Review Sheet – Chemistry, Level 3 – Ch. 15: Solutions
Name: ____________________
Date:___________
Period:___
This test covers materials in the textbook in Chapters 15 (pp. 500-529). You are always responsible for the topics on the
previous tests.
Test Expectations
The student should be able:





Use the basic vocabulary of solutions
Describe how a solution of an ionic solid in a polar solvent forms
Calculate concentrations in units of mass percent, molarity, molality, and mole fraction
Use colligative properties such as freezing point depression and boiling point elevation to
calculate molalities, changes in freezing or boiling points, and molar masses for both
electrolytes and nonelectrolyte solutes in nonvolatile solvents
Analyze the factors affecting solubility, including temperature, pressure, and nature of the
solvent (“likes dissolves like”)
1. Define each of the following terms clearly and completely.
(a) solute –
(k) molarity –
(b) solvent –
(l) molality—
(c) alloy –
(m) mass percent –
(d) miscible –
(n) electrolyte –
(e) aqueous –
(o) mole fraction –
(f) saturated –
(p) mass percent –
(g) solvation –
(q) supersaturated –
(h) hydration –
(p) vapor pressure –
(i) soluble –
(m) van’t Hoff factor –
(j) colligative property –
(n) volatile –
2. What is the molarity of the solution formed by mixing 0.20 mol of sodium hydroxide (NaOH) with enough
water to make 150 mL of solution?
Molarity 
moles of solute
0.20 mol

 1.33 M
liters of solution
0.150 L
3. What is the molality of an alloy containing 1.3 g of silver and 40.8 g of iron? What is the mass percent?
 1 mol 

1.3 g
107.9 g 
moles of solute

molality 

 0.295 m
kg of solution
0.0408 kg
mass of component
1.3 g
mass percent 

x 100  3.1%
total mass of solution 1.3 g  40.8 g
4. A gas mixture contains 26.3 g of nitric oxide (NO) and 36.2 g of oxygen gas. What is the mole fraction of
the nitric oxide? What is the mass percent?
mole fraction  X NO
mass percent 
 1 mol 

26.3 g
30.0 g 
moles of component



 0.437
total moles of solution
 1 mol 
 1 mol 
  36.2 g

26.3 g
30
.
0
g
32
.
0
g




mass of component
26.3 g

x 100  42.1%
total mass of solution
26.3 g  36.2 g
5. An evil villain attempts to create 250 mL of a 4.7 M solution of potassium cyanide (KCN) to use against his
archenemy (the mutated sea bass are on vacation). The evil genius never finished all six years of evil medical
school and cannot complete the calculations without help. How many grams of KCN does the evil genius need?
moles of solute
moles

 4.7 M
liters of solution
0.250 L
 65.1 g 
moles  4.7 M * 0.250 L  1.2 moles 
  78.1 g
 1 mole 
Molarity 
6. What is the boiling point elevation of a solution containing 125 g of the nonelectrolyte ethylene glycol
(C2H6O2) in 1200 g of water. Kb for water is .52 C/m.
 1 mol 

125 g
62.0 g 
moles of solute

molality 

 1.68 m
kg of solution
1.200 kg
ΔTb  K b m  (1.68m)(0. 52  C/m)  0.87  C
7. How many kilograms of ethylene glycol (C2H6O2) should be added to 25 kg of water to lower the freezing
point by 4.0 C. Kf for water is 1.86 C/m.
ΔTb  K b m
m
Tb
4.0

 2.15m
K b 1.86
2.15m 
 1kg 
moles
 62.0 g 
  3.33kg
 moles  2.15m * 25 kg  53.8moles * 
  3332 g 
25kg
 1mole 
 1000 g 
8. What is the boiling point elevation of a solution containing 45.9 g of the electrolyte CaCl2 in 1300 g of water.
Kb for water is 0.52 C/m.
 1 mol 

45.9 g
111.1g 
moles of solute

molality 

 0.318 m
kg of solution
1.300 kg
ΔTb  iK b m  (3)(0.318m)(0 .52  C/m)  0.50  C
9. A 25.0 g sample of an unknown nonelectrolyte is dissolved in 150 g of benzene (C6H6). The boiling pont is
raised 3.18 C above the boiling point of the pure benzene. What is the molar mass of the sample? Kb for
benzene is 2.67 C/m.
ΔTb  K b m
m
Tb 3.18

 1.19m
Kb
2.67
moles
 moles  1.19m * 0.150 kg  0.179moles
0.150kg
25.0 g
MolarMass 
 140.g / mole
0.179moles
1.19m 
10. Check the box for the solvent that is most likely to dissolve the solute in each row.
Solute\Solvent
C12H24
CaCl2
Engine grease
Pentane (C5H10)
X
Ethyl Alcohol (C2H5OH)
X
X
Like dissolve like: the polar solvent (ethyl alcohol) dissolves the ionic compound (CaCl2). The nonpolar solvent
(C5H10) dissolves the nonpolar substances (C12H24) and engine grease.