Version 042 – Make up Exam 1 – Laude – (52390)
1
This
print-out
should
have
28
questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
6.0 points
Rank the following liquids in increasing
order of solubility in water (H
2
O):
CH
3
F,
(CH
3
)
2
NH, CH
3
Cl, CBr
4
.
1.
(CH
3
)
2
NH
<
CBr
4
<
CH
3
Cl
<
CH
3
F
2.
CBr
4
<
CH
3
Cl
<
CH
3
F
<
(CH
3
)
2
NH
correct
3.
CH
3
Cl
<
CH
3
F
<
(CH
3
)
2
NH
<
CBr
4
4.
CH
3
F
<
(CH
3
)
2
NH
<
CBr
4
<
CH
3
Cl
Explanation:
According the axiom that ”like dissolves
like,” the least miscible will be the non-polar
molecule CBr
4
.
The most miscible will be
(CH
3
)
2
NH, which exhibits hydrogen bonding
just as water does.
Between the two polar
molecules, CH
3
F is the more polar and thus
the more miscible because of the greater elec-
tronegativity difference.
002
6.0 points
Which of the following statements is/are
always
true concerning
K
w
?
I) It gets larger as the temperature in-
creases
II) It equals 10
−
14
III)
K
w
= [H
+
][OH
−
]
1.
I, II, III
2.
II, III
3.
I only
4.
II only
5.
I, III
correct
6.
III only
7.
I, II
Explanation:
All equilibrium processes are temperature
dependent, and because auto-protolysis is en-
dothermic,
K
w
increases as temperature in-
crease.
Thus statement I is true, but state-
ment II is only true at room temperature.
Statement III is the definition of
K
w
.
003
6.0 points
An unknown liquid has a vapor pressure of
88 mmHg at 45
◦
C and a heat of vaporization
of 32 kJ
/
mol, what is its vapor pressure at
55
◦
C?
1.
156 mmHg
2.
127 mmHg
correct
3.
61 mmHg
4.
89
.
4 mmHg
Explanation:
Use
the
Clausius-Clapeyron
Equation.
Here, the only thing we don’t know is
P
2
.
ln
parenleftbigg
P
2
P
1
parenrightbigg
=
Δ
H
vap
R
parenleftbigg
1
T
1
−
1
T
2
parenrightbigg
ln
parenleftbigg
P
2
88
parenrightbigg
=
32
,
000
8
.
31
parenleftbigg
1
318
−
1
328
parenrightbigg
ln
P
2
=
32
,
000
8
.
31
parenleftbigg
1
318
−
1
328
parenrightbigg
+ ln 88
P
2
=
e
4
.
84
= 127 mmHg
004
6.0 points
What would be the pH of a solution of hy-
poiodous acid (HOI) prepared by dissolving
144 grams of the acid in 200 mL of pure
water (H
2
O)? The Ka of hypoiodous acid is
2
×
10
−
11
1.
7
2.
13
3.
1
This
preview
has intentionally blurred sections.
Sign up to view the full version.
Version 042 – Make up Exam 1 – Laude – (52390)
2
4.
10
5.
5
correct
Explanation:
144 g HOI
×
1 mol
144 g
= 1 mol HOI
1 mol HOI
0
.
2 L H
2
O
= 5 M HOI
[H
+
] = (K
a
·
C
a
)
1
/
2
= (2
×
10
−
11
·
5)
1
/
2
= (10
−
10
)
1
/
2
= 10
−
5
pH =
−
log[H
+
] =
−
log(10
−
5
) = 5
005
6.0 points
Which of the following would lower the va-
por pressure of a sample of water in a closed
container?
I) decreasing the size of the container
II) lowering the container temperature
III) removing water from the container
1.
I only
2.
II and III only
3.
I and III only
4.
III only
5.
I and II only
6.
I, II and III
7.
II onlt
correct
Explanation:
The size of a container and the amount of
the volatile species contained in it are irrel-
evant to vapor pressure.
Temperature is a
strong determinant vapor pressure.

This is the end of the preview.
Sign up
to
access the rest of the document.
- Spring '09
- Laude
-
Click to edit the document details