molality of particles in solution. Since P4 is a covalent compound and does not ionize in water,
i = 1. Once Tf is calculated, the freezing point is determined by subtracting it from the freezing point of pure CS2
(–111.5°C). The boiling point of a solution is increased relative to the pure solvent by the relationship Tb = iKbm
where Kb is 2.34°C/m when CS2 is the solvent, i is the van’t Hoff factor, and m is the molality of particles in
solution. P4 is a nonelectrolyte (it is a molecular compound) so i = 1. Once Tb is calculated, the boiling point is
determined by adding it to the boiling point of pure CS2 (46.2°C).
Solution:
Molality of P4 solution:
Moles of P4 = (8.44 g P4)
Molality of P4 =

1 mol P4

= 0.0681 mol

123.88 g P4
moles solute
0.0681 mol P4
kg solvent

=

60.0 g CS2

1000 g
1 kg

= 1.14 m P4

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13-5


Freezing point:
Tf = iKfm = (1)(3.83°C/m)(1.14 m) = 4.37°C
The freezing point is: –111.5°C – 4.37°C = –115.9°C.
Boiling point:
Tb = iKbm = (1)(2.34°C/m)(1.14 m) = 2.67°C
The boiling point is 46.2°C + 2.67°C = 48.9°C.
13.8B

Plan: The question asks for the concentration of ethylene glycol that would prevent freezing at 0.00°F. First,
convert 0.00°F to °C. The change in freezing point of water will then be this temperature subtracted from the
freezing point of pure water, 0.00°C. Use this value for T in Tf = 1.86°C/m  molality of solution and solve
for molality.
Solution:
Temperature conversion:
 5C 
 5C 
T (°C) = T  F  32F  
=  0.00F  32F  

 = –17.7778°C
 9F 
 9F 
Tf = m Kf

17.7778 C
Tf
=
= 9.557956 = 9.56 m
Kf
1.86 C/m
The minimum concentration of ethylene glycol would have to be 9.56 m in order to prevent the water from
freezing at 0.00°F.

m =

13.9A

Plan: Use the osmotic pressure, the temperature, and the gas constant to calculate the molarity of the solution with
the equation  = MRT. Multiply the molarity of the solution by the volume of the sample to calculate the number
of moles in the sample. The molar mass is calculated by dividing the mass of the sample by the number of moles
in the sample.
Solution:
 = MRT , so M =
M=

Π

RT
1 atm
8.98 torr x 760 torr

0.0821

atm • L
mol •K

27°C +273.15 K

= 4.79x10–4 M

Amount (mol) protein = (4.79x10–4 mol/L) (12.0 mL) (1 L/1000 mL) = 5.75x10–6 mol
M=
13.9B

0.200 g
5.75x10

–6

mol

= 3.48x104 g/mol

Plan: The osmotic pressure is calculated as the product of molarity, temperature, and the gas constant.
Solution:
L • atm 
 0.30 mol 
 = MRT = 
 0.0821 mol • K    273  37  K  = 7.6353 = 7.6 atm
L




13.10A Plan: Write the formula of potassium phosphate, determine if the compound is soluble and, if soluble, how many
cations and anions result when one unit of the compound is placed in water. Calculate the molality of the solution.
The boiling point of a solution is increased relative to the pure solvent by the relationship Tb = iKbm where Kb is
0.512°C/m for water, i is the van’t Hoff factor (equal to the number of particles or ions produced when one unit of
the compound is dissolved in the solvent), and m is the molality of particles in solution. Once Tb is calculated,
the boiling point is determined by adding it to the boiling point of pure H2O (100.00°C).
Solution:
a) The formula for potassium phosphate is K3PO4. When it is placed in water, 3 K+ ions and 1 PO43– ion are
formed. Scene B is the only scene that shows separate ions in a 3K+/1PO43– ratio.

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13-6


b)
Molality of K3PO4 solution:
Moles of K3PO4 = (31.2 g K3PO4)
Molality of K3PO4 =

moles solute
kg solvent

=

1 mol K3 PO4
212.27 g K3 PO4

= 0.147 mol K3PO4

0.147 mol K3 PO4

1000 g

85.0 g H2 O

1 kg

= 1.73 m K3PO4

Because 4 ions total are formed when each unit of K3PO4 is placed in water, i = 4.
Tb = iKbm = (4)(0.512 °C/m)(1.73 m) = 3.54°C
The boiling point is 100.00°C + 3.54°C = 103.54°C.
13.10B Plan: Calculate the molarity of the solution, then calculate the osmotic pressure of the magnesium chloride
solution. Do not forget that MgCl2 is a strong electrolyte, and ionizes to yield three ions per formula unit. This will
result in a pressure three times as great as a nonelectrolyte (glucose) solution of equal concentration.
Solution:
 1 mL   103 L 
a) Volume (L) of solution = 100. + 0.952 g solution  
 = 0.1003499 L
 
 1.006 g   1 mL 

 1 mol MgCl2 
–3
Moles of MgCl2 =  0.952 g MgCl2  
 = 9.9989x10 mol MgCl2
95.21
g
MgCl
2 

mol MgCl2
0.0099989 mol
Molarity of MgCl2 =
=
= 0.099640 M MgCl2
0.1003499 L
volume solution
L • atm 
 0.099640 mol 
 = iMRT = 3 
 0.0821 mol • K    273.2  20.0  K  = 7.1955 = 7.20 atm
L



b) The magnesium chloride solution has a higher osmotic pressure, thus, solvent will diffuse from the glucose
solution into the magnesium chloride solution. The glucose side will lower and the magnesium chloride side will
rise. Scene C represents this situation.

CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B13.1

a) The colloidal particles in water generally have negatively charged surfaces and so repel each other, slowing the
settling process. Cake alum, Al2(SO4)3, is added to coagulate the colloids. The Al3+ ions neutralize the negative
surface charges and allow the particles to aggregate and settle.
b) Water that contains large amounts of divalent cations (such as Ca2+ and Mg2+) is called hard water. During
cleaning, these ions combine with the fatty-acid anions in soaps to produce insoluble deposits.
c) In reverse osmosis, a pressure greater than the osmotic pressure is applied to the solution, forcing the water
back through the membrane and leaving the ions behind.
d) Chlorine may give the water an unpleasant odor, and can form carcinogenic chlorinated compounds.
e) The high concentration of NaCl displaces the divalent and polyvalent ions from the ion-exchange resin.

B13.2

Plan: Osmotic pressure is calculated from the molarity of particles, the gas constant, and temperature. Convert the
mass of sucrose to moles using the molar mass, and then to molarity. Sucrose is a nonelectrolyte so i = 1.
Solution:
T = 273 + 20.°C = 293 K
 1 mol sucrose 
Moles of sucrose =  3.55 g sucrose  
 = 0.01037102 mol sucrose
 342.30 g sucrose 
moles of sucrose
0.01037102 mol
Molarity =
=
= 1.037102x10–2 M sucrose
volume of solution
1.0 L
 = iMRT = (1)(1.037102x10–2 mol/L)(0.0821 L•atm/mol•K)(293 K) = 0.2494780 = 0.249 atm
A pressure greater than 0.249 atm must be applied to obtain pure water from a 3.55 g/L solution.

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13-7


END–OF–CHAPTER PROBLEMS
13.1

The composition of seawater, like all mixtures, is variable. The components of seawater (water and various
ions) have not been changed and thus retain some of their properties. For example, seawater has a salty taste due
to the presence of salts such as NaCl.

13.2

When a salt such as NaCl dissolves, ion-dipole forces cause the ions to separate, and many water molecules cluster
around each of them in hydration shells. Ion-dipole forces hold the first shell. Additional shells are held by
hydrogen bonding to inner shells.

13.3

In CH3(CH2)nCOOH, as n increases, the hydrophobic (CH) portion of the carboxylic acid increases and the
hydrophilic part of the molecule stays the same, with a resulting decrease in water solubility.

