Question

Two liquids having vapour pressures $P_{1}^{\circ}$ and $P_{2}^{\circ}$ in pure state in the ratio of $2: 1$ are mixed in the molar ratio of $1: 2$. The ratio of their moles in the vapour state would be

• A. $1: 1$
• B. $1: 2$
• C. $2: 1$
• D. $3: 2$

Answer 1

Answer: (A)

$P_{1}^{\circ}: P_{2}^{\circ}=\frac{2}{1}$
$\frac{n_{1}}{n_{2}}=\frac{1}{2}$
$\therefore P _{1}= P _{0}^{1}\, x _{1}$
$P _{2}= P _{2}^{\circ} \,x _{2}$
$x _{1}=\frac{ n _{1}}{ n _{1}+ n _{2}}=\frac{1}{1+2}=\frac{1}{3}$
$P_{1}^{0} \,x_{1} ; P_{2}^{0} \,x_{2}$
$\Rightarrow 2 \times \frac{1}{3} ; 1 \times \frac{2}{3}$
$\Rightarrow \frac{2}{3} ; \frac{2}{3}$
$x _{2}=\frac{ n _{2}}{ n _{1}+ n _{2}}=\frac{2}{1+3}=\frac{2}{3}$
$\therefore P _{1}^{0} \,x _{1}=\frac{2}{3}$
$P _{2}^{0} x _{2}=\frac{2}{3}$
$\frac{y_{1}}{y_{2}}=\frac{P_{1} \times P_{\text {total }}}{P_{\text {total }} \times P_{2}}$
$y=$ mole fraction in vapor phase
$\Rightarrow \frac{ P _{1}^{0} x _{1}}{ P _{2}^{0} x _{2}} \Rightarrow \frac{\frac{2}{3}}{\frac{2}{3}} $
$\Rightarrow 1: 1$