Name: ___KEY___________
Chem142 Exam II W2009 Chapters 12 & 14 – 100 points
*Important equations and values on last page
1. __F__T/F Temperature and gas solubility in solution are directly proportional. (2
points)
2. __F__T/F Detergents have a long polar tail and a non-polar head. (2 points)
3. __F__T/F The van’t Hoff factor of HgI2 is 2. (2 points)
4. __T__T/F According to Raoult’s Law, the vapor pressure of a solution is less than
that of the pure solvent. (2 points)
5. __F__T/F Given a constant temperature, mountain rivers carry greater dissolved
oxygen than rivers at lower elevation. (2 points)
6. Given the following reaction:
A(s) + 2BC(aq)  AC2(aq) + B2(g); H° is exothermic
a. If the flask containing the reactants and products suffers from a decreased
volume the reaction will ____________. (4 points)
i. Shift to the right
ii. Shift to the left
iii. Not shift either way
b. If more A is added the concentration of AC2 will ___________. (4 points)
i. Increase
ii. Decrease
iii. Remain the same
c. If the reaction flask is heated the amount of A will _________. (4 points)
i. Increase
ii. Decrease
iii. Stay the same
d. The reaction will ____________ if B2 is depleted from the system. (4
points)
i. Shift to the right
ii. Shift to the left
iii. Not shift either way
________________/26
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CHEM142 EXAM II W2009
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7. Arrange the following substances in order of increasing miscibility in water:
HOCH2OH,CH3OH, CH3OCH3, and CH3CH2CH3. (8 points)
CH3CH2CH3, CH3OCH3, CH3OH, HOCH2OH
8. Define, in your own words, the concept of equilibrium. (4 points)
When the rate of the product formation = the rate of the reactant formation
9. A solution of hydrobromic acid is 30.0% acid by weight. State the molarity and
molality of the solution in addition to the mole fraction of hydrobromic acid, and
the mole percent, given the MW of HBr = 80.91 g/mol, the MW of H2O = 18.02
g/mol, a density of solution = 1.50 g/mL. (20 points)
1molHBr 1.50 g soln 1000mL solution
1
30.0 gHBr 



 5.56M
80.91gHBr 1mL soln
1L solution
100.0 g soln
1molHBr
1
1000 g water
30.0 gHBr 

 5.30m
80.91gHBr 70.0g water 1kg water
1molHBr
30.0 gHBr 
 0.371 mol HBr
80.91gHBr
1mol water
70.0 g water 
 3.88 mol water
18.02g water
0.371mol
X HBr 
 0.0873
0.371mol  3.88mol water
mol %  8.73%
_______________/32
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CHEM142 EXAM II W2009
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10. A mass of 0.150 g of protein dissolved in 100.0 mL of H2O at 300.0 K supports
an osmotic pressure of 768 mm Hg. Compute the molecular mass of this protein.
(10 points)
  MRT
M= 
(768mmHg 
RT

1
)
760mmHg
(0.08206 L  atm
mol  K
)(300.0 K )
 0.0410 mol
0.0410 mol  0.1000 L  0.00410mol
L
0.150 g
g
 36.6
0.00410mol
mol
11. A 0.171 g sample of an unknown organic compound is dissolved in ether. The
solution has a total mass of 2.470 g. The boiling point of the solution is found to
be 36.43 °C. What is the molar mass of the organic compound? 34.60 °C =
boiling point of ether & Kbp = 2.02 °C/m (10 points)
T  K  m  i
(36.43 C  34.60 C )  2.02
0.906
C
 m 1
m
mol
m
kg
2.470 g solution-0.171g solute = 2.299g solvent = 2.299 10-3kg
mol
 2.299 10-3kg=0.00208 mol solute
kg
0.171g
 82.2 g
mol
0.00208mol
0.906
__________/20
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CHEM142 EXAM II W2009
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L
12. Consider the following equation:
HX(aq) + MB(aq)  HB(aq) + MX(s)
If a solution initially contains 0.0054 M HX and 0.0068 M MB, what is the
equilibrium concentration of the HB at STP, given Kc = 1.9 x 10-9? (12 points)
I
0.0054 0.0068
0
C
-x
-x
+x
E
[0.0054-x]
[0.0068-x]
x
[ x]
[ x]
K c  1.9 109 

[0.0054  x][0.0068  x] [0.0054][0.0068]
x  7.0 1014 M
5%rule=
7.0 1014
100  1.3  109 %
0.0054
13. For the following reaction at equilibrium
1st reaction: A(g) + 3B(g)  AB3(g); K1 = 0.060
2nd reaction: AB3(g) + C(g)  AB2(g) + BC(g); K2 = 4.9 x 108
Net reaction: 2AB2(g) + 2BC(g)  2A(g) + 6B(g) + 2C(g); Knet = ?
K net
a. State the numeric value for Knet.(10 points)
1
1
 ( ) 2  ( ) 2  1.2 1015
K1
K2
________/22
=MRT; R = 0.08206 (L x atm)/(mol x K)
T=K x m x i
Psolution = Xsolvent x P°solvent
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