Q68 / 180 · AIIMS 2009 · Chemistry

A mixture of two miscible liquids $A$ and $B$ is distilled under equilibrium conditions at $1 \mathrm{~atm}$ pressure. The mole fraction of $A$ in solution and vapour phase are 0.30 and 0.60 respectively. Assuming ideal behaviour of the solution and the vapour, calculate the ratio of the vapour pressure of pure $A$ to that of pure $B$

  • A
    $4.0$
  • Correct answer
    B
    $3.5$
  • C
    $2.5$
  • D
    $1.85$

Solution

In solution, $x_A=0.30 ; x_B=0.70$ In vapour phase, $x_A^{\prime}=0.60 ; x_B^{\prime}=0.40$ Using Dalton's law and Raoult's law $\begin{aligned} & x_A^{\prime}=0.60=\frac{p_A}{P}=\frac{p_A}{p_A+p_B}=\frac{0.30 p_A^{\circ}}{0.30 p_A^{\circ}+0.70 p_B^{\circ}} \\ & x_B^{\prime}=0.40=\frac{p_B}{P}=\frac{p_B}{p_A+p_B}=\frac{0.70 p_B^{\circ}}{0.30 p_A^{\circ}+0.70 p_B^{\circ}} \\ & \frac{x_A^{\prime}}{x_B^{\prime}}=\frac{0.60}{0.40}=\frac{0.30 p_A^{\circ}}{0.70 p_B^{\circ}} \\ & \frac{p_A^{\circ}}{p_B^{\circ}}=\frac{0.60 \times 0.70}{0.40 \times 0.30}=\frac{7}{2}=3.5\end{aligned}$

Hence, the correct answer is option: 'B'.

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