Question

An ideal solution contains two volatile liquids $A(p^{\circ }=100$ torr $)$ and $B(p^{\circ }=200$ torr). If mixture contains $1$ mole of $A$ and $4$ moles of $B$, then total vapour pressure of the distillate is

• A. 150 torr
• B. 180 torr
• C. 188.88 torr
• D. 198.88 torr

Answer 1

Answer: (C)

$p_{\text{Total }}=p_A^{\circ }{X}_A+p_B^{\circ }{X}_B$
$=100\times \frac{1}{5}+200\times \frac{4}{5}$
$=20+160=180$
(Vapours are condensed and again brought in equilibrium
with vapours).
i.e for distillate
$y_A\to$ to be replaced by $x_A$
$y_B\to$ to be replaced by $X_B$.
$y_A=\frac{p_A^{\circ }{x}_A}{p_T}=\frac{20}{180}$
$y_B=\frac{p_B^{\circ }{x}_B}{p_T}=\frac{160}{180}$
$P_{\text{distillate }}=\frac{100\times 20}{180}+\frac{200\times 160}{180}$
$=188.88$