Question

A solution containing ethyl alcohol and propyl alcohol has a vapour pressure of $290mmHg$ at ${30}^{\circ }C$. Find the vapour pressure of pure ethyl alcohol if its mole fraction in the solution is 0.65. The vapour pressure of propyl alcohol is $210mmHg$ at the same temperature.

• A. $396.12mmHg$
• B. $313.12mmHg$
• C. $233.12mmHg$
• D. $333.12mmHg$

Answer 1

The correct option is D $333.12mmHg$

Partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.
Given,
$P_{solution}=290mmHg$
$p_{P.A.}^{\circ }=210\text{mm Hg}[P.A\Rightarrow \text{Propyl alcohol}]X_{E.A}=0.65[E.A\Rightarrow \text{Ethyl alcohol}]\therefore X_{P.A}=1-X_{E.A}=1-0.65=0.35$
By Raoult's Law:

$p_{E.A}=X_{E.A}\times p_{E.A}^o$
$p_{P.A}=X_{P.A}\times p_{P.A}^o$
$P_{total}=X_{E.A}\times p_{E.A}^o+X_{P.A}\times p_{P.A}^o$
$\therefore P_{solution}=X_{E.A}\times p_{E.A}^{\circ }+X_{P.A}\times p_{P.A}^{\circ }290=0.65\times p_{E.A}^{\circ }+0.35\times 210p_{E.A}^o=\frac{290-73.5}{0.65}=333.12mmHg$