13.4

Sodium stearate would be a more effective soap because the hydrocarbon chain in the stearate ion is longer than
the chain in the acetate ion. A soap forms suspended particles called micelles with the polar end of the soap
interacting with the water solvent molecules and the nonpolar ends forming a nonpolar environment inside the
micelle. Oils dissolve in the nonpolar portion of the micelle. Thus, a better solvent for the oils in dirt is a more
nonpolar substance. The long hydrocarbon chain in the stearate ion is better at dissolving oils in the micelle than
the shorter hydrocarbon chain in the acetate ion.

13.5

Hexane and methanol, as gases, are free from any intermolecular forces and can simply intermix with each other.
As liquids, hexane is a nonpolar molecule, whereas methanol is a polar molecule. “Like dissolves like.”

13.6

Hydrogen chloride (HCl) gas is actually reacting with the solvent (water) and thus shows a higher solubility than
propane (C3H8) gas, which does not react.

13.7

Plan: A more concentrated solution will have more solute dissolved in the solvent. Determine the types of
intermolecular forces in the solute and solvents. A solute tends to be more soluble in a solvent whose
intermolecular forces are similar to its own.
Solution:
Potassium nitrate, KNO3, is an ionic compound and can form ion-dipole forces with a polar solvent like water,
thus dissolving in the water. Potassium nitrate is not soluble in the nonpolar solvent CCl4. Because potassium
nitrate dissolves to a greater extent in water, a) KNO3 in H2O will result in the more concentrated solution.

13.8

b) Stearic acid in CCl4. Stearic acid will not dissolve in water. It is nonpolar while water is very polar. Stearic
acid will dissolve in carbon tetrachloride, as both are nonpolar.

13.9

Plan: To identify the strongest type of intermolecular force, check the formula of the solute and identify the forces
that could occur. Then look at the formula for the solvent and determine if the forces identified for the solute
would occur with the solvent. Ionic forces are present in ionic compounds; dipole-dipole forces are present in
polar substances, while nonpolar substances exhibit only dispersion forces. The strongest force is ion-dipole
followed by dipole-dipole (including H bonds). Next in strength is ion–induced dipole force and then dipole–
induced dipole force. The weakest intermolecular interactions are dispersion forces.
Solution:
a) Ion-dipole forces are the strongest intermolecular forces in the solution of the ionic substance cesium chloride
in polar water.
b) Hydrogen bonding (type of dipole-dipole force) is the strongest intermolecular force in the solution of polar
propanone (or acetone) in polar water.
c) Dipole–induced dipole forces are the strongest forces between the polar methanol and nonpolar carbon
tetrachloride.

13.10

a) metallic bonding
b) dipole-dipole
c) dipole–induced dipole

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13-8


13.11

Plan: To identify the strongest type of intermolecular force, check the formula of the solute and identify the forces
that could occur. Then look at the formula for the solvent and determine if the forces identified for the solute
would occur with the solvent. Ionic forces are present in ionic compounds; dipole-dipole forces are present in
polar substances, while nonpolar substances exhibit only dispersion forces. The strongest force is ion-dipole
followed by dipole-dipole (including H bonds). Next in strength is ion–induced dipole force and then dipole–
induced dipole force. The weakest intermolecular interactions are dispersion forces.
Solution:
a) Hydrogen bonding occurs between the H atom on water and the lone electron pair on the O atom in dimethyl
ether (CH3OCH3). However, none of the hydrogen atoms on dimethyl ether participates in hydrogen bonding
because the CH bond does not have sufficient polarity.
b) The dipole in water induces a dipole on the Ne(g) atom, so dipole–induced dipole interactions are the
strongest intermolecular forces in this solution.
c) Nitrogen gas and butane are both nonpolar substances, so dispersion forces are the principal attractive forces.

13.12

a) dispersion forces
b) hydrogen bonding
c) dispersion forces

13.13

Plan: CH3CH2OCH2CH3 is polar with dipole-dipole interactions as the dominant intermolecular forces. Examine
the solutes to determine which has intermolecular forces more similar to those in diethyl ether. This solute is the
one that would be more soluble.
Solution:
a) HCl would be more soluble since it is a covalent compound with dipole-dipole forces, whereas NaCl is an ionic
solid. Dipole-dipole forces between HCl and diethyl ether are more similar to the dipole forces in diethyl ether
than the ion-dipole forces between NaCl and diethyl ether.
b) CH3CHO (acetaldehyde) would be more soluble. The dominant interactions in H2O are hydrogen bonding, a
stronger type of dipole-dipole force. The dominant interactions in CH3CHO are dipole-dipole. The solute-solvent
interactions between CH3CHO and diethyl ether are more similar to the solvent intermolecular forces than the
forces between H2O and diethyl ether.
c) CH3CH2MgBr would be more soluble. CH3CH2MgBr has a polar end (–MgBr) and a nonpolar end
(CH3CH2–), whereas MgBr2 is an ionic compound. The nonpolar end of CH3CH2MgBr and diethyl ether would
interact with dispersion forces, while the polar end of CH3CH2MgBr and the dipole in diethyl ether would interact
with dipole-dipole forces.

13.14

a) CH3CH2-O-CH3(g), due to its smaller size (smaller molar mass).
b) CH2Cl2, because it is more polar than CCl4.
c) Tetrahydropyran is more water soluble due to hydrogen bonding between the oxygen atom and water
molecules.

13.15

No, river water is a heterogeneous mixture, with its composition changing from one segment to another.

13.16

Plan: Determine the types of intermolecular forces present in the two compounds and in water and hexane.
Substances with similar types of forces tend to be soluble while substances with different type of forces tend to be
insoluble.
Solution:
Gluconic acid is a very polar molecule because it has –OH groups attached to every carbon. The abundance of
–OH bonds allows gluconic acid to participate in extensive H bonding with water, hence its great solubility in
water. On the other hand, caproic acid has a five carbon, nonpolar, hydrophobic (“water hating”) tail that does not
easily dissolve in water. The dispersion forces in the nonpolar tail are more similar to the dispersion forces in
hexane, hence its greater solubility in hexane.

13.17

There may be a disulfide linkage (a covalent disulfide bridge) between the ends of two cysteine side chains
that bring together parts of the chain. There may be salt links between ions –COO– and –NH3+ groups. There may
be hydrogen bonding between the C=O of one peptide bond and the N–H of another.

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13-9


13.18

The nitrogen bases hydrogen bond to their complimentary bases. The flat, N-containing bases stack above each
other, which allow extensive interaction through dispersion forces. The exterior negatively charged sugarphosphate chains form ion-dipole and hydrogen bonds to the aqueous surroundings, but this is of minor
importance to the structure.

13.19

While an individual hydrogen bond is not too strong, there are very large numbers of hydrogen bonds present in
DNA. The energy of so many hydrogen bonds keeps the chains together. But the hydrogen bonds are weak
enough that a few are easily broken when the two chains in DNA must separate.

13.20

The more carbon and hydrogen atoms present, the more soluble the substance is in nonpolar oil droplets.
Therefore, sodium propanoate is not as effective a soap as sodium stearate with the longer hydrocarbon chain.

13.21

Dispersion forces are present between the nonpolar portions of the molecules within the bilayer. Polar groups are
present to hydrogen bond or to form ion-dipole interactions with the aqueous surroundings.

13.22

In soluble proteins, polar groups are found on the exterior and nonpolar groups on the interior. In proteins
embedded in a membrane, the exterior of the protein that lies within the bilayer consists of nonpolar amino acid
side chains, whereas the portion lying outside the bilayer has polar side chains.

13.23

Amino acids with side chains that may be ionic are necessary. Two examples are lysine and glutamic acid.

13.24

The Hsolvent and Hmix components of the heat of solution combined together represent the enthalpy change
during solvation, the process of surrounding a solute particle with solvent particles. When the solvent is water,
solvation is called hydration.

13.25

For a general solvent, the energy changes needed to separate solvent into particles (Hsolvent), and that needed to
mix the solvent and solute particles (Hmix) would be combined to obtain Hsolution.

13.26

a) Charge density is the ratio of the ion’s charge to its volume. An ion’s charge and size affect its charge density.
b) – < + < 2– < 3+
c) The higher the charge density, the more negative the Hhydration. Hhydration increases with increasing charge and
decreases with increasing size.

