Question

A solution containing 27.20 mg of an unknown protein per 29.5 mL solution was found to have an osmotic pressure of 3.70 torr at 27 ∘C. What is the molar mass of the protein?

Answer 1

Expression for osmotic pressure (P) can be written as follows:

P = iMRT

Where, i = 1 (protein does not dissociate, hence i = 1)

M = molarity

R = gas constant = 0.082067 L atm K^-1 mol^-1

P = osmotic pressure = 3.70 torr = 3.70 torr x 1 atm / 760 torr = 4.868 x 10^-3 atm

T = temperature = 27 ∘C = (27 + 273.15) K = 300.15K

Therefore, molarity can be calculated as follows:

M = P / iRT

M = 4.868 x 10^-3 atm / (1 x 0.082067 L atm K^-1 mol^-1 x 300.15K)

M = 1.976 x 10^-4 mol/L

Hence, number of moles of protein in given 29.5 mL of solution can be calculated as follows:

No. of moles = 29.5 mg x (1g/1000mg) x (1.976 x 10^-4 mol/L)

No. of moles = 5.829 x 10^-6 mol

Thus, molar mass of the 27.20mg of protein can be calculated as follows:

Molar mass = Mass/ moles

Molar mass = 27.20 mg x (1g/1000mg) / (5.829 x 10^-6 mol)

Molar mass = 4666 g/mol

Therefore, molar mass of given protein is 4666 g/mol