Question

An ideal solution contains two volatile liquids $A\left(p^{\circ}=100\right.$ torr $)$ and $B\left(p^{\circ}=200\right.$ 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
$y _{ A } \rightarrow$ to be replaced by $x _{ A }$
$y _{ B } \rightarrow$ 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} $
$=188.88$