13.27

The solution cycle for ionic compounds in water consists of two enthalpy terms: the lattice energy, and the
combined heats of hydration of the cation and anion.
Hsolution = Hlattice +Hhydration of ions
For a heat of solution to be zero (or very small)
Hlattice  Hhydration of ions, and they would have to have opposite signs.

13.28

a) Endothermic
b) The lattice energy term is much larger than the combined ionic heats of hydration.
c) The increase in entropy outweighs the increase in enthalpy, so ammonium chloride dissolves.

13.29

This compound would be very soluble in water. A large exothermic value in Hsolution (enthalpy of solution)
means that the solution has a much lower energy state than the isolated solute and solvent particles, so the system
tends to the formation of the solution. Entropy that accompanies dissolution always favors solution formation.
Entropy becomes important when explaining why solids with endothermic Hsolution values (and higher energy
states) are still soluble in water.

13.30

Plan: Hsolution = Hlattice +Hhydration. Lattice energy values are always positive as energy is required to
separate the ions from each other. Hydration energy values are always negative as energy is released when
intermolecular forces between ions and water form. Since the heat of solution for KCl is endothermic, the lattice
energy must be greater than the hydration energy for an overall input of energy.

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13-10


Solution:
K+(g) + Cl(g)

Hlattice
Enthalpy

Hhydration
K+(aq) + Cl(aq)
Hsolution

KCl(s)

Hsolution > 0 (endothermic)
13.31

Lattice energy values are always positive as energy is required to separate the ions from each other.
Hydration energy values are always negative as energy is released when intermolecular forces between
ions and water form. Since the heat of solution for NaI is exothermic, the negative hydration energy must be
greater than the positive lattice energy.

Na+(g) + I (g)

Enthalpy

Hlattice
Hhydration

NaI(s)
Na+(aq) + I(aq)

Hsolution

13.32

Plan: Charge density is the ratio of an ion’s charge (regardless of sign) to its volume. An ion’s volume is
related to its radius. For ions whose charges have the same sign (+ or –), ion size decreases as a group in
the periodic table is ascended and as you proceed from left to right in the periodic table. Charge density increases
with increasing charge and increases with decreasing size.
Solution:
a) Both ions have a +1 charge, but the volume of Na+ is smaller, so it has the greater charge density.
b) Sr2+ has a greater ionic charge and a smaller size (because it has a greater Zeff), so it has the greater charge
density.
c) Na+ has a smaller ion volume than Cl–, so it has the greater charge density.
d) O2– has a greater ionic charge and similar ion volume, so it has the greater charge density.
e) OH– has a smaller ion volume than SH– (O is smaller than S), so it has the greater charge density.
f) Mg2+ has the higher charge density because it has a smaller ion volume.
g) Mg2+ has the higher charge density because it has both a smaller ion volume and greater charge.
h) CO32– has the higher charge density because it has both a smaller ion volume and greater charge.

13.33

a) I – has a smaller charge density (larger ion volume) than Br–.

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13-11


b) Ca2+ has a lower ratio than Sc3+, due to its smaller ion charge.
c) Br– has a lower ratio than K+, due to its larger ion volume.
d) Cl– has a lower ratio than S2–, due to its smaller ion charge.
e) Sc3+ has a lower ratio than Al3+, due to its larger ion volume.
f) ClO4– has a lower ratio due to its smaller ion charge.
g) Fe2+ has a lower ratio due to its smaller ion charge.
h) K+ has a lower ratio due to its smaller ion charge.
13.34

Plan: The ion with the greater charge density will have the larger Hhydration.
Solution:
a) Na+ would have a larger Hhydration than Cs+ since its charge density is greater than that of Cs+.
b) Sr2+ would have a larger Hhydration than Rb+.
c) Na+ would have a larger Hhydration than Cl–.
d) O2– would have a larger Hhydration than F–.
e) OH– would have a larger Hhydration than SH–.
f) Mg2+ would have a larger Hhydration than Ba2+.
g) Mg2+ would have a larger Hhydration than Na+.
h) CO32– would have a larger Hhydration than NO3–.

13.35

a) I–

13.36

Plan: Use the relationship Hsolution = Hlattice +Hhydration. Given Hsolution and Hlattice, Hhydration can be
calculated. Hhydration increases with increasing charge density, and charge density increases with increasing
charge and decreasing size.
Solution:
a) The two ions in potassium bromate are K+ and BrO3–.
Hsolution = Hlattice +Hhydration
Hhydration = Hsolution – Hlattice = 41.1 kJ/mol – 745 kJ/mol = –703.9 = –704 kJ/mol
b) K+ ion contributes more to the heat of hydration because it has a smaller size and, therefore, a greater charge
density.

13.37

a) Hhydration = Hsolution – Hlattice
Hhydration = 17.3 kJ/mol –763 kJ/mol
Hhydration = –745.7 = – 746 kJ/mol
b) Na+ ion contributes more to the heat of hydration due to its smaller size (larger charge density).

13.38

Plan: Entropy increases as the possible states for a system increase, which is related to the freedom of motion of
its particles and the number of ways they can be arranged.
Solution:
a) Entropy increases as the gasoline is burned. Gaseous products at a higher temperature form.
b) Entropy decreases as the gold is separated from the ore. Pure gold has only the arrangement of gold atoms next
to gold atoms, while the ore mixture has a greater number of possible arrangements among the components of the
mixture.
c) Entropy increases as a solute dissolves in the solvent.

13.39

a) Entropy increases.
b) Entropy decreases.
c) Entropy increases.

13.40

Hsolution = Hlattice + Hhydration
Hsolution = 822 kJ/mol + (– 799 kJ/mol)
Hsolution = 23 kJ/mol

b) Ca2+ c) Br–

d) Cl–

e) Sc3+

f) ClO4–

g) Fe2+

h) K+

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13-12


13.41

Add a pinch of the solid solute to each solution. A saturated solution contains the maximum amount of dissolved
solute at a particular temperature. When additional solute is added to this solution, it will remain undissolved. An
unsaturated solution contains less than the maximum amount of dissolved solute and so will dissolve added solute.
A supersaturated solution is unstable and addition of a “seed” crystal of solute causes the excess solute to
crystallize immediately, leaving behind a saturated solution.

13.42

KMnO4(s) + H2O(l) + heat → KMnO4(aq)
Prepare a mixture of more than 6.4 g KMnO4/100 g H2O and heat it until the solid completely dissolves. Then
carefully cool it, without disturbing it or shaking it, back to 20°C. If no crystals form, you would then have a
supersaturated solution.

13.43

An increase in temperature produces an increase in kinetic energy; the gaseous solute molecules overcome the
weak intermolecular forces, which results in a decrease in solubility of any gas in water. In nearly all cases, gases
dissolve exothermically (Hsoln < 0).

13.44

Plan: The solubility of a gas in water decreases with increasing temperature and increases with increasing
pressure.
Solution:
a) Increasing pressure for a gas increases the solubility of the gas according to Henry’s law.
b) Increasing the volume of a gas causes a decrease in its pressure (Boyle’s law), which decreases the solubility
of the gas.

13.45

a) increase

13.46

Plan: Solubility for a gas is calculated from Henry’s law: Sgas = kH  Pgas. We know kH and Pgas, so Sgas can be
calculated with units of mol/L. To calculate the mass of oxygen gas, convert moles of O2 to mass of O2 using the
molar mass.
Solution:
a) Sgas = kH  Pgas
mol 

Sgas =  1.28x10 3
1.00 atm  = 1.28x10–3 mol/L

L•atm 

 1.28x103 mol O 2   32.0 g O 2 
Mass (g) of O2 = 
 
  2.50 L  = 0.1024 = 0.102 g O2

L

  1 mol O 2 
b) The amount of gas that will dissolve in a given volume decreases proportionately with the partial pressure of
the gas.
mol 

Sgas =  1.28x10 3
 0.209 atm  = 2.6752x10–4 mol/L
L•atm 


b) stay the same

 2.6752x104 mol O 2
Mass (g) of O2 = 

L


  32.0 g O 2 
 
  2.50 L  = 0.0214016 = 0.0214 g O2
  1 mol O 2 

13.47

mol 
0.93% 

Solubility =  1.5x10 3
= 1.395x10–5 = 1.4x10–5 mol/L
1.0 atm  


L•atm 

 100% 

13.48

The solution is saturated.

