 ```Chapter 12 Exam Practice Problems
(1) In which solvent would chromium acetate be most soluble?
(a) ethanol, CH3CH2OH
(b) methanol, CH3OH
(c) benzene, C6H6
(d) hexane, C6H14
Chromium acetate is a polar solute and
would therefore dissolve best in a polar
solvent.
(2) In which solvent would vegetable oil be most soluble?
(a) ethanol, CH3CH2OH
Vegetable oil is a non-polar solute
(b) methanol, CH3OH
and would therefore dissolve best
(c) benzene, C6H6
in a non-polar solvent.
(d) water, H2O
(3) A solution of potassium hydroxide is made by dissolving 2.40 g of KOH in 8.72 g of
water. The percent, by mass, of KOH in the solution is:
% 𝑚𝑎𝑠𝑠 =
𝑚𝑎𝑠𝑠 𝐾𝑂𝐻
2.40 𝑔
&times; 100% =
&times; 100% = 𝟐𝟏. 𝟔%
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠
2.40 𝑔 + 8.72 𝑔
(4) A glucose solution is prepared by dissolving 3.40 g of glucose, C6H12O6, in 115.5 g of
water. What is the molality of the glucose solution?
First calculate the number of moles of glucose that are present:
1 𝑚𝑜𝑙𝑒 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
3.40 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 &times;
= 0.01887 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
180.2 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
Calculate the kg of solvent:
1 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟
115.5 𝑔 𝑤𝑎𝑡𝑒𝑟 &times;
= 0.1155 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟
1000 𝑔 𝑤𝑎𝑡𝑒𝑟
Use these numbers to calculate the molality:
𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒 0.01887 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 =
=
= 𝟎. 𝟏𝟔𝟑 𝒎
𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
0.1155 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟
(5) Would you expect the heat of solution for CaCl2 to be positive or negative? Explain your
The heat of solution should be positive (i.e. dissolution would be an endothermic reaction).
You have to break very strong (ionic) bonds and are forming considerably weaker iondipole forces.
(6) The vapor pressure of pure water is 92.5 mm Hg at 50 oC. If 32.78 g of glucose (C6H12O6)
is stirred into 1.32 kg of water at 50 oC, what will the vapor pressure of the solution be?
Step one is to calculate the mole fraction of the solvent (water). In order to do this we first
calculate the number of moles of each component:
1 𝑚𝑜𝑙𝑒 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
32.78 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 &times;
= 0.1819 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
180.2 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
1.32 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟 &times;
1000 𝑔 𝑤𝑎𝑡𝑒𝑟
1 𝑚𝑜𝑙𝑒 𝑤𝑎𝑡𝑒𝑟
&times;
= 73.25 𝑚𝑜𝑙 𝑤𝑎𝑡𝑒𝑟
1 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟
18.02 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
Now we can calculate the mole percent:
𝑚𝑜𝑙 % 𝑤𝑎𝑡𝑒𝑟
73.25 𝑚𝑜𝑙
𝑋𝑤𝑎𝑡𝑒𝑟 =
=
= 0.9975
𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙
73.25 𝑚𝑜𝑙 + 0.1819 𝑚𝑜𝑙
Finally, calculate the vapor pressure of the solution using Raoult’s Law:
0
0
𝑃𝑠𝑜𝑙𝑛
= 𝑋𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑃𝑠𝑜𝑙𝑣𝑒𝑛𝑡
= (0.9975)(92.5 𝑚𝑚 𝐻𝑔) = 92.3 𝑚𝑚 𝐻𝑔
(7) The density of a 6.10 M aqueous solution of NaOH is 1.22 g/mL. What is the molality of
this solution?
Remember that molality is defined as the number of moles of solute divided by the mass
of the solvent in kg. Begin by assuming that you have 100.0 mL of solution.
Calculate the number of moles of NaOH: (0.100 L) (6.10 M) = 0.610 moles NaOH
This is the number of moles of solute.
Now we need to calculate the mass of the solvent. We can calculate the total mass of the
solution using the density:
(100.0 mL)(1.22 g/mL) = 122 g
Some of this 122 g that we’ve calculated is solute (NaOH), and some of it is solvent (water).
We can calculate the mass of NaOH by multiplying the number of moles by the molar mass:
𝑔
0.610 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻 &times; 40.0
= 24.4 𝑔 𝑁𝑎𝑂𝐻
𝑚𝑜𝑙
The rest of the solution (122 g -24.4 g = 97.6 g= 0.976 kg) must be water.
Now we have enough information to calculate the molality.
0.610 𝑚𝑜𝑙
𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 =
= 6.25 𝑚
0.976 𝑘𝑔
(8) At 20.0 oC, the vapor pressure of acetone (CH3COCH3) is 181.7 mm Hg and the vapor
pressure of water is 17.5 mm Hg. A solution is made by mixing 128.4 g water and 97.60 g
acetone. What is the vapor pressure of the solution?
