The osmotic pressure of a solution prepared by dissolving 50 mg of H_…

The correct answer is Option (2) → 0.0187 atm

Osmotic pressure of a dilute solution is given by the van’t Hoff equation, with the van’t Hoff factor i for an electrolyte:

    \( \pi = iCRT \)

\( \ce{H2SO4} \) in such a dilute solution is taken as fully dissociated: \( \ce{H2SO4 -> 2H+ + SO4^2-} \), three ions per formula unit, so i = 3.

    \( \displaystyle n = \frac{50 \times 10^{-3}\ \text{g}}{98\ \text{g mol}^{-1}} = 5.10 \times 10^{-4}\ \text{mol} \)

    \( \displaystyle C = \frac{5.10 \times 10^{-4}}{2\ \text{L}} = 2.55 \times 10^{-4}\ \text{mol L}^{-1} \)

    \( \pi = 3 \times 2.55 \times 10^{-4} \times 0.0821 \times 298 \)

    \( \pi = 0.0187\ \text{atm} \)

So the osmotic pressure is 0.0187 atm. (The volume of water, 2 L, is taken as the volume of the very dilute solution.)

Note 1: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Exercises – “1.41 Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25° C, assuming that it is completely dissociated.”

Note 2: The same/similar question has been asked in different years by NTA as per the details given below: