Question

A mixture of two miscible liquids $A$ and $B$ is distilled under equilibrium conditions at $1$ 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. $4.0$
• B. $3.5$
• C. $2.5$
• D. $1.85$

Answer 1

The correct option is B $3.5$

In solution $x_A$ $=0.30;$ $x_B=0.70$

In vapour phase, $y_A=0.60;y_B=0.40$
Using Dalton's law and Raoult's law

$y_A=0.60=\frac{P_A}{P_T}=\frac{P_A}{P_A+P_B}=\frac{0.30p_A^o}{0.30p_A^o+0.70p_B^o}$

$y_B=\frac{0.70p_B^o}{0.30p_A^o+0.70p_B^o}$

$\frac{y_A}{y_B}=\frac{0.60}{0.40}=\frac{0.30p_A^o}{0.70p_B^o}$

$\frac{p_A^o}{p_B^o}=\frac{0.60\times 0.70}{0.40\times 0.30}=3.5$

Option B is correct.