Calculate the vapour pressure (mm Hg) of the solution formed by mixing 25.5 g of A and 40.0 g of B at 298 K. The vapour pressure of A and B at 298 K are 200 mm Hg and 415 mm Hg, respectively. (Given: Molar mass of A =119.5 g/mol and B are 85.0 g/mol)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 347.9
For a solution of two volatile liquids, Raoult’s law states that the partial vapour pressure of each component equals its vapour pressure in the pure state multiplied by its mole fraction in the solution. The total vapour pressure is the sum of the partial pressures:
\( \displaystyle p_{total}=p_A^{\circ}\,x_A+p_B^{\circ}\,x_B \)
So the masses must first be converted to moles and then to mole fractions.
\( \displaystyle n_A=\frac{25.5}{119.5}=0.2134\ \mathrm{mol} \)
\( \displaystyle n_B=\frac{40.0}{85.0}=0.4706\ \mathrm{mol} \)
\( \displaystyle x_A=\frac{0.2134}{0.2134+0.4706}=\frac{0.2134}{0.6840}=0.312 \)
\( \displaystyle x_B=1-0.312=0.688 \)
\( \displaystyle p_{total}=200\times0.312+415\times0.688 \)
\( \displaystyle p_{total}=62.4+285.5=347.9\ \mathrm{mm\,Hg} \)
So the vapour pressure of the solution is 347.9 mm Hg.
Shortcut: Since xB = 1 − xA, Raoult’s law can be written with one mole fraction only: ptotal = pB° − (pB° − pA°) xA = 415 − 215 × 0.312 = 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 – …