13.49

Plan: Solubility for a gas is calculated from Henry’s law: Sgas = kH  Pgas. We know kH and Pgas, so Sgas can be
calculated with units of mol/L.
Solution:
Sgas = kH  Pgas = (3.7x10–2 mol/L•atm)(5.5 atm) = 0.2035 = 0.20 mol/L

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13-13


13.50

Solubility of gases increases with increasing partial pressure of the gas, and the goal of these devices is to increase
the amount of oxygen dissolving in the bloodstream.

13.51

Molarity is defined as the number of moles of solute dissolved in one liter of solution. Molality is defined as the
number of moles of solute dissolved in 1000 g (1 kg) of solvent. Molal solutions are prepared by measuring
masses of solute and solvent, which are additive and not changed by temperature, so the concentration in molality
does not change with temperature and is the preferred unit when the temperature of the solution may change.

13.52

Plan: Refer to the table of concentration definitions for the different methods of expressing concentration.
Solution:
a) Molarity and parts-by-volume (% w/v or % v/v) include the volume of the solution.
b) Parts-by-mass (% w/w) include the mass of solution directly. (Others may involve the mass indirectly.)
c) Molality includes the mass of the solvent.

13.53

No, 21 g solute/kg of solvent would be 21 g solute/1.021 kg solution.

13.54

Converting between molarity and molality involves conversion between volume of solution and mass of solution.
Both of these quantities are given so interconversion is possible. To convert to mole fraction requires that the
mass of solvent be converted to moles of solvent. Since the identity of the solvent is not given, conversion to mole
fraction is not possible if the molar mass is not known.

13.55

% w/w, mole fraction, and molality are weight-to-weight relationships that are not affected by changes in
temperature. % w/v and molarity are affected by changes in temperature, because the volume is temperature
dependant.

13.56

Plan: The molarity is the number of moles of solute in each liter of solution: M =

mol of solute
. Convert the
V(L) of solution

masses to moles and the volumes to liters and divide moles by volume.
Solution:
 32.3 g C12 H 22 O11   1 mL   1 mol C12 H 22 O11 
a) Molarity = 
 = 0.943617 = 0.944 M C12H22O11
  3  
100. mL

  10 L   342.30 g C12 H 22 O11 

 5.80 g LiNO3   1 mL   1 mol LiNO3 
b) Molarity = 
 = 0.166572 = 0.167 M LiNO3
  3  
 505 mL   10 L   68.95 g LiNO3 
13.57

 0.82 g C 2 H 5 OH   1 mL   1 mol C 2 H 5 OH 
a) Molarity = 
 = 1.69514 = 1.7 M C2H5OH
  3  
10.5 mL

  10 L   46.07 g C 2 H 5 OH 

 1.27 g NH3   1 mL   1 mol NH3 
b) Molarity = 
 = 2.2261 = 2.23 M NH3
  3  
 33.5 mL   10 L   17.03 g NH3 
13.58

Plan: Dilution calculations can be done using MconcVconc = MdilVdil.
Solution:
a) Mconc = 0.240 M NaOH
Vconc = 78.0 mL
Mdil = ?
Vdil = 0.250 L
 M conc Vconc   0.240 M  78.0 mL   103 L 
Mdil =
=

 = 0.07488 = 0.0749 M
Vdil 
 0.250 L 
 1 mL 
Vconc = 38.5 mL
Mdil = ?
Vdil = 0.130 L
b) Mconc = 1.2 M HNO3

3
 M conc Vconc  1.2 M  38.5 mL   10 L 
Mdil =
=

 = 0.355385 = 0.36 M
Vdil 
 0.130 L   1 mL 

13.59

Dilution calculations can be done using MconcVconc = MdilVdil

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13-14


a) Mconc = 6.25 M HCl
Vconc = 25.5 mL
Mdil = ?
Vdil = 0.500 L
 M conc Vconc   6.25 M  25.5 mL   103 L 
=
Mdil =

 = 0.31875 = 0.319 M
Vdil 
 0.500 L 
 1 mL 
b) Mconc = 2.00x10–2 M KI Vconc = 8.25 mL
Mdil =
13.60

 M conc Vconc 
Vdil 

=

 2.00x10

2



Mdil = ?

M 8.25 mL 

12.0 mL 

Vdil = 12.0 mL
= 0.01375 = 0.0138 M

Plan: For part a), find the number of moles of KH2PO4 needed to make 365 mL of a solution of this molarity.
Convert moles to mass using the molar mass of KH2PO4. For part b), use the relationship MconcVconc = MdilVdil to
find the volume of 1.25 M NaOH needed.
Solution:
 103 L   8.55x102 mol KH 2 PO 4 
a) Moles of KH2PO4 =  365 mL  
 = 0.0312075 mol
 1 mL  
L




 136.09 g KH 2 PO4 
Mass (g) of KH2PO4 =  0.0312075 mol KH2 PO4  
 = 4.24703 = 4.25 g KH2PO4
 1 mol KH 2 PO4 
Add 4.25 g KH2PO4 to enough water to make 365 mL of aqueous solution.
b) Mconc = 1.25 M NaOH
Vconc = ?
Mdil = 0.335 M NaOH
Vdil = 465 mL
M
V
0.335
M
465
mL
 dil  dil  


=
= 124.62 = 125 mL
Vconc =
1.25
V
M


 conc 
Add 125 mL of 1.25 M NaOH to enough water to make 465 mL of solution.
13.61

a) Find the number of moles NaCl needed to make 2.5 L of this solution. Convert moles to mass using the molar
mass of NaCl.
 0.65 mol NaCl   58.44 g NaCl 
Mass (g) of NaCl =  2.5 L  
  1 mol NaCl  = 94.965 = 95 g NaCl
L



Add 95 g NaCl to enough water to make 2.5 L of aqueous solution.
b) Use the relationship MconcVconc = MdilVdil to find the volume of 2.1 M urea needed.
Mconc = 2.1 M urea
Vconc = ?
Mdil = 0.3 M urea
Vdil = 15.5 L
M
V
0.3
M
15.5
L
 dil  dil  


=
= 2.21429 = 2 L
Vconc =
2.1
M
V


 conc 
Add 2 L of 2.1 M urea to enough water to make 15.5 L of solution.
Note because of the uncertainty in the concentration of the dilute urea (0.3 M), only one significant figure
is justified in the answer.

13.62

Plan: To find the mass of KBr needed in part a), find the moles of KBr in 1.40 L of a 0.288 M solution and
convert to grams using the molar mass of KBr. To find the volume of the concentrated solution that will be diluted
to 255 mL in part b), use MconcVconc = MdilVdil and solve for Vconc.
Solution:
 0.288 mol KBr 
a) Moles of KBr = 1.40 L  
 = 0.4032 mol
L



 119.00g KBr 
Mass (g) of KBr =  0.4032 mol  
 = 47.9808 = 48.0 g KBr
 1 mol KBr 
To make the solution, weigh 48.0 g KBr and then dilute to 1.40 L with distilled water.
Vconc = ?
Mdil = 0.0856 M LiNO3 Vdil = 255 mL
b) Mconc = 0.264 M LiNO3
 M dil Vdil   0.0856 M  255 mL 
=
= 82.68182 = 82.7 mL
Vconc =
 0.264 M 
Vconc 
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13-15


To make the 0.0856 M solution, measure 82.7 mL of the 0.264 M solution and add distilled water to make a total
of 255 mL.
13.63

a) To find the mass of Cr(NO3)3 needed, find the moles of Cr(NO3)3 in 57.5 mL of a 1.53x10–3 M solution and
convert to grams using molar mass of Cr(NO3)3.
 103 L   1.53x103 mol Cr(NO3 )3   238.03 g Cr(NO3 )3 
Mass (g) of Cr(NO3)3 =  57.5 mL  
 
 


L
 1 mL  
  1 mol Cr(NO3 )3 
= 0.020941 = 0.0209 g Cr(NO3)3
To make the solution, weigh 0.0209 g Cr(NO3)3 and then dilute to 57.5 mL with distilled water.
b) To find the volume of the concentrated solution that will be diluted to 5.8x103 m3 use MconcVconc = MdilVdil and
solve for Vconc.
Mconc = 2.50 M NH4NO3 Vconc = ?
Mdil = 1.45 M NH4NO3
Vdil = 5.8x103 m3
Vconc =

