Question

If the total vapour pressure of the liquid mixture A and B is given by the equation: $P = 180 X_{A} + 90$ then the ratio of the vapour pressure of the pure liquids A and B is given by:

• A. $3 : 2$
• B. $4 : 1$
• C. $3 : 1$
• D. $6 : 2$

Answer 1

Answer: (C)

As we know
$PT=P_{A}^{o}X_{s}+P_{B}^{o}X_{A}$
$=P_{A}^{o}X_{A}+P_{B}^{o}\left(1-X_{A}\right)$
$=\left(P_{A}^{o}-P_{B}^{o}\right)X_{A}+P_{B}^{o} \ldots\left(1\right)$
But from question $P_{T}=180\,X_{A}+90\ldots\left(2\right)$
Equating equation $\left(1\right)$ and $\left(2\right)$
$\therefore P_{A}^{o}-P^{o}_{B}=180$
$P_{B}^{o}=90; P_{A}^{o}=180+90=270$
$\frac{P_{A}^{o}}{P_{B}^{o}}=\frac{270}{90}=3:1$