Question

At $T(K)$, the vapour pressures of pure liquids $A$ and $B$ are $100mm$ and $160mm$ respectively. An ideal solution is formed by mixing $2$ moles of $A$ and $3$ moles of $B$ at the same temperature. The mole fraction of $A$ and $B$ in the vapour state respectively are

• A. $0.706,0.294$
• B. $0.294,0.706$
• C. $0.40,0.60$
• D. $0.60,0.40$

Answer 1

Answer: (B)

Vapour pressure of solution,

$p_{\text{total }}=p_A+p_{B^{\prime }}={\chi }_A{p}_A^{\circ }+{\chi }_B{p}_B^{\circ }$

$d\left[\because p_A={\chi }_A{p}_A^{\circ }\right]$

Also vapour pressure of component $1,p_1=y_1{p}_{\text{total }}$

where $y$, is the mole fraction of component 1 in vapour phase.

Given,

Vapour pressure of pure liquid $A,p_A^{\circ }=100mm$

Vapour pressure of pure liquid $B,p_B^{\circ }=160mm$

$\therefore$ Total vapour pressure of solution $=p_A+p_B$

$p_{\text{total }}={\chi }_A{p}_A^{\circ }+{\chi }_B{p}_B^{\circ }=\frac{2}{5}\times 100+\frac{3}{5}\times 160$

$=40+96=136mm$

Also $p_A=y_A{p}_{\text{total }}$

where, $y_A$ is the mole fraction of $A$.

Mole fraction of $A,y_A=\frac{p_A}{p_{\text{totinl }}}=\frac{40}{136}=0.294$

$\therefore y_B=1-y_A=1-0.294=0.706$