Chapter 10 Study Guide - KEY
1.
What is the molarity of a 2.05 L solution that is made by dissolving 27.5 g of sodium
chloride in water?
27.5 g NaCl 1 mol

 0.470 mol
58.5 g
M
mol solute 0.470 mol

 0.229M
L sol'n
2.05 L
2. What is the normality of a 6.0 M solution of phosphoric acid?
N = (M)(total pos. ox) = (6.0 M)(3) = 18 N
3. What is the molality of a solution that is made by dissolving 98.7 g of calcium chloride in
755 g of water?
98.7 g CaCl2 1 mol

 0.888 mol
111.1 g
m
mol solute 0.888 mol

 1.18 m
kg solvent
0.755 kg
4. How many grams of Kl are needed to prepare 325 mL of a 1.33 M solution?
n mol
1.33 M 
0.325 L
n  0.432 mol KI
mass 
0.432 mol KI

166.1 g
 71.8 g KI
1 mol
5. How many grams of NaCl are needed to prepare a 4.15 m solution, using 176 g of water?
n mol
4.15 m 
0.176 kg
n  0.7304 mol NaCl
mass 
0.7304 mol NaCl

58.5 g
 42.7 g NaCl
1 mol
6. Ethyl iodide (C2H5I) boils at 72.5 C and has a density of 1.933 g/mL.
a. A solution prepared by dissolving 0.300 mol of a nonelectrolyte in 750.0 mL of
ethyl iodide boils at 73.5 C. What is the boiling point constant (Kb) for ethyl
iodide?
C2H5I = 156.0 g/mol
750.0 mL 1.933 g
mass solvent 

 1449.75 g solvent
1 mL
m
0.300 mol
 0.207 m
1.44975 kg
Tb  kb  m
1.0  C  (kb )(0.207m)
kb  4.83  C/m
b. Another solution is prepared by dissolving 12.5 g of an unknown nonelectrolyte in
100.0 mL of ethyl iodide. The resulting solution boils at 74.9 C. What is the
molar mass (molecular weight) of the compound?
100.0 mL 1.933 g
mass solvent 

 193.3 g solvent
1 mL
Tb  kb  m
2.4C  ( 4.83C/m)(m)
m  0.497 m
0.497 m 
n mol
0.1933 kg
n  0.096 mol
MW 
12.5 g
 130.g/mol
0.096mol
7. What is the freezing point of a solution of 74.2 g of sucrose (C 12H22O11) in 883 g of
water?
74.2g C12H22O11 1 mol

 0.217 mol
342 g
m
0.217 mol
 0.246 m
0.883 kg
Tf  kf  m
Tf  (1.86C/m)(0.246 m)  0.458 C
f.p.  0C - 0.458 C   0.458 C
8. What is the boiling point of a solution of 9.75 g of sodium chloride, an electrolyte, in
115 g of water?
9.57g NaCl
1 mol

 0.167 mol
58.5 g
m
0.167 mol
 1.449 m
0.115 kg
Tb  kb  m  n
Tb  (0.52C/m)(1.449 m)(2)  1.51C
b.p.  100C  1.51C  101.5 C  102C
9. How many grams of sodium chloride, an electrolyte, should be dissolved in 700.0 g of
water to produce a solution that freezes at –3.75°C?
Tf  kf  m  n
3.75C  (1.86C/m)(m)(2)
m  1.00 mol
1.00mol
n mol

kg
0.700 kg
n  0.700 mol
mass 
0.700 mol

58.5 g
 41.3 g NaCl
1 mol
10. A solution was made by dissolving 4.32 g of an unknown covalent solute in 108.5 g of
acetone. The solution boiled at 58.75°C. The boiling point of pure acetone is 55.95°C
and Kb= 1.71 °C/m. Calculate the molar mass of the solute.
Tb  58.75C  55.95C  2.8C
Tb  kb  m  n
2.8C  (1.71C/m)(m)
m  1.64 m
1.64mol
n mol

kg
0.1085 kg
n  0.178 mol
MW 
4.32 g
 24.3 g/mol
0.178 mol
11. Describe how you would prepare 225.0 mL of a 0.775 M solution of ammonium nitrite
starting with
a. solid ammonium nitrite
NH4NO2
0.775 m
n mol

1L
0.225L
n  0.174 mol
0.174375 mol NH4 NO2

64.0 g
 11.2 g NH4 NO2
1 mol
Measure out 11.2 g NH4NO2 and add it to enough water to make 225 mL of
solution
b. 2.25 M ammonium nitrite solution
M1V1  M2V2
(2.25M)(V1 )  (0.775M)(2 25.0mL)
V1  77.5 mL
Measure out 77.5 mL of 2.25 M solution; dilute to a final volume of 225 mL