The vapour pressure of pure benzene at a certain temperature is 0.850…

The correct answer is Option (2) → 340 g mol⁻¹

A non-volatile solute lowers the vapour pressure of the solvent, because fewer solvent molecules are at the surface. By Raoult’s law, the relative lowering of vapour pressure equals the mole fraction of the solute:

    \( \displaystyle \frac{p^0-p}{p^0}=x_2=\frac{n_2}{n_1+n_2} \)

For a dilute solution, n₂ is much smaller than n₁, so n₁ + n₂ ≈ n₁. Writing n = w/M:

    \( \displaystyle \frac{p^0-p}{p^0}=\frac{n_2}{n_1}=\frac{w_2\times M_1}{M_2\times w_1} \)

    \( \displaystyle M_2=\frac{w_2\times M_1\times p^0}{w_1\times(p^0-p)} \)

The data give p⁰ = 0.850 bar and p = 0.845 bar, so the lowering p⁰ − p = 0.005 bar. Also w₂ = 1.0 g, w₁ = 39.0 g and M₁ = 78 g mol⁻¹.

    \( \displaystyle M_2=\frac{1.0\times78\times0.850}{39.0\times0.005} \)

    \( \displaystyle M_2=\frac{66.3}{0.195}=340\ \mathrm{g\,mol^{-1}} \)

So the molar mass of the solute is 340 g mol⁻¹. Using the exact expression (with n₁ + n₂) gives about 338 g mol⁻¹, which is why the question says “around”.

Shortcut: 78/39 = 2, so M₂ = 2 × 0.850/0.005 = 2 × 170 = 340.

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):
Relative Lowering of Vapour Pressure – “… Example 1.6: The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol^–1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?”