The vapour pressures of chloroform \((CHCl_3)\) and dichloromethane \((CH_2Cl_2)\) at 298 K are 400 mm Hg and 615 mm Hg, respectively. What is the mole fraction of \(CHCl_3\) in the vapour phase of the solution prepared by mixing 30 g of \(CHCl_3\) and 45 g of \(CH_2Cl_2\)?
[Consider the mass of the given atoms as: C = 12.01 u, H = 1.01 u and Cl = 35.45 u]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.23
By Raoult’s law the partial vapour pressure of each component is pi = xipi°; by Dalton’s law the mole fraction of a component in the vapour is yi = pi/ptotal.
Molar mass of \( \ce{CHCl3} \) = 12.01 + 1.01 + 3 × 35.45 = 119.37 g mol−1
Molar mass of \( \ce{CH2Cl2} \) = 12.01 + 2 × 1.01 + 2 × 35.45 = 84.93 g mol−1
Moles of \( \ce{CHCl3} \) = 30/119.37 = 0.2513 mol
Moles of \( \ce{CH2Cl2} \) = 45/84.93 = 0.5298 mol
Total = 0.7811 mol
x(CHCl3) = 0.2513/0.7811 = 0.322; x(CH2Cl2) = 0.678
p(CHCl3) = 0.322 × 400 = 128.7 mm Hg
p(CH2Cl2) = 0.678 × 615 = 417.1 mm Hg
ptotal = 128.7 + 417.1 = 545.8 mm Hg
y(CHCl3) = 128.7/545.8 = 0.236 ≈ 0.23
The vapour is richer in the more volatile component, \( \ce{CH2Cl2} \), so the mole fraction of \( \ce{CHCl3} \) in the vapour (0.236) is lower than in the liquid (0.322).
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):
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.”