If a 500 mL solution contains 2 g of protein at 400 K has an osmotic…

The correct answer is Option (1) → 80,000 g mol⁻¹

Osmotic pressure (π) is a colligative property. For a dilute solution, it follows the van’t Hoff equation, where C is the molar concentration. Writing C as moles of solute (w/M) per litre of solution (V):

    \( \displaystyle \pi=CRT=\frac{w}{M}\times\frac{RT}{V} \)

    \( \displaystyle M=\frac{wRT}{\pi V} \)

Here w = 2 g, V = 500 mL = 0.5 L, T = 400 K, π = 1.66 × 10⁻³ bar, and R = 0.083 L bar K⁻¹ mol⁻¹.

    \( \displaystyle M=\frac{2\times0.083\times400}{1.66\times10^{-3}\times0.5} \)

    \( \displaystyle M=\frac{66.4}{8.3\times10^{-4}} \)

    \( \displaystyle M=80000\ \mathrm{g\,mol^{-1}} \)

So the molar mass of the protein is 80,000 g mol⁻¹.

Shortcut: 2 × 0.083 = 0.166, which is 100 times 1.66 × 10⁻³. The rest is 400/0.5 = 800, so M = 100 × 800 = 80,000 g mol⁻¹.

Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Osmosis and Osmotic Pressure – “… 200 cm³ of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10^–3 bar. Calculate the molar mass of the protein.”