 M dil Vdil 
Vconc 

=

1.45 M   5.8x103 m3 
 2.50 M 

= 3.364x103 = 3.4x103 m3

To make the 1.45 M solution, measure 3.4x103 m3 of the 2.50 M solution and add distilled water to make
5.8x103 m3.
13.64

Plan: Molality, m, =

moles of solute
. Convert the mass of solute to moles and divide by the mass of solvent in
kg of solvent

units of kg.
Solution:

 1 mol glycine 
a) Moles of glycine = 85.4 g glycine 
 = 1.137605 mol
 75.07 g glycine 
1.137605 mol glycine
m glycine =
= 0.895752 = 0.896 m glycine
1.270 kg
 1 mol glycerol 
b) Moles of glycerol = 8.59 g glycerol 
 = 0.093278 mol
 92.09 g glycerol 
 1 kg 
Volume (kg) of solvent = 77.0 g  3  = 0.0770 kg
 10 g 


0.093278 mol glycerol
m glycerol =
= 1.2114 = 1.21 m glycerol
0.0770 kg

13.65

Molality = moles solute/kg solvent.
 1 mol HCl 
174 g HCl 

3
 36.46 g HCl   10 g  = 6.3043 = 6.30 m HCl
a) m HCl =


 757 g 
 1 kg 

 1 mol naphthalene 
16.5 g naphthalene 

3
 128.16 g naphthalene   10 g  = 2.41548 = 2.42 m naphthalene
b) m naphthalene =


 53.3 g 
 1 kg 
13.66

moles of solute
. Use the density of benzene to find the mass and then the moles of
kg of solvent
benzene; use the density of hexane to find the mass of hexane and convert to units of kg. Divide the moles of
benzene by the mass of hexane.

Plan: Molality, m, =

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13-16


Solution:

 0.877 g 
Mass (g) of benzene =  44.0 mL C6 H6  
 = 38.588 g benzene
 1 mL 

 1 mol C6 H6 
Moles of benzene =  38.588 g C6 H6  
 = 0.49402 mol benzene
 78.11 g C6 H6 
 0.660 g   1kg 
Mass (kg) of hexane = 167 mL C6 H14  
 = 0.11022 kg hexane

 mL   103 g 

m=
13.67

 0.49402 mol C6 H6 
moles of solute
=
= 4.48213 = 4.48 m C6H6
kg of solvent
 0.11022 kg C6 H14 

Molality = moles solute/kg solvent.
1.59 g   1 mol CCl4

 mL   153.81 g CCl4

 2.66 mL CCl4  
Molality of CCl4 =

13.68

 76.5 mL CH 2 Cl2  

1.33 g 

 mL 



3
  10 g  = 0.2702596 = 0.270 m CCl


4
 1 kg 

Plan: In part a), the total mass of the solution is 3.10x102 g, so masssolute + masssolvent = 3.10x102 g.
Assume that you have 1000 g of the solvent water and find the mass of C2H6O2 needed to make a 0.125 m
solution. Then a ratio can be used to find the mass of C2H6O2 needed to make 3.10x102 g of a 0.125 m solution.
Part b) is a dilution problem. First, determine the amount of solute in your target solution and then determine the
amount of the concentrated acid solution needed to get that amount of solute.
Solution:
 0.125 mol C2 H6 O2   62.07 g C2 H6 O2 
a) Mass (g) of C2H6O2 in 1000 g (1 kg) of H2O = 1 kg H2 O  


1 kg H2 O

 1 mol C2 H6 O2 
= 7.75875 g C2H6O2 in 1000 g H2O
Mass (g) of this solution = 1000 g H2O + 7.75875 g C2H6O2 = 1007.75875 g
 7.75875 g C2 H 6O2 
2
Mass (g) of C2H6O2 for 3.10x102 g of solution = 
 3.10x10 g solution
1007.75875
g
solution


= 2.386695 g C2H6O2
Masssolvent = 3.10x102 g – masssolute = 3.10x102 g – 2.386695 g C2H6O2 = 307.613305 = 308 g H2O
Therefore, add 2.39 g C2H6O2 to 308 g of H2O to make a 0.125 m solution.
 2.20% 
b) Mass (kg) of HNO3 in the 2.20% solution = 1.20 kg  
 = 0.0264 kg HNO3 (solute)
 100% 



Mass % =

mass of solute
100
mass of solution

Mass of 52.0% solution containing 0.0264 kg HNO3 =

mass of solute 100 



=

0.0264 kg 100 

mass %
52.0%
= 0.050769 = 0.0508 kg
Mass of water added = mass of 2.2% solution – mass of 52.0% solution
= 1.20 kg – 0.050769 kg = 1.149231 = 1.15 kg
Add 0.0508 kg of the 52.0% (w/w) HNO3 to 1.15 kg H2O to make 1.20 kg of 2.20% (w/w) HNO3.

13.69

a) The total weight of the solution is 1.50 kg, so
masssolute + masssolvent = 1.50 kg

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13-17


 0.0355 mol C2 H5 OH   46.07 g C2 H5OH 
Mass (g) of C2H5OH in 1000 g (1 kg) of H2O = 1 kg H2 O  


1 kg H2 O

 1 mol C2 H5 OH 
= 1.635485 g C2H5OH
Mass (g) of this solution = 1000 g H2O + 1.635485 g C2H5OH = 1001.635485 g
 1.635485 g C2 H5 OH 
Mass (g) of C2H5OH for 1.50 kg of solution = 
 1500 g solution 
 1001.635485 g solution 
= 2.449222 = 2.45 g C2H5OH
Masssolvent = 1500 g – masssolute = 1500 g – 2.449222 g C2H5OH = 1497.551 = 1498 g H2O
Therefore, add 2.45 g C2H5OH to 1498 g of H2O to make a 0.0355 m solution.
b) This is a disguised dilution problem. First, determine the amount of solute in your target solution:
 13.0% 
Mass (kg) of HCl in the 13.0% solution =  445 g  
 = 57.85 g HCl (solute)
 100% 
Then determine the amount of the concentrated acid solution needed to get 57.85 g solute:
mass of solute
Mass % =
100
mass of solution
mass of solute 100 
57.85 g 100 
=
Mass of 34.1% solution containing 57.85 g HCl =
= 169.6481 = 170. g
mass %
34.1%
Mass of water added = mass of 13.0% solution – mass of 34.1% solution
= 445 g – 169.6481 g = 275.35191 = 275 g
Add 170. g of the 34.1% (w/w) HCl to 275 g H2O
13.70

Plan: You know the moles of solute (C3H7OH) and the moles of solvent (H2O). Divide moles of C3H7OH by the
total moles of C3H7OH and H2O to obtain mole fraction. To calculate mass percent, convert moles of solute and
solvent to mass and divide the mass of solute by the total mass of solution (solute + solvent). For molality, divide
the moles of C3H7OH by the mass of water expressed in units of kg.
Solution:
a) Mole fraction is moles of isopropanol per total moles.
moles of isopropanol
0.35 mol isopropanol
Xisopropanol =
=
= 0.2916667 = 0.29
moles of isopropanol + moles of water
 0.35  0.85 mol
(Notice that mole fractions have no units.)
mass of solute
b) Mass percent =
100  . From the mole amounts, find the masses
mass of solution
of isopropanol and water:
 60.09 g C3 H7 OH 
Mass (g) of isopropanol =  0.35 mol C3 H7 OH  
 = 21.0315 g isopropanol
 1 mol C3 H7 OH 

 18.02 g H 2 O 
Mass (g) of water =  0.85 mol H2 O  
 = 15.317 g water
 1 mol H 2 O 
mass of solute
21.0315 g isopropanol
Mass percent =
100  =
100  = 57.860710 = 58%
mass of solution
 21.0315  15.317  g
c) Molality of isopropanol is moles of isopropanol per kg of water.
moles of solute
0.35 mol isopropanol  103 g 
m=
=

 = 22.85043 = 23 m isopropanol
kg of solvent
15.317 g water
 1 kg 
13.71

a) Mole fraction is moles of NaCl per total moles.

