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AIIMS 2009
Exam
AIIMS
Year
2009
Subject
Chemistry
Chapter
Solutions

AIIMS 2009 · Chemistry · Solutions

A mixture of two miscible liquids A and B is distilled under...

A mixture of two miscible liquids AA and BB is distilled under equilibrium conditions at 1 atm1 \mathrm{~atm} pressure. The mole fraction of AA 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 AA to that of pure BB

Options

  1. A4.04.0
  2. B3.53.5
  3. C2.52.5
  4. D1.851.85

Correct answer

B. 3.53.5

Step-by-step solution

In solution, xA=0.30;xB=0.70x_A=0.30 ; x_B=0.70 In vapour phase, xA′=0.60;xB′=0.40x_A^{\prime}=0.60 ; x_B^{\prime}=0.40 Using Dalton's law and Raoult's law xA′=0.60=pAP=pApA+pB=0.30pA∘0.30pA∘+0.70pB∘xB′=0.40=pBP=pBpA+pB=0.70pB∘0.30pA∘+0.70pB∘xA′xB′=0.600.40=0.30pA∘0.70pB∘pA∘pB∘=0.60×0.700.40×0.30=72=3.5\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}
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