JEE Main 2012 — Chemistry Physical Chemistry

2012mcqmedium

Liquids A and B form an ideal solution. At $30^{\circ} \mathrm{C}$ , the total vapour pressure of a solution containing $1 \mathrm{~mol}$ of A and $2 \mathrm{~mol}$ of B is $250 \mathrm{~mm} \mathrm{Hg}$ . The total vapour pressure becomes $300 \mathrm{~mm} \mathrm{Hg}$ when 1 more mol of $\mathrm{A}$ is added to the first solution. The vapour pressures of pure $\mathrm{A}$ and $\mathrm{B}$ at the same temperature are

Official previous-year question

Held on 19 May 2012 · Verified 6 Jul 2026.

Options

A
$150,450 \mathrm{mmHg}$
B
$125,150 \mathrm{~mm} \mathrm{Hg}$
C
$450,150 \mathrm{~mm} \mathrm{Hg}$
D
$250,300 \mathrm{~mm} \mathrm{Hg}$

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Solution

Let vapour pressure of $A=P_A^0$ Vapour pressure of $B=P_B^0$ In first solution, Mole fraction of $A\left(x_A\right)=\frac{1}{1+2}=\frac{1}{3}$ Mole fraction of $B\left(x_B\right)=\frac{2}{1+2}=\frac{2}{3}$ According to Raoult's law, Total vapour pressure $\begin{aligned} & =250=P_A^0 x_A+P_B^0 x_B \\ & 250=\frac{1}{3} P_A^0+\frac{2}{3} P_B^0 \end{aligned}$ In second solution Mole fraction of $A\left(x_A\right)=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}$ Mole fraction of $B\left(x_B\right)=\frac{2}{4}=\frac{1}{2}$ $\therefore$ Total vapour pressure $\begin{aligned} & =300=P_A^0 x_A+P_B^0 x_B \\ & 300=\frac{1}{2} P_A^0+\frac{1}{2} P_B^0 \end{aligned}$ Multiplying equation (i) by $\frac{1}{2}$ and equation (ii) by $\frac{1}{3}$ $\begin{aligned} & \frac{1}{6} P_A^0+\frac{2}{6} P_B^0=125 \\ & \frac{1}{6} P_A^0+\frac{1}{6} P_B^0=100 \\ & \frac{1}{6} P_B^0=25 \\ & P_B^0=25 \times 6=150 \mathrm{~mm} \mathrm{Hg} \end{aligned}$ On substituting value of $P_B^0$ in equation (ii) we get $\begin{aligned} & 300=P_A^0 \times \frac{1}{2}+150 \times \frac{1}{2} \\ & P_A^0=450 \mathrm{~mm} \mathrm{Hg} \end{aligned}$

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JEE Main 2012 — Chemistry Physical Chemistry

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JEE Main 2012 — Chemistry Physical Chemistry

Options

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Solution

Your note

JEE Main Physical Chemistry in other years

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