Like problem #6, this is a Raoult’s Law problem. It differs from #6 in that both of the
components of the system have a vapor pressure; we need to calculate the vapor pressure
for each of the components. Start by calculating the number of moles of each:
128.4 𝑔 𝐻2 𝑂 &times;
1 𝑚𝑜𝑙 𝐻2 𝑂
= 7.129 𝑚𝑜𝑙 𝐻2 𝑂
18.02 𝑔 𝐻2 𝑂
97.60 𝑔 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 &times;
1 𝑚𝑜𝑙 𝑎𝑐𝑒𝑡𝑜𝑛𝑒
= 1.680 𝑚𝑜𝑙 𝑎𝑐𝑒𝑡𝑜𝑛𝑒
58.09 𝑔 𝑎𝑐𝑒𝑡𝑜𝑛𝑒
Now we can calculate the mole fraction of each component:
𝑋𝑤𝑎𝑡𝑒𝑟 =
𝑚𝑜𝑙 % 𝑤𝑎𝑡𝑒𝑟
7.129 𝑚𝑜𝑙
=
= 0.8093
𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙
7.129 𝑚𝑜𝑙 + 1.680 𝑚𝑜𝑙
𝑋𝑎𝑐𝑒𝑡𝑜𝑛𝑒 =
𝑚𝑜𝑙 % 𝑎𝑐𝑒𝑡𝑜𝑛𝑒
1.680 𝑚𝑜𝑙
=
= 0.1907
𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙
7.129 𝑚𝑜𝑙 + 1.680 𝑚𝑜𝑙
The vapor pressure of the solution is equal to the sum of the partial pressures of each
component:
0
0
𝑃𝑠𝑜𝑙𝑛 = 𝑃𝑤𝑎𝑡𝑒𝑟 + 𝑃𝑎𝑐𝑒𝑡𝑜𝑛𝑒 = 𝑋𝑤𝑎𝑡𝑒𝑟 𝑃𝑤𝑎𝑡𝑒𝑟
+ 𝑋𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑃𝑎𝑐𝑒𝑡𝑜𝑛𝑒
𝑃𝑠𝑜𝑙𝑛 = (0.8093)(17.5 𝑚𝑚 𝐻𝑔) + (0.1907)(181.7 𝑚𝑚 𝐻𝑔)
= 14.16 𝑚𝑚 𝐻𝑔 + 34.65 𝑚𝑚 𝐻𝑔 = 𝟒𝟖. 𝟖𝟏 𝒎𝒎 𝑯𝒈
(9) A 5% w/w aqueous solution of ammonia (NH3) has a density of 0.9597 g/mL. For this
solution, calculate:
(a) molarity of ammonia
(b) molality of ammonia
(c) mole fraction of ammonia
Assume that you have have 100.0 g of solution. By definition, for a 5% w/w solution of
ammonia, 5% of the 100.0 g (= 5.000 g) is ammonia, and the other 95.00 g must be water.
a. molarity is defined as moles of solute divided by volume of solution.
Moles of ammonia can be easily calculated using the mass of ammonia:
1 𝑚𝑜𝑙 𝑁𝐻3
5.000 𝑔 𝑁𝐻3 &times;
= 0.293 𝑚𝑜𝑙 𝑁𝐻3
17.04 𝑔 𝑁𝐻3
To calculate the volume of solution, use the density of the solution in conjunction with the
mass:
1 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
100.0 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 &times;
= 104.2 𝑚𝐿 = 0.1042 𝐿
0.9597 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Now calculate the molarity:
0.293 𝑚𝑜𝑙
𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
= 𝟐. 𝟖𝟐 𝑴
0.1042 𝐿
b. molality is defined as moles of solute divided by mass of solvent
At the beginning, we deduced that 95.00 g of the solution is water (the solvent). This is
0.095 kg.
𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 =
0.293 𝑚𝑜𝑙
= 𝟑. 𝟎𝟖𝟖 𝒎
0.095 𝑘𝑔
c. To calculate the mole fraction you first need to know the total number of moles. The
number of moles of ammonia was calculated earlier so you just need to calculate the
number of moles of water:
95.00 𝑔 𝑤𝑎𝑡𝑒𝑟 &times;
1 𝑚𝑜𝑙 𝑤𝑎𝑡𝑒𝑟
= 5.271 𝑚𝑜𝑙 𝑤𝑎𝑡𝑒𝑟
18.02 𝑔 𝑤𝑎𝑡𝑒𝑟
Now you can calculate the mole fraction of ammonia:
𝑛𝑁𝐻3
0.293
𝑋𝑁𝐻3 =
=
= 𝟎. 𝟎𝟓𝟎𝟎
𝑛𝑁𝐻3 + 𝑛𝐻2𝑂 0.293 + 5.271
(10) When 124 g of a sugar is added to 1.10 kg of water, the freezing point is lowered by
2.4 oC. Kf for water is 1.86 oC/m. What is the approximate molecular weight of the
compound?
You can use the freezing point change to calculate the molality of the solution. (Since this is
a sugar it will not dissociate in solution, so i=1.)
