At 27°C, a solution containing 5 g of protein sample in 100 mL of solution exerts an osmotic pressure 0.4 bar. The molecular mass (in g mol−1, nearest integer) of the solute is (Given R = 0.083 L bar K−1 mol−1):
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3112
Osmotic pressure (π) of a dilute solution is given by the van’t Hoff equation. Writing the molar concentration as moles of solute (w/M) per volume V gives a direct route to the molar mass:
\( \displaystyle \pi=CRT=\frac{w}{M}\cdot\frac{RT}{V}\quad\Rightarrow\quad M=\frac{wRT}{\pi V} \)
Data: w = 5 g, T = 27 + 273 = 300 K, π = 0.4 bar, V = 100 mL = 0.1 L, R = 0.083 L bar K−1 mol−1.
\( \displaystyle M=\frac{5\times0.083\times300}{0.4\times0.1} \)
\( \displaystyle M=\frac{124.5}{0.04} \)
\( \displaystyle M=3112.5\approx3112\ \mathrm{g\,mol^{-1}} \)
So the molecular mass of the protein is about 3112 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 – …
