Comprehension:
Solutions
Let us consider a binary solution of two volatile liquids and denote the two components as 1 and 2. When taken in a closed vessel, both the components would evaporate and eventually an equilibrium would be established between vapour phase and the liquid phase. Let the total vapour pressure at this stage be ptotal and p1 and p2 be the partial vapour pressures of the two components 1 and 2 respectively. These partial pressures are related to the mole fractions x1 and x2 of the two components 1 and 2 respectively. The French chemist, Francois Marie Raoult (1886) gave the quantitative relationship between them. The relationship is known as the Raoult's law which states that for a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction (x1) present in solution.
The vapour pressure of pure liquids P and R are 957 and 379.5 mm Hg, respectively at 350 K. The composition of the P and R mixture if total vapour pressure is 760 mm Hg will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → xP = 0.66, xR = 0.34
For a solution of two volatile liquids, Raoult’s law gives the partial vapour pressure of each as its pure vapour pressure times its mole fraction. The total vapour pressure is the sum, and since xR = 1 − xP, it can be written with one mole fraction only:
\( \displaystyle p_{total}=p_P^{\circ}\,x_P+p_R^{\circ}\,x_R \)
\( \displaystyle p_{total}=p_R^{\circ}+(p_P^{\circ}-p_R^{\circ})\,x_P \)
Here pP° = 957 mm Hg, pR° = 379.5 mm Hg and ptotal = 760 mm Hg.
\( \displaystyle 760=379.5+(957-379.5)\,x_P \)
\( \displaystyle 380.5=577.5\,x_P \)
\( \displaystyle x_P=\frac{380.5}{577.5}=0.66 \)
\( \displaystyle x_R=1-0.66=0.34 \)
So the composition is xP = 0.66, xR = 0.34.
Shortcut: 760 mm Hg is much closer to 957 than to 379.5, so the mixture must be richer in P. Only Options (1) and (4) have xP > 0.5, and 957 × 0.66 + 379.5 × 0.34 = 760 mm Hg confirms Option (1).
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Intext Questions – “The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.”