Chemistry II Molar Mass of Polymer

Question

A 276 mL aqueous solution containing 0.400 g of an unknown polymer has an osmotic pressure of 26.9 torr at 30.0 °C. What is the molar mass of the polymer?

Answer 1

Colligative Properties:

Osmotic Pressure

π = iMRT

π = osmotic pressure = 26.9 torr x 1 atm / 760 torr = 0.03539 atm

i = van't Hoff factor = 1 for a polymer

M = molarity = moles of polymer/liter of solution = ?

R = gas constant = 0.0821 Latm/Kmol (this is why we changed torr to atm above)

T = temperature in K = 30.0ºC + 273 = 303K

Solving for M we have...

M = (0.03539) / (1)(0.0821)(303)

M = 0.001423 moles/L

Since we have only 276 ml = 0.276 L we can find the moles of polymer...

moles polymer = 0.001423 mol/L x 0.276 L = 0.0003927 moles

The mass that was used = 0.400 g, and so we can now determine the molar mass

molar mass = g / moles = 0.400 / 0.0003927 moles = 1019 g/mol = 1020 g/mol (3 sig. figs.)