Question

A certain liquid mixture of two liquids A and B (behaving ideally) has a