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dang (rtd356) – H03: Colligative Properties – Mccord – (52450) 1
This print-out should have 16 questions.
Multiple-choice questions may continue on
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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

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 .