Question

A 0.40g sample of a polypeptide dissolved in 1.0 L of an aqueous solution at 27 degrees C gave rise to an osmotic pressure of 3.74 Torr. What is the molar mass of polypeptide?

Answer 1

P= 3.74 torr
= (3.74/760) atm
= 0.0049 atm
T= 27.0 oC
= (27.0+273) K
= 300 K


use:
P = C*R*T
0.0049 = C*0.0821*300.0
C =0.0002 M

volume , V = 1.0 L


use:
number of mol,
n = Molarity * Volume
= 0.0002*1
= 1.998*10^-4 mol

mass(solute)= 0.40 g


use:
number of mol = mass / molar mass
1.998*10^-4 mol = (0.4 g)/molar mass
molar mass = 2.002*10^3 g/mol
Answer: 2.00*10^3 g/mol