UNIT VIII – SOLUTIONS
REVIEW
I. Good things to know:
• Mixtures: homogeneous and heterogeneous
• Types of colloids: aerosol, foam, emulsion, sol
• types of solutions
• solution terms: solute, solvent, insoluble, soluble
• molecular view of solution process
• saturated, unsaturated, supersaturated; fractional crystallization
• solubility of solid solutes vs. temperature
• gas solubility vs. temperature and pressure (Henry’s Law)
• colligative properties, van’t Hoff Factor
II. Essays/Short Answer
1. Explain why ethanol (CH3CH2OH) is soluble in both water and carbon tetrachloride.
The CH3CH2 portion is nonpolar, therefore it will be more soluble in a nonpolar solvent, such as carbon
tetrachloride. The OH portion of the molecule is more polar, thus making it also soluble in a polar
solvent like water.
2. Explain why in a dilute solution, molarity and molality will have the same value.
In a dilute solution, the density will be very close to that of pure water (1.0 g/mL). As a result, a
kilogram of solvent (from molality) will be occupying a liter of space (from molarity). Since the number
of moles of solute is the same, the molality and molarity values will be roughly the same.
3. Give the estimated van’t Hoff factor for the following substances:
2
a. NaCl
1
b. CH3CH2OH (ethanol)
3
c. K2SO4
III. Problems
1. A solution is made by dissolving 49g of NaCl in enough water to make 1.00L of solution. Assume that the
density of the solution is 1.05 g/cm3. Calculate the mass percent, molarity, molality, and mole fraction
of NaCl.
1.00 L = 1000mL ∗
1.05 g
= 1050 g solution
1mL
1050 - 49 = 1001g H 2 O = 1.001kg H 2 O → 1001g H 2 O ∗
1mol H 2 O
= 55.611mol H 2 O
18 g H 2 O
1 mol NaCl
= 0.838mol NaCl → 0.838 + 55.611 = 56.449mol total
58.5 g NaCl
49g NaCl
0.838mol NaCl
%=
x100% = 4.7% → M =
= 0.84 M
1050 g sol' n
1.00 L
0.838mol NaCl
0.838mol NaCl
m=
= 0.84m → X =
= 0.015
1.001kg H 2 O
56.449mol
49g NaCl ∗
UNIT VIII – SOLUTIONS
REVIEW
2. Using the chart shown
a. what is the solubility of: HCl at 30oC?
67 g HCl
100 g H 2 O
b. which of the examples below is the most
soluble at 20oC?
NH3
KNO3
NaCl
NaNO3
c. Explain why the solubility of HCl, NH3, and
SO2 decrease as temperature increases.
because they are all gases
d. If you have 25θg of KNO3 how much water
would you need to make a saturated solution at
room temperature (25oC) ?
40 g KNO 3 250 g KNO 3
=
→ x = 625g H 2 O
100 g H 2 O
x
e. If you added 50.0g of KCl to 10θg of water at
40oC, would the solution be saturated,
supersaturated, or unsaturated? If it is saturated,
how much is left over?
40 g KCl
→ 40 < 50, so saturated → 50 - 40 = 10g left
100 g H 2 O
3. Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25oC the vapor pressures of pentane and
hexane are 511 and 15θ Torr, respectively. A solution is prepared by mixing 25 mL pentane
(density = 0.63 g/mL) with 45 mL hexane (density = 0.66 g/mL). What is the total vapor pressure of
the resulting mixture of gases?
0.63g 1mol
∗
= 0.21875mol pentane
1 mL 72 g
0.219 + 0.345 = 0.564mol
0.66g 1mol
45ml hexane ∗
∗
= 0.34535mol hexane
1 mL 86 g
25ml pentane ∗
0.219
= 0.388 → Ppen tan e = (0.388)(511Torr ) = 198Torr
0.564
198 + 92 = 290Torr
0.345
=
= 0.612 → Phexane = (0.612)(150Torr ) = 92Torr
0.564
X pen tan e =
X hexane
UNIT VIII – SOLUTIONS
REVIEW
4. A solution of sodium chloride in water has a vapor pressure of 21.9 Torr at 25oC. What is the mole
fraction of NaCl in this solution? What would be the vapor pressure of this solution at 45oC?
The vapor pressure of pure water is 23.8 Torr at 25oC and 71.9 Torr at 45oC.
21.9 = X solvent (23.8) → X solvent = 0.920 → X NaCl = 1 − 0.920 = 0.080
P45 = (0.920)(71.9Torr ) = 66.1Torr
5. You form a solution by dissolving 50.00g of the nonelectrolyte benzoic acid (C7H6O2) in 200.0 mL of carbon
tetrachloride (CCl4) (D = 1.594 g/mL). Using the table on the bottom of p. 8 for any reference you need:
a. calculate the molality of the solution
1 mol
= 0.410mol BA
0.410mol
122g BA
m=
= 1.29m
0.3188kg
1.594g
200.0mL ∗
= 318.8 g = 0.3188kg
1mL
50.00 g BA ∗
b. calculate the boiling point of the solution
∆Tb = (5.03
o
C
m
)(1.29m) = 6.49 o C → 76.5 + 6.5 = 83.0 o C
c. calculate the freezing point of the solution
∆T f = (30
o
C
m
)(1.29m) = 38.7 o C → −23.0 − 38.7 = −61.7 o C
6. An aqueous solution is created by dissolving 25.03g of iron(III) chloride (FeCl3) (i=3.4) in 300.0 mL of water.
Calculate the boiling and melting points of the solutions.
25.03 g FeCl3 ∗
1mol
= 0.1545mol ÷ 0.3000kg H 2O = 0.515m
162 g
∆Tb = (3.4)(0.51
o
∆Tf = (3.4)(1.86
o
C
C
m
m
)(0.515m) = 0.89o C → Tb = 100 + 0.89 = 100.89o C
)(0.515m) = 3.26o C → Tb = 0 − 3.26 = −3.26o C
UNIT VIII – SOLUTIONS
REVIEW
7. The molecular weight and formula of a hydrocarbon are to be determined through the use of the freezing point
depression method. The hydrocarbon is known to be 86% carbon and 14% hydrogen by mass. In the experiment,
3.72g of the unknown hydrocarbon were placed into 50.0g of liquid benzene (C6H6). The freezing point of the
solution was measured to be 0.06oC. The normal freezing point of benzene is 5.50oC, and the freezing point
depression constant for benzene is 5.12 oC/m.
a. What is the molecular weight of the compound?
∆T f = 5.50 − 0.06 = 5.44 o C = (5.12
o
C
m
)m
x
→ x = 0.053125mol
0.050kg
3.72 g
= 70.0 g mol
FM =
0.053125mol
m = 1.06 =
b. What is the molecular formula of the hydrocarbon?
1C
= 7.16 C
12g C
÷ 7.16 → CH 2 ( FM = 14)
1H
14g H ∗
= 14 H
1g H
86 g C ∗
70
= 5 → mf = C5 H10
14
c. What is the mole fraction of benzene in the solution?
0.0531mol C5 H10
50.0 g benzene ∗
X benzene =
0.0531 + 0.641 = 0.694mol
1 mol
= 0.641mol benzene
78 g
0.641mol
= 0.924
0.694mol
d. If the density of the solution is 875 g/L, what is the molarity of the solution?
50.0 g + 3.72 g = 53.72 g sol' n ∗
M=
1L
= 0.0614 L sol' n
875g
0.0531mol C5 H10
= 0.865M
0.0614 L