Question

A saturated solution is made by dissolving 0.325 g of a polypeptide (a substance formed by joining together in a chainlike fashion some number of amino acids) in water to give 1.37 L of solution. The solution has an osmotic pressure of 4.13 torr at 28 °C. What is the approximate molecular mass of the polypeptide? g/mol

Answer 1

pi = MRT
pi = 4.13 torr/760 = ? atm
M = unknown
R = 0.08205
T = 28 + 273 = ?
Solve for M in mols/L
You know M and L, solve for mols.
Then mol = grams/molar mass. Youj know grams and mols, solve for molar mass.

NOTE: In the future it would help if you would show what you know about the problem or how you think it should be approached. If you've done some calculations show that. It helps us do a better job of helping you.

Answer 2

So would you do M=RT/Pi
M=(301 x .08205)/.005
4939.41?

Answer 3

No, if pi = MRT then M = pi/RT
M = 0.00543/(0.08205*301) = about 0.000220M or mols/L.
M = mols/L
mols = M x L = 0.000220*1.37L = about 0.000301

mols = grams/molar mass or
molar mass = grams/mols = 0.325/0.000301 = ?
Note, however, that even if what you proposed were correct that you threw away two perfectly good numbers with that 0.005.
4.13/760 = 0.00543