Q9 / 90 · JEE Main 2012 (19 May Online) · Chemistry
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
- A$150,450 \mathrm{mmHg}$
- B$125,150 \mathrm{~mm} \mathrm{Hg}$
- Correct answerC$450,150 \mathrm{~mm} \mathrm{Hg}$
- D$250,300 \mathrm{~mm} \mathrm{Hg}$
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
$$
\begin{aligned}
& 300=P_A^0 \times \frac{1}{2}+150 \times \frac{1}{2} \\
& P_A^0=450 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.