Vapour pressure of CHCl_3 and CH_2Cl_2 at 298 K are 200 mmHg and 415…

The correct answer is Option (4) → 347.9 mmHg

For an ideal solution of two volatile liquids, Raoult’s law gives the total vapour pressure from the mole fractions:

    \( \displaystyle p_{\text{total}} = x_1 p_1^\circ + x_2 p_2^\circ \)

    \( \displaystyle M(\ce{CHCl3}) = 12 + 1 + 3 \times 35.5 = 119.5\ \mathrm{g\,mol^{-1}} \)

    \( \displaystyle M(\ce{CH2Cl2}) = 12 + 2 + 2 \times 35.5 = 85\ \mathrm{g\,mol^{-1}} \)

    \( \displaystyle n(\ce{CHCl3}) = \frac{25.5}{119.5} = 0.213\ \mathrm{mol},\quad n(\ce{CH2Cl2}) = \frac{40}{85} = 0.47\ \mathrm{mol} \)

    \( \displaystyle x(\ce{CH2Cl2}) = \frac{0.47}{0.683} = 0.688,\quad x(\ce{CHCl3}) = 1 - 0.688 = 0.312 \)

    \( \displaystyle p_{\text{total}} = 0.312 \times 200 + 0.688 \times 415 = 62.4 + 285.5 = 347.9\ \text{mm Hg} \)

So the vapour pressure of the solution is 347.9 mm Hg.

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):
Vapour Pressure of Liquid-Liquid Solutions – “… Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K and, (ii) mole fractions of each component in vapour phase.”