Question

A 200 cm³ aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure at 300 K was found to be \(2.57 \times 10^{-3}\) bar. Calculate the molar mass of the protein. (R = 0.083 L bar K^-1 mol^-1)

Hint:

Use \(M = \dfrac{wRT}{\pi V}\) with R = 0.083 L bar K^-1 mol^-1 and V in litres (0.200 L).

Answer 1

Step 1: First find the molar concentration. Osmotic pressure obeys \(\pi = CRT\), so \(C = \dfrac{\pi}{RT}\), where C is concentration in mol L^-1.
Step 2: \(C = \dfrac{2.57 \times 10^{-3}}{0.083 \times 300} = \dfrac{2.57 \times 10^{-3}}{24.9} = 1.032 \times 10^{-4}\) mol L^-1.
Step 3: Moles of protein in 0.200 L: \(n = C \times V = 1.032 \times 10^{-4} \times 0.200 = 2.064 \times 10^{-5}\) mol.
Step 4: Molar mass is mass divided by moles: \(M = \dfrac{1.26}{2.064 \times 10^{-5}}\).
Step 5: \(M = 6.10 \times 10^{4}\) g mol^-1.
\[\boxed{M \approx 61040 \ \text{g mol}^{-1}}\]