∆𝑇𝑓 = 𝑖𝐾𝑓 𝑚
so
∆𝑇
2.4 ℃
𝑚=
=
= 1.29 𝑚
𝑖𝐾𝑓 (1) (1.86 ℃ )
𝑚
Remember that molality is defined as moles of solute divided by mass of solvent. Given
the molality and the mass of solvent), the number of moles of solute can be calculated:
(molality)(mass solvent) = (1.29 m)(1.10 kg) = 1.42 moles compound
Use the number of moles, in conjunction with the given mass of sugar, to calculate the
molar mass:
124 g/1.42 moles = 87.3 g/mol
(11) The Henry’s law constant for hydrogen gas (H2) is 7.8 x 10-4 M/atm at 25 oC. A sealed
container of water has a partial pressure of 2.0 atm H2 in the headspace above water. What
is the concentration of H2 in the water?
Henry’s Law states that the concentration of a gas in solution is the Henry’s law constant
times the pressure above the solvent:
𝐶𝐻2 = 𝑘𝐻𝐻2 𝑃𝐻2 = (7.8 &times;
10−4 𝑀
) (2.0 𝑎𝑡𝑚) = 𝟏. 𝟔 &times; 𝟏𝟎−𝟑 𝑴
𝑎𝑡𝑚
(12) The density of apple juice is 1.04 g/mL. A 16-oz. (480 mL) bottle contains 52 g of
sucrose, C11H22O11 . Calculate the percentage by mass of sugar of apple juice.
By definition, the mass percent of a substance is the mass of that substance divided by the
total mass of the solution. Here, the mass of sucrose is given, but the total mass of the
solution needs to be calculated by using the density:
msoln =(1.04 g/mL)(480 mL) = 499.2 g
𝑚𝑠𝑢𝑐𝑟𝑜𝑠𝑒
52 𝑔
𝑚𝑎𝑠𝑠 % =
&times; 100% =
&times; 100% = 𝟏𝟎. 𝟒% 𝒔𝒖𝒄𝒓𝒐𝒔𝒆
𝑚𝑠𝑜𝑙𝑛
499.2 𝑔
(13) Cyclohexane has a normal boiling point of 80.74oC and a Kb value of 2.79 oC/m.
If 2.60 g of glucose (C6H12O6) is added to 284 g of cyclohexane, what would the new boiling
point be?
Start by calculating the number of moles of glucose added to the solution:
2.60 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 &times;
1 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
= 0.0144 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
180.1 𝑔𝑔𝑙𝑢𝑐𝑜𝑠𝑒
Now calculate the molality of the solution; remember that molality is defined as moles of
solute divided by mass of solvent:
𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 =
0.0144 𝑚𝑜𝑙
= 0.0507 𝑚
0.284 𝑘𝑔
We can use this to calculate the change in boiling point:
∆𝑇𝑏 = 𝑖𝐾𝑏 𝑚 = (1) (2.79
℃
) (0.0507) = 0.141 ℃
𝑚
The new boiling point would be: 80.74 oC + 0.141 oC = 80.88 oC
(14) Helium has a solubility of 5.55 x 10-4 M at 1.50 atm. What would its solubility be at
9.76 atm?
𝐶1 𝐶2
=
𝑃1 𝑃2
5.55 &times; 10−4 𝑀
𝐶2
=
1.50 𝑎𝑡𝑚
9.76 𝑎𝑡𝑚
C2 = 3.61 x 10-3 M
(15) What is the molarity of an 0.80 m solution of CH3COOH? The density of the solution is
1.0055 g/mL.
Remember that molality is defined as moles of solute divided by mass of solvent; thus, an
0.80 m solution is 0.80 moles in 1.0 kg of solvent.
Molarity is defined as moles of solute divided by volume of solution. You don’t know the
volume of the solution but since you are given the density, you can calculate the volume
using the mass. The mass of the solution would be the mass of the solvent (here, 1.0 kg or
1.0 x 103 g) plus the mass of the solute, which you need to calculate:
0.80 𝑚𝑜𝑙 𝐶𝐻3 𝐶𝑂𝑂𝐻 &times;
60.06 𝑔𝐶𝐻3 𝐶𝑂𝑂𝐻
= 48.048 𝑔 𝐶𝐻3 𝐶𝑂𝑂𝐻
1 𝑚𝑜𝑙 𝐶𝐻3 𝐶𝑂𝑂𝐻
The total mass of the solution will be:
mtot= msolvent+ msolute=1000g + 48.048 g = 1048.048 g
You can use this to calculate the volume of the solution:
1 𝑚𝐿
1𝐿
𝑉 = 1048.048 𝑔 (
)(
) = 1.042 𝐿
1.0055 𝑔 1000 𝑚𝐿
Use this to calculate the molarity:
𝑀=
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 0.80 𝑚𝑜𝑙
=
= 𝟎. 𝟕𝟕 𝑴
𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1.042 𝐿
```