Colligative Properties – Worked Examples and Practice Problems
1.
The vapor pressure of pure benzene (C
6
H) is 100 torr at 26.1 oC. Calculate the vaporpressure of a solution containing 24.6 g of camphor (C
10
H
16
O) dissolved in 100 mL of benzene. The density of benzene is 0.877 g/mL.
Solution:
X ben
= n ben
/(n ben
+ n cam
) n ben
= 100. mL x (0.877 g/mL) x (1 mol/78.1 g) = 1.12 mol n cam
= 24.6 g x (1 mol/152.2 g) = 0.162 mol
X ben
= 1.12 mol/(1.12 mol + 0.162 mol) = 0.874
P ben
= (X ben
)(P o ben
) = (0.874)(100 torr) = 87.4 torr
2.
Ethylene glycol (EG), CH2(OH)CH2(OH), is a common automobile antifreeze.
Calculate the freezing point of a solution containing 651 g of EG in 2505 g of water.
Would you keep the substance in your car radiator during the summer? The molar mass of EG is 62.01 g/mol. K f
= 1.86 o
C/m and K b
= 0.52 o
C for water.
Solution: mol of EG = 651 g x (1 mol/62.10 g) = 10.5 mol molality of solution = 10.5 mol EG/2.505 kg of H
2
ΔT = K f
m = (1.86
o
C/m)(4.19 m) = 7.79
o
C
O = 4.19 m
Since pure water freezes at 0 oC, the solution will freeze at -7.79 oC.
The boiling point elevation can be calculated in the same way.
ΔT = K b
m = (0.52
o
C/m)(4.19 m) = 2.2
o
C
Because the solution will boil at 102.2
o
C, it would be preferable to the antifreeze in the car radiator in summer to prevent the solution from boiling.
3.
A 7.85 g sample of a compound with empirical formula C
5
H
4 is dissolved in 301 g of benzene. The freezing point of the solution is 1.05
o
C below that of the pure benzene.
What are the molar mass and molecular formula of this compound? K f for benzene is
5.12
o
C/m.
Solution:
Molality =
ΔT
/K f
= 1.05
o
C/(5.12
o
C/m) = 0.205 m
The number of moles of solute in 301 g or 0.301 kg of solvent is given by
(0.205 mol/1 kg solvent) x 0.310 kg solvent = 0.0617 mol
Molar mass of solute = 7.85 g/0.0617 mol = 127 g/mol
Empirical formula mass of solute = 64 g/mol
Molar mass of solute/ Empirical formula mass of solute = (127 g/mol)/(64 g/mol) = 2
Empirical formula is 2 x C
5
H
4
= C
10
H
8
(naphthalene)
4.
Calculate the concentration (in moldm
-3
) of urea (NH
2
CONH
2
) that has an osmotic pressure of 30.0 atm at 25 o
C.
Solution:
Π
= c RT
⇒ c =
Π
/RT = (30.0 x 101325 Pa)/[(8.314 JK
-1 mol
-1
)(25 + 273.15 K)] c = 1.23 x 10
-3
mol/L = 1.23 M

5.
A 50.00 mL sample of an aqueous solution contains 1.08 g of human serum albumin, blood-plasma protein. The solution has an osmotic pressure of 5.85 mmHg at 298 K.
What is the molar mass of albumin? Answer: 6.86 x 10
4
g/mol
6.
A chemist (like you!) is trying to identify a human hormone that controls metabolism by determining its molar mass. A sample weighing 0.546 g was dissolved in 15.0 g benzene, and the freezing point depression was determined to be 0.240
o
C. Calculate the molar mass of the hormone. Answer: 776 g/mol
7.
What concentration (in mol/L) of sodium chloride in water is needed to produce an aqueous solution isotonic with blood (osmotic pressure,
Π
= 7.70 atm at 25 o
C)?
Answer: 0.158 M
8.
The observed osmotic pressure of a 0.10 M solution of Fe(NH
4
)
2
(SO
4
)
2
at 25 o
C is
10.8 atm. Compare the expected and experimental values for i .
Answer:
Expected i value is 5
Experimental I value is 4.4
9.
Calculate the freezing point and the boiling point of each of the following aqueous solutions. Assume complete dissociation. K f
= 1.86
o
C/m; K b
= 0.51
o
C/m
(a) 0.050 m MgCl
2
(b) 0.050 m FeCl
3
Answers:
(a) T f
= - 0.28
o
C; T b
= 100.077
o
C
(b) T f
= - 0.37
o
C; T b
= 100.10
o
C
10.
From the following:
Pure water
Solution of C
12
H
22
O
11
(m = 0.01 molkg
-1
) in water
Solution of NaCl (m = 0.01 molkg
-1
Solution of CaCl2 (m = 0.01 molkg
) in water
-1
)
Choose the one with the
(a) Highest freezing point
(b) Lowest freezing point
(c) Highest boiling point
(d) Lowest boiling point
(e) Highest osmotic pressure