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: Write the osmotic pressure formula. For a dilute solution \(\pi = \dfrac{n}{V}RT = \dfrac{wRT}{MV}\), where w is mass of solute, M its molar mass, V the volume in litres, R the gas constant and T the temperature.
Step 2: Rearrange for molar mass: \(M = \dfrac{wRT}{\pi V}\).
Step 3: List the data: w = 1.26 g, R = 0.083 L bar K^-1 mol^-1, T = 300 K, \(\pi = 2.57 \times 10^{-3}\) bar, V = 200 cm³ = 0.200 L.
Step 4: Substitute: \(M = \dfrac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.200}\).
Step 5: Arithmetic. Numerator = 1.26 x 0.083 x 300 = 31.374. Denominator = \(2.57 \times 10^{-3} \times 0.200 = 5.14 \times 10^{-4}\).
\(M = \dfrac{31.374}{5.14 \times 10^{-4}} = 6.10 \times 10^{4}\) g mol^-1.
\[\boxed{M \approx 61039 \ \text{g mol}^{-1}}\]