Question

6) 1.23 g of a polymer is dissolved in 175 mL of water. The resulting osmotic pressure is 16.9 torr at 33 C.
a) Determine the molecular weight of the polymer.
b) Then, assuming the density of water if 1.0 g/mL, determine by how much the freezing point of the solution has been depressed.

Answer 1

a)
P = 16.9 torr = 16.9/760 atm = 0.0222 atm
T = 33 oC = (33+273) K = 306 K
V = 175 mL = 0.175 L

use:
P = C*R*T
0.0222 = C*0.0821*306.0
C = 8.837*10^-4 M

Now use:
C = (mass)/(molar mass * volume)
8.837*10^-4 = 1.23/(MM*0.175)
MM = 7953.9 g/mol
Answer: 7953.9 g/mol

b)
Kf = 1.86 oC/m

number of mol = mass/molar mass
= 1.23 g / (7953.9 g/mol)
=1.55*10^-4 mol

since density of water is 1g/mL, mass of water = 175 g = 0.175 kg

molality = number of mol / mass of solute in Kg
= (1.55*10^-4 mol)/(0.175 Kg)
= 8.86*10^-4 m

ΔTf = Kf*m
= 1.86 * 8.86*10^-4
= 1.65*10^-3 oC

Answer: 1.65*10^-3 oC