A solution of 0.5 g solute 'X' in 39 g benzene records the vapor pres…

The correct answer is Option (3) → 170

By Raoult’s law, the relative lowering of vapour pressure of a solvent caused by a non-volatile solute equals the mole fraction of the solute:

    \( \displaystyle \frac{p_1^\circ-p_1}{p_1^\circ}=x_2=\frac{n_2}{n_1+n_2} \)

For a dilute solution n₂ is very small compared with n₁, so:

    \( \displaystyle \frac{p_1^\circ-p_1}{p_1^\circ}=\frac{n_2}{n_1}=\frac{w_2\times M_1}{M_2\times w_1} \)

Here w₂ = 0.5 g, w₁ = 39 g of benzene (M₁ = 78 g mol⁻¹), p°₁ = 0.850 bar and p₁ = 0.845 bar.

    \( \displaystyle \frac{0.850-0.845}{0.850}=\frac{0.5\times78}{M_2\times39} \)

    \( \displaystyle \frac{0.005}{0.850}=\frac{1}{M_2} \)

    \( \displaystyle M_2=\frac{0.850}{0.005}=170\ \mathrm{g\,mol^{-1}} \)

So the molar mass of the solute X is 170 g mol⁻¹.

Shortcut: 0.5 g of solute and 39 g (half a mole) of benzene make w₂M₁/w₁ equal to 1. The molar mass is then p°/(p° − p) = 0.850/0.005 = 170.

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Relative Lowering of Vapour Pressure – …