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13-18


XNaCl =

0.100 mol NaCl
= 0.01149425 = 0.0115 (Notice that mole fractions have no units.)
 0.100  8.60 mol

b) Mass percent is the mass of NaCl per 100 g of solution.
Mass (g) of NaCl = (0.100 mol NaCl)(58.44 g/mol) = 5.844 g NaCl
Mass (g) of water = (8.60 mol water)(18.02 g/mol) = 154.972 g water
 5.844 g NaCl 
x100% = 3.63396677 = 3.63% NaCl
Mass percent NaCl =
 5.844  154.972  g
c) Molality of NaCl is moles of NaCl per kg of solvent.
0.100 mol NaCl  103 g 
Molality NaCl =

 = 0.645277856 = 0.645 m NaCl
154.972 g water  1 kg 
13.72

moles of solute
. Use the density of water to convert the volume of water to mass. Multiply the
kg of solvent
mass of water in kg by the molality to find moles of cesium bromide; convert moles to mass. To find the mole
fraction, convert the masses of water and cesium bromide to moles and divide moles of cesium bromide by the
total moles of cesium bromide and water. To calculate mass percent, divide the mass of cesium bromide by the
total mass of solution and multiply by 100.
Solution:
The density of water is 1.00 g/mL. The mass of water is:
 1 mL   1.00 g   1 kg 
Mass (g) of water =  0.500 L   3  
 = 0.500 kg

 10 L   1 mL   103 g 
moles of solute
m=
kg of solvent

Plan: Molality =

moles CsBr
0.500 kg H 2 O
Moles of CsBr = (0.400 m)(0.500 kg H2O) = 0.200 mol
 212.8 g CsBr 
Mass (g) of CsBr =  0.200 mol CsBr  
 = 42.56 = 42.6 g CsBr
 1 mol CsBr 

0.400 m CsBr =

 103 g 
Mass (g) of water =  0.500 kg  
= 500. g water
 1 kg 


Mass of solution = mass (g) H2O + mass of CsBr = 500. g H2O + 42.56 g CsBr = 542.56 g
 1 mol H2 O 
Moles of H2O =  500. g H2 O  
 = 27.74695 mol H2O
 18.02 g H2 O 

XCsBr =

0.2000 mol CsBr
mol CsBr
=
= 7.156x10–3 = 7.16x10–3
mol CsBr + mol H 2 O
 0.2000  27.74695 mol

Mass percent CsBr =
13.73

 42.56 g CsBr 
mass of CsBr
x 100% = 7.84429 = 7.84% CsBr
100  =
mass of solution
 42.56  500. g

The density of water is 1.00 g/mL. The mass of water is:
Mass (g) of water = (0.400 L)(1 mL/10–3 L)(1.00 g/mL) = 4.00 102 g H2O
Moles of H2O = (400. g H2O)(1 mol H2O/18.02 g H2O) = 22.197558 mol H2O
Moles of KI = (0.30 g KI)(1 mol KI /166.0 g KI) = 1.80723x10–3 mol KI
XKI =

1.80723x103 mol KI

1.80723x10

3



 22.197558 mol

= 8.14091x10–5 = 8.1x10–5

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13-19


Mass percent KI =
13.74

 0.30 g KI 
 0.30  400. g

x 100% = 0.07494 = 0.075% KI

Plan: You are given the mass percent of the solution. Assuming 100. g of solution allows us to
express the mass % as the mass of solute, NH3. To find the mass of solvent, subtract the mass of NH3 from the
mass of solution and convert to units of kg. To find molality, convert mass of NH3 to moles and divide by the
mass of solvent in kg. To find molarity, you will need the volume of solution. Use the density of the solution to
convert the 100. g of solution to volume in liters; divide moles of NH3 by volume of solution. To find the mole
fraction, convert mass of solvent to moles and divide moles of NH3 by the total moles.
Solution:
Determine some fundamental quantities:
 8.00% NH3 
Mass (g) of NH3 = 100 g solution  
 = 8.00 g NH3
 100% solution 
Mass (g) H2O = mass of solution – mass NH3 = (100.00 – 8.00) g = 92.00 g H2O
 1 kg 
Mass (kg) of H2O =  92.00 g H 2 O   3  = 0.09200 kg H2O
 10 g 



 1 mol NH3 
Moles of NH3 =  8.00 g NH3  
 = 0.469759 mol NH3
 17.03 g NH3 
 1 mol H 2 O 
Moles of H2O =  92.00 g H2 O  
 = 5.1054 mol H2O
 18.02 g H 2 O 
 1 mL solution   103 L 
Volume (L) of solution = 100.00 g solution  
 = 0.103616 L
 
 0.9651 g solution   1 mL 
Using the above fundamental quantities and the definitions of the various units:
 0.469759 mol NH3 
moles of solute
= 
Molality = m =
 = 5.106076 = 5.11 m NH3
kg of solvent
 0.09200 kg H 2 O 

 0.469759 mol NH3 
moles of solute
= 
 = 4.53365 = 4.53 M NH3
0.103616 L
L of solution


moles of NH3
0.469759 mol NH3
Mole fraction = X =
=
= 0.084259 = 0.0843
total moles
 0.469759  5.1054  mol

Molarity = M =

13.75

The information given is 28.8 mass % FeCl3 solution with a density of 1.280 g/mL.
For convenience, choose exactly 100.00 g of solution.
Determine some fundamental quantities:
Mass (g) of FeCl3 = (100.00 g solution)(28.8% FeCl3/100%) = 28.8 g FeCl3
Mass (g) of H2O = mass of solution – mass FeCl3 = (100.00 – 28.8) g = 71.20 g H2O
Moles of FeCl3 = (28.80 g FeCl3)(1 mol FeCl3/162.20 g FeCl3) = 0.1775586 mol FeCl3
Moles of H2O = (71.20 g H2O)(1 mol H2O/18.02 g H2O) = 3.951165 mol H2O
Volume of solution = (100.00 g solution)(1 mL/1.280 g)(10–3 L/1 mL) = 0.078125 L
Using the above fundamental quantities and the definitions of the various units:
 0.1775586 mol FeCl3   103 g 
Molality = M = moles solute/kg solvent = 
 = 2.49380 = 2.49 m FeCl3
 
71.20 g H 2 O

  1 kg 
0.1775586 mol FeCl3
Molarity = m = moles solute/L solution =
= 2.272750 = 2.27 M FeCl3
0.078125 L
Mole fraction = X = moles substance/total moles =

0.1775586 mol FeCl3
= 0.043005688 = 0.0430
 0.1775586  3.951165 mol

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13-20


13.76

Plan: Use the equation for parts per million, ppm. Use the given density of solution to find the mass of solution;
divide the mass of each ion by the mass of solution and multiply by 1x106.
Solution:
 1 mL  1.001 g 
5
Mass (g) of solution is 100.0 L solution   3 
 = 1.001x10 g
 10 L  1 mL 

 mass solute 
6
ppm = 
 x 10
mass
solution




0.25 g Ca 2+
ppm Ca2+ = 
x 106 = 2.49750 = 2.5 ppm Ca2+
 1.001x105 g solution 




0.056 g Mg 2+
ppm Mg2+ = 
x 106 = 0.5594406 = 0.56 ppm Mg2+
5
 1.001x10 g solution 



13.77

The information given is that ethylene glycol has a density of 1.114 g/mL and a molar mass of 62.07 g/mol. Water
has a density of 1.00 g/mL. The solution has a density of 1.070 g/mL.
For convenience, choose exactly 1.0000 L as the equal volumes mixed. Ethylene glycol will be
designated EG.
Determine some fundamental quantities:
Mass (g) of EG = (1.0000 L EG)(1mL/10–3 L)(1.114 g EG/mL) = 1114 g EG
Mass (g) of H2O = (1.0000 L H2O)(1mL/10–3 L)(1.00 g H2O/mL) = 1.00x103 g H2O
Moles of EG = (1114 g EG)(1 mol EG/62.07 g EG) = 17.94747865 mol EG
Moles of H2O = (1.00x103 g H2O)(1 mol H2O/18.02 g H2O) = 55.49389567 mol H2O
Volume (L) of solution = (1114 g EG + 1.00x103 g H2O)(1 mL/1.070 g)(10–3 L/1 mL)
= 1.97570 L
Using the above fundamental quantities and the definitions of the various units:
a) Volume percent = (1.0000 L EG/1.97570 L)100% = 50.61497 = 50.61% v/v
b) Mass percent = [(1114 g EG)/(1114 + 1.00x103) g]100% = 52.6963 = 52.7% w/w
17.94747865 mol EC
c) Molarity = moles solute/L solution =
= 9.08411 = 9.08 M ethylene glycol
1.97570 L
d) Molality = moles solute/kg solvent =

17.94747865 mol EG  103 g 


1.00 x103 g H 2 O  1 kg 

= 17.94747865 = 17.9 m ethylene glycol
e) Mole fraction = XEG = moles substance/total moles =

17.94747865 mol EG
17.94747865
 55.49389567  mol


= 0.244378 = 0.244
13.78

Colligative properties of a solution are affected by the number of particles of solute in solution. The density of a
solution would be affected by the chemical formula of the solute.

