kamalska (mk23835) – HW02 – Quigley – (104001)
1
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30
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Multiplechoice questions may continue on
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001
10.0 points
What is the maximum possible value of thevan’t Hoff factor for a solution of the strongelectrolyte FeCl3that is so dilute that no ionassociation occur?
002
(part 1 of 3) 10.0 points
For the next three questions, consider thefollowing pressure diagram for a binary liquidcontaining A and B.01xA90 torr60 torrIf 1760 g of A (110 g/mol) and 1995 g ofB (105 g/mol) are mixed, what is the vaporpressure of B?1.90 torr2.27 torrcorrect3.60 torr4.48 torr5.76 torrExplanation:
4.
90 torr
5.
60 torr
Explanation:
2What is the corresponding total vapor pressure?
2.
60 torr
3.
48 torr
4.
76 torr
correct
5.
27 torr
Explanation:
0
1
0
.
46
60 torr
76 torr
005
10.0 points
Chemical processes, like dissolving solids, are
favored by (decreased, increased) potential
energy and (decreased, increased, unchanged)
disorder.
1.
increased; increased
2.
decreased; decreased
3.
increased; decreased
4.
10.0 points
A wellbehaved (nonvolatile, nonionic) so
lute is dissolved in water to yield a solution
that freezes at

0
.
930
◦
C. At what temper
ature would this solution boil under 1.0000
atm pressure?
Note that
K
f
= 1
.
86
◦
C/
m
,
K
b
= 0
.
512
◦
C
/m
.
1.
99.744
◦
C
2.
101.394
◦
C
3.
100.930
◦
C
4.
100.256
◦
C
correct
5.
100.000
◦
C
6.
99.070
◦
C
Explanation:
T
f
=

0
.
9306
◦
C
K
f
= 1
.
866
◦
C
/m
K
b
= 0
.
512
◦
C
/m
First, we have to determine the molality of
this solution. Knowing the freezing point will
lead us to the molality. Using the equation
Δ
T
f
=
K
f
m
and knowing that the freezing point of water is
0
◦
C, we know that the freezing point has been
depressed 0.930
◦
C. This means the molality
of this solution is
0
.
930
◦
C = (1
.
86
◦
C
/m
)(
m
)
m
= 0
.
500
m
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 Fall '08
 QUIGLEY
 Solubility, Vapor pressure, Freezingpoint depression