Question

At $T( K )$, the vapour pressures of pure liquids $A$ and $B$ are $100\, mm$ and $160 \,mm$ 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 $A, p_{A}^{\circ}=100 \,mm$

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

$\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 $

$=40+96=136 \,mm$

Also $p_{A}=y_{A} p_{\text {total }}$

where, $y_{A}$ is the mole fraction of $A$.

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

$\therefore \, y_{B}=1-y_{A}=1-0.294=0.706$