13.79

A nonvolatile nonelectrolyte is a covalently bonded molecule that does not dissociate into ions or evaporate when
dissolved in a solvent. In this case, the colligative concentration is equal to the molar concentration, simplifying
calculations.

13.80

The “strong” in “strong electrolyte” refers to the ability of an electrolyte solution to conduct a large current. This
conductivity occurs because solutes that are strong electrolytes dissociate completely into ions when dissolved in
water.

13.81

Raoult’s law states that the vapor pressure of solvent above the solution equals the mole fraction of the solvent
times the vapor pressure of the pure solvent. Raoult’s law is not valid for a solution of a volatile solute in solution.
Both solute and solvent would evaporate based upon their respective vapor pressures.

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13-21


13.82

The boiling point temperature is higher and the freezing point temperature is lower for the solution compared to
the solvent because the addition of a solute lowers the freezing point and raises the boiling point of a liquid.

13.83

Yes, the vapor at the top of the fractionating column is richer in content of the more volatile component.

13.84

The boiling point of a 0.01 m KF solution is higher than that of 0.01 m glucose. KF dissociates into ions in water
(K+ and F–) while the glucose does not, so the KF produces more particles.

13.85

A dilute solution of an electrolyte behaves more ideally than a concentrated one. With increasing concentration,
the effective concentration deviates from the molar concentration because of ionic attractions. Thus, the more
dilute 0.050 m NaF solution has a boiling point closer to its predicted value.

13.86

Univalent ions behave more ideally than divalent ions. Ionic strength (which affects “activity” concentration) is
greater for divalent ions. Thus, 0.01 m NaBr has a freezing point that is closer to its predicted value.

13.87

Cyclohexane, with a freezing point depression constant of 20.1°C/m, would make calculation of molar mass of a
substance easier, since Tf would be greater.

13.88

Plan: Strong electrolytes are substances that produce a large number of ions when dissolved in water; strong acids
and bases and soluble salts are strong electrolytes. Weak electrolytes produce few ions when dissolved in water;
weak acids and bases are weak electrolytes. Nonelectrolytes produce no ions when dissolved in water. Molecular
compounds other than acids and bases are nonelectrolytes.
Solution:
a) Strong electrolyte When hydrogen chloride is bubbled through water, it dissolves and dissociates completely
into H+ (or H3O+) ions and Cl– ions. HCl is a strong acid.
b) Strong electrolyte Potassium nitrate is a soluble salt and dissociates into K+ and NO3– ions in water.
c) Nonelectrolyte Glucose solid dissolves in water to form individual C6H12O6 molecules, but these units are not
ionic and therefore do not conduct electricity. Glucose is a molecular compound.
d) Weak electrolyte Ammonia gas dissolves in water, but is a weak base that forms few NH4+ and OH– ions.

13.89

a) NaMnO4
b) CH3COOH
c) CH3OH
d) Ca(C2H3O2)2

13.90

Plan: To count solute particles in a solution of an ionic compound, count the number of ions per mole and
multiply by the number of moles in solution. For a covalent compound, the number of particles equals the number
of molecules.
Solution;
 0.3 mol KBr  2 mol particles 
a) 
 1 mol KBr  1 L  = 0.6 mol of particles
L



Each KBr forms one K+ ion and one Br– ion, two particles for each KBr.
 0.065 mol HNO3   2 mol particles 
b) 
 1 L  = 0.13 mol of particles

L

  1 mol HNO3 

strong electrolyte
weak electrolyte
nonelectrolyte
strong electrolyte

HNO3 is a strong acid that forms H+(H3O+) ions and NO3– ions in aqueous solution.
 104 mol KHSO 4   2 mol particles 
–4
c) 
 
 1 L  = 2x10 mol of particles

L

  1 mol KHSO 4 
Each KHSO4 forms one K+ ion and one HSO4– ion in aqueous solution, two particles for each KHSO4.
 0.06 mol C2 H5 OH   1 mol particles 
d) 
 1 L  = 0.06 mol of particles

L

  1 mol C2 H5 OH 

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13-22


Ethanol is not an ionic compound so each molecule dissolves as one particle. The number of moles of particles is
the same as the number of moles of molecules, 0.06 mol in 1 L.
13.91

a) (0.02 mol CuSO4/L)(2 mol particles/mol CuSO4)(10–3 L/1 mL)(1 mL) = 4x10–5 mol of particles
b) (0.004 mol Ba(OH)2/L)(3 mol particles/mol Ba(OH)2)(10–3 L/1 mL)(1 mL)
= 1.2x10–5 = 1x10–5 mol of particles
c) (0.08 mol C5H5N/L)(1 mol particles/mol C5H5N)(10–3 L/1 mL)(1 mL) = 8x10–5 mol of particles
d) (0.05 mol (NH4)2CO3/L)(3 mol particles/mol (NH4)2CO3)(10–3 L/1 mL)(1 mL)
= 1.5x10–4 = 2x10–4 mol of particles

13.92

Plan: The magnitude of freezing point depression is directly proportional to molality. Calculate the molality of
solution by dividing the moles of solute by the mass of solvent in kg. The solution with the larger molality will
have the lower freezing point.
Solution:
11.0 g CH3OH   1 mol CH3OH   103 g  = 3.4332085 = 3.43 m CH OH
a) Molality of CH3OH =

3


100. g H 2 O   32.04 g CH3OH   1 kg 
Molality of CH3CH2OH =

 22.0 g CH3CH 2 OH   1 mol CH3CH 2 OH   103 g 



 200. g H 2 O   46.07 g CH3CH 2 OH   1 kg 

= 2.387671 = 2.39 m CH3CH2OH
The molality of methanol, CH3OH, in water is 3.43 m whereas the molality of ethanol, CH3CH2OH, in water is
2.39 m. Thus, CH3OH/H2O solution has the lower freezing point.
 20.0 g H2O   1 mol H2 O 
b) Molality of H2O =

 = 1.10988 = 1.11 m H2O
1.00 kg CH3OH   18.02 g H2O 
 20.0 g CH3CH2OH   1 mol CH3CH2OH 
Molality of CH3CH2OH =

 = 0.434122 = 0.434 m CH3CH2OH
1.00 kg CH3OH   46.07 g CH3CH2OH 
The molality of H2O in CH3OH is 1.11 m, whereas CH3CH2OH in CH3OH is 0.434 m. Therefore, H2O/CH3OH
solution has the lower freezing point.
13.93

The magnitude of boiling point elevation is directly proportional to molality.
 38.0 g C3 H8 O3   1 mol C3 H8 O3   103 g  = 1.650559 = 1.65 m C H O
a) Molality of C3H8O3 =

3 8 3


 250. g ethanol   92.09 g C3 H8 O3   1 kg 
Molality of C2H6O2 =

 38.0 g C2 H 6 O 2   1 mol C2 H 6 O 2   103 g  = 2.44885 = 2.45 m C H O

2 6 2


 250. g ethanol   62.07 g C2 H 6 O 2   1 kg 

The molality of C2H6O2, in ethanol is 2.45 m whereas the molality of C3H8O3, in ethanol is 1.65 m. Thus,
C2H6O2/ethanol solution has the higher boiling point.
15 g C2 H6O2   1 mol C2 H6O2 
b) Molality of C2H6O2 =

