331 mg of an unknown protein (non electrolyte) is dissolved in enough water to make 5.00 ml of aqueous solution at 300K, with an osmotic pressure of 35.0 torr. What is the molar mass of the protein?

Solution: The equation for osmotic pressure π= iMRT where

π is osmotic pressure measured in atm,

i= number of ions of solute (1 for non electrolytes)

M = molarity which is equal to moles of solute/ L of solution

R = ideal gas constant (0.0821 atm•L/mol•K)

T = temperature in K

Let's first convert pressure into atm

35.0 torr * __1 atm = __0.04605atm

760 torr

i= 1 since it is a nonelectrolyte

Plugging into our equation, we get

0.04605atm = 1(M)(0.0821 atm•L/mol•K)(300K)

M = 0.00187mol/L

5.00 ml * __1L __ = 0.005L

1000ml

0.00187mol/L (0.005L) = 9.36*10^-6mol

Molar mass is measured in grams over moles. We have 331 mg = 0.331g

Now all we need to do I divide grams by moles

0.331g/9.36*10^-6mol = 35366g/mol = **3.54*10^4 g/mol**