Question

A mixture of (a) and (b), which are two miscible liquids, is distilled under equilibrium conditions at atmospheric pressure. The mole fraction of (a) in solution is 0.3 and in vapour phase is 0.6. If the solution behaves ideally the ratio of $p_a^{\circ }$ to $p_b^{\circ }$ is :

• A. $1.5$
• B. $3.5$
• C. $2.5$
• D. $4.0$
• E. $7$

Answer 1

The correct option is B $3.5$

A/q in solution phase $X_a=0.3,X_b=1-0.3=0.7$

A/q in gaseous phase $Y_a=0.6,Y_b=1-0.6=0.4$

Now, using daltons law,

$Y_a=P_a/P_{total}$ and $Y_b=P_b/P_{total}$

and $P=P^{\circ }\times X$

Therefore, $Y_a=(X_a\times P_a^{\circ })/P_{total}and{Y}_b=(X_b\times P_b^{\circ })/P_{total}$

or $Y_a/Y_b=\left(P_a^{\circ }{X}_a\right)/\left(P_b^{\circ }{X}_b\right)$

$\Rightarrow 6/4=ratio\times 3/7$ where, $ratio=\frac{P_a^{\circ }}{P_b^{\circ }}$

$\Rightarrow ratio=14/4=3.5$

Option B is correct.