Molar Mass from osmotic pressure

Question

When 2.63 g of a polypeptide is dissolved in 237 mL of water, the resulting solution is found to have an osmotic pressure of 0.125 atm at 37.0 °C. What is the molar mass of the polypeptide? (Assume the volume doesn't change when the polypeptide is added.)

Normally I try to include what work I already did to get started, but this time I'm really and completely lost.

Answer 1

So, you can begin with the standard equation for osmotic pressure:

osmotic pressure (π) = iMRT

π = 0.125 atm

i = van't Hoff factor = 1 for a polypeptide since it does not ionize or dissociate

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

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 37ºC + 273 = 298K

We now solve for M (moles/liter):

M = π / (i)(R)(T) = (0.125) / (1)(0.0821)(298)

M = 0.00511 moles/liter

Since we have 237 ml, that is the same as 0.237 L, so we can find the moles of polypeptide:

moles = 0.00511 moles/L x 0.237 L = 0.00121 moles

Molar mass = g/mol = 2.63 g / 0.00121 moles

Molar mass = 2174 g/mol