Question

The vapour pressure of a pure liquid A is 70 torr at 300 K. It forms an ideal solution with another liquid B. The mole fraction of B in the solution is 0.2 and total pressure of solution is 84 torr at 300 K. The vapour pressure of pure liquid B at 27C is ?

• A. 0.14 torr
• B. 560 torr
• C. 140 torr
• D. 70 torr

Answer 1

The correct option is C 140 torr

Solution:- (C) $140\text{torr.}$

Mole fraction of $B=0.2$

As we know that,

$x_A+x_B=1$

$\Rightarrow x_A=1-0.2=0.8$

${p^{\circ }}_A=70\text{torr.}$

${p^{\circ }}_B=?$

$p_{total}=84\text{torr.}$

From Dalton's law of partial pressure,

$p_{total}=p_A+p_B.....\left(1\right)$

From Raoult's law of vapour pressure,

$p_A={p^{\circ }}_A\cdot x_A$

$\Rightarrow p_A=70\times 0.8=56$

Again, from Raoult's law of vapour pressure,

$p_B={p^{\circ }}_B\cdot x_B$

$p_B={p^{\circ }}_B\times 0.2$

Substituting these values in equation $\left(1\right)$, we have

$84=56+0.2\times {p^{\circ }}_B$

$\Rightarrow {p^{\circ }}_B=\frac{84-56}{0.2}=140\text{torr.}$

Hence the vapour pressure of pure liquid $B$ is $140\text{torr.}$.