Question

If the total vapour pressure of the liquid mixture A and B is given by the equation: $P=180X_A+90$ 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
$PT=P_A^o{X}_s+P_B^o{X}_A$
$=P_A^o{X}_A+P_B^o\left(1-X_A\right)$
$=\left(P_A^o-P_B^o\right)X_A+P_B^o\dots \left(1\right)$
But from question $P_T=180X_A+90\dots \left(2\right)$
Equating equation $\left(1\right)$ and $\left(2\right)$
$\therefore P_A^o-P_B^o=180$
$P_B^o=90;P_A^o=180+90=270$
$\frac{P_A^o}{P_B^o}=\frac{270}{90}=3:1$