Osmotic pressure

Question

A solution of 0.5grams of polypeptide is 1dm³ of aqueous solution has an osmotic pressure of 3.74torr at 27•c. What is the molar mass of the poly mass

Answer 1

π = iMRT

π = osmotic pressure = 3.74 torr x 1 atm / 760 torr = 0.004921 atm

i = van't Hoff factor = 1 for a polypeptide (non electrolyte)

M = molarity = moles polypeptide / liter = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 27C + 273 = 300K

0.004921 atm = (1)(M)(0.0821 Latm/Kmol)(300K)

M = 0.004921 / 24.63

M = 1.998x10^-4 moles / L

Since we have 1dm³ (same as 1 L), we have 1.998x10^-4 moles of the polypeptide

Since this is equal to the 0.5 grams used, we can find the molar mass

Molar mass = grams / moles = 0.5 g / 1.998x10^-4 moles = 2503 g / mole