dang (rtd356) – H03: Colligative Properties – Mccord – (52450)
1
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16
questions.
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before answering.
001
10.0 points
We dissolve 7.5 grams of urea (a nonelec-
trolyte with MW 60 g/mol) in 500 grams of
water. At what temperature would the solu-
tion boil?
1.
100.46
◦
C
2.
99.54
◦
C
3.
99.87
◦
C
4.
0.13
◦
C
5.
100.13
◦
C
correct
Explanation:
m
urea
= 7.5 g
m
H
2
O
= 500 g
Δ
T
b
=
K
b
m
=
K
b
mol urea
kg water
= (0
.
515
◦
C
/
m)
7
.
5
60
mol urea
0
.
5000 kg water
= 0
.
12
◦
C
T
b
=
T
0
b
+ Δ
T
= 100
.
12
◦
C
002
10.0 points
50 grams of NaNO
3
is dissolved in 538 grams
of water. What is the boiling point elevation
in degrees Celsius? Note that
K
b
for water is
0.512
◦
C
/m
. Assume complete dissociation of
the salt and ideal behavior of the solution.
Correct answer: 1
.
11962
◦
C.
Explanation:
m
NaNO
3
= 50 g
m
water
= 538 g
K
b
water = 0
.
512
◦
C
/m
NaNO
3
→
Na
+
+ NO
−
3
i
= 2
m
= (50 g NaNO
3
)
parenleftBig
1 mol NaNO
3
85
.
0 g NaOH
parenrightBig
= 0
.
588235 mol NaNO
3
Δ
T
b
=
K
b
·
m
·
i
=
parenleftbigg
0
.
512
◦
C
m
parenrightbigg parenleftbigg
0
.
588235
0
.
538
m
parenrightbigg
(2)
= 1
.
11962
◦
C
003
10.0 points
Which
of
the
following
aqueous
solutions
would exhibit the highest boiling point?
1.
0
.
1
m
urea
2.
0
.
1
m
CaCl
2
correct
3.
0
.
1
m
NaCl
Explanation:
Here the solution with the highest
effec-
tive
molality would have the highest boiling
point.
Urea does not ionize in water, so its
stated molality and effective molality should
be the same.
When NaCl dissolves, 2 ions
are formed, but when CaCl
2
dissolves, 3 ions
are formed. The effective molality for CaCl
2
would then be (approximately) 3 times the
stated molality, while for NaCl the effective
molality would only be (approximately) twice
the stated molality.
Thus, we would expect
that CaCl
2
would exhibit the highest boiling
point.
004
10.0 points
The freezing point of an aqueous solution con-
taining a nonelectrolyte dissolved in 240 g of
water is
-
1
.
6
◦
C. How many moles of solute
are present? Given that
k
f
is 1
.
86 K
·
kg
/
mol.
Correct answer: 0
.
206452 mol.
Explanation:
k
f
= 1
.
86 K
·
kg
/
mol
Δ
T
f
=
-
1
.
6
◦
C = 1
.
6 K
m
solvent
= 240 g = 0
.
24 kg
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dang (rtd356) – H03: Colligative Properties – Mccord – (52450)
2
Δ
T
f
=
k
f
m
=
k
f
n
solute
m
solvent
n
solute
=
Δ
T
f
m
solvent
k
f
=
(1
.
6 K) (0
.
24 kg)
1
.
86 K
·
kg
/
mol
= 0
.
206452 mol
.

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- Spring '07
- Holcombe
- Chemistry, Colligative properties, Freezing-point depression
-
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