EXAM 2: LB 172 PRACTICE PROBLEMS
JULIAN THWAINEY
1. The boiling point elevation of a solution of 0.177 mol of a sugar in 53
g of water is 1.713 °C. Calculate kb?
Δ
T
b
=
miK
b
K
b
=
molal boiling point elevation constant
−
different for every every liquid
−
different IMFs in each case!
K
b
>
0
m
=
mol of solute
kg of solvent
Calculate K
b
Δ
T
b
=
1.713
°
C
m
=
.177mol
.053kg
=
3.34
m
i
=
1 (sugar is molecular)
K
b
=
?
Δ
T
b
mi
=
K
b
1.713
1(3.34)
=
K
b
=
.513
°C
m
2.Cyclohexane has a freezing point of 6.50 °C and a Kf of 20.0 °C/m .
What is the freezing point of a solution made by dissolving 0.694g of
biphenyl (C12H10) in 25.0g of cyclohexane?
Δ
T
f
=
T
f
−
T
f
o
Δ
T
f
freezing point depression, must be negative,depends on # of dissolved particles
T
f
=freezing point of solution
T
f
o
=freezing point of pure solvent
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=
mol of solute
kg of solvent
Δ
T
f
=
miK
f
K
f
=
molal freezing point elevation constant
−
different for every every liquid
−
different IMFs in each case!
K
f
<
0
Calculate T
f
T
f
o
=
6.50
°
C
K
f
=
−
20.0
°
C/m
mol biphenyl (C
12
H
10
)=0.694g
×
1mol
154.2g
=
.00450mol biphenyl
m
=
.00450mol
.025kg
=
.180
m
T
f
=?
i
=
1 (biphenyl is molecular, contains only H and C)
Δ
T
f
=
1(.180)(
−
20)
=
−
3.60
−
3.60
=
T
f
−
6.50
T
f
= 2.90°C
3. An osmotic pressure device was used to determine the molar mass of
a protein. A mass of 0.150 g of protein dissolved in 100 mL of H2O
at 300 K supports a column of water 5.00 cm high by osmotic
pressure. It takes 10.0 m of this water to equal 1.00 atm. What is the
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 Spring '08
 all
 Rate equation, ΔTf, liquid −different IMFs

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