 = 0.4833253 = 0.48 m C2H6O2
 0.50 kg H2 O   62.07 g C2 H6O2 
15 g NaCl   1 mol NaCl 
= 0.513347 = 0.51 m NaCl
Molality of NaCl =
 0.50 kg H2O   58.44 g NaCl 
Since the NaCl is a strong electrolyte, the molality of particles would be:
(2 particles/NaCl)(0.513347 mol NaCl/kg) = 1.026694 = 1.0 m particles
The molality of C2H6O2 in H2O is 0.48 m, whereas NaCl in H2O is 1.0 m. Therefore, NaCl/H2O solution has the
higher boiling point.
13.94

Plan: To rank the solutions in order of increasing osmotic pressure, boiling point, freezing point, and vapor
pressure, convert the molality of each solute to molality of particles in the solution. The higher the molality of
particles, the higher the osmotic pressure, the higher the boiling point, the lower the freezing point, and the lower
the vapor pressure at a given temperature.

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13-23


Solution:

 2 mol particles 
(I)  0.100 m NaNO3  
 = 0.200 m ions
 1 mol NaNO3 
NaNO3 consists of Na+ ions and NO3– ions, two particles for each NaNO3.
 1 mol particles 
(II)  0.100 m glucose  
 = 0.100 m molecules
 1 mol glucose 
Glucose is not an ionic compound so each molecule dissolves as one particle. The number of moles of particles is
the same as the number of moles of molecules.
 3 mol particles 
(III)  0.100 m CaCl2  
 = 0.300 m ions
 1 mol CaCl2 
CaCl2 consists of Ca+2 ions and Cl– ions, three particles for each CaCl2.
a) Osmotic pressure:
II < I < III
b) Boiling point:
bpII < bpI < bpIII
c) Freezing point:
fpIII < fpI < fpII
d) Vapor pressure at 50°C:
vpIII < vpI < vpII
13.95

I 0.04 m (H2N)2CO x 1 mol particles/1 mol (H2N)2CO = 0.04 m molecules
II 0.01 m AgNO3 x 2 mol particles/1 mol AgNO3 = 0.02 m ions
III 0.03 m CuSO4 x 2 mol particles/1 mol CuSO4 = 0.06 m ions
a) Osmotic pressure:
III > I > II
b) Boiling point:
bpIII > bpI > bpII
c) Freezing point:
fpII > fpI > fpIII
d) Vapor pressure at 298 K:
vpII > vpI > vpIII

13.96

Plan: The mole fraction of solvent affects the vapor pressure according to the equation Psolvent = XsolventP°solvent.
Convert the masses of glycerol and water to moles and find the mole fraction of water by dividing moles of water
by the total number of moles. Multiply the mole fraction of water by the vapor pressure of water to find the vapor
pressure of the solution.
Solution:
 1 mol C3 H8O3 
Moles of C3H8O3 =  34.0 g C3 H8 O3  
 = 0.369204 mol C3H8O3
 92.09 g C3 H8 O3 

 1 mol H 2 O 
Moles of H2O =  500.0 g H 2 O  
 = 27.7469 mol H2O
 18.02 g H 2 O 
mol H 2 O
27.7469 mol H 2 O
Xsolvent =
=
= 0.9868686
mol H 2 O + mol glycerol
27.7469 mol H 2 O + 0.369204 mol glycerol
Psolvent = XsolventP°solvent = (0.9868686)(23.76 torr) = 23.447998 = 23.4 torr
13.97

The mole fraction of solvent affects the vapor pressure according to the equation Psolvent = XsolventP°solvent.
Xsolvent = (5.4 mol toluene)/[(0.39) + (5.4)] mol = 0.93264
Psolvent = XsolventP°solvent = (0.93264)(41 torr) = 38.2382 = 38 torr

13.98

Plan: The change in freezing point is calculated from Tf = iKfm, where Kf is 1.86°C/m for aqueous solutions, i is
the van’t Hoff factor, and m is the molality of particles in solution. Since urea is a covalent compound and
does not ionize in water, i = 1. Once Tf is calculated, the freezing point is determined by subtracting it from
the freezing point of pure water (0.00°C).
Solution:
Tf = iKfm = (1)(1.86°C/m)(0.251 m) = 0.46686°C
The freezing point is 0.00°C – 0.46686°C = –0.46686 = –0.467°C.

13.99

Tb = iKbm = (1)(0.512°C/m)(0.200 m) = 0.1024°C

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13-24


The boiling point is 100.00°C + 0.1024°C = 100.1024 = 100.10°C.
13.100 Plan: The boiling point of a solution is increased relative to the pure solvent by the relationship Tb = iKbm.
Vanillin is a nonelectrolyte (it is a molecular compound) so i = 1. To find the molality, convert mass of vanillin to
moles and divide by the mass of solvent expressed in units of kg. Kb is given (1.22°C/m).
Solution:
 1 mol vanillin 
Moles of vanillin =  6.4 g vanillin  
 = 0.0420665 mol
 152.14 g vanillin 
Molality of vanillin =

moles of vanillin
0.042065 mol vanillin  103 g 
=


kg of solvent (ethanol)
50.0 g ethanol
 1 kg 

= 0.8413 m vanillin
Tb = iKbm = (1)(1.22°C/m)(0.8413 m) = 1.026386°C
The boiling point is 78.5°C + 1.026386°C = 79.5264 = 79.5°C.
13.101 Moles of C10H8 = (5.00 g C10H8)(1 mol C10H8/128.16 g C10H8) = 0.0390137 mol C10H8
C10H8 is a nonelectrolyte so i = 1.
Mass = (444 g benzene)(1 kg/103 g) = 0.444 kg benzene
Molality = (0.0390137 mol C10H8)/(0.444 kg) = 0.08786869 m
Tf = iKfm = (1)(4.90°C/m)(0.08786869 m) = 0.43056°C
Freezing point = (5.5 – 0.43056)°C = 5.06944 = 5.1°C
13.102 Plan: The molality of the solution can be determined from the relationship Tf = iKf m with the value 1.86°C/m
inserted for Kf and i = 1 for the nonelectrolyte ethylene glycol (ethylene glycol is a covalent compound that will
form one particle per molecule when dissolved). Convert the freezing point of the solution to °C and find Tf by
subtracting the freezing point of the solvent from the freezing point of the solution. Once the molality of the
solution is known, the mass of ethylene glycol needed for a solution of that molality can be found.
Solution:
.
°C = (5/9)(°F – 32.0) = (5/9)(–12.0°F – 32.0) = –24.44444°C
Tf = Tf(solution) – Tf(solvent) = (0.00 – (–24.44444))°C = 24.44444°C
Tf = iKf m

24.44444 C
Tf
=
= 13.14217 m
Kf
1.86 C/m
Ethylene glycol will be abbreviated as EG.
moles of EG
Molality of EG =
kg of solvent (water)

m=

 13.14217 mol EG 
Moles of EG = molality x kg of solvent = 
 14.5 kg H 2 O  = 190.561465 mol EG
1 kg H 2 O


 62.07 g EG 
Mass (g) of ethylene glycol = 190.561465 mol EG  

 1 mol EG 
= 1.18282x104 = 1.18x104 g ethylene glycol
To prevent the solution from freezing, dissolve a minimum of 1.18x104 g ethylene glycol in 14.5 kg water.
13.103 The molality of the solution can be determined from the relationship Tf = iKfm with the value 1.86°C/m
inserted for Kf, i = 1 for the nonelectrolyte glycerol, and the given Tf of –15°C.
m = Tf/Kf = (15°C/(1.86°C/m) = 8.06452 m
Glycerol will be abbreviated as GLY.
 103 g   1 kg   92.09 g GLY 
 8.06452 mol GLY 
Mass (g) of glycerol = 
  3  
 11.0 mg H 2 O  

1 kg H 2 O


 1 mg   10 g   1 mol GLY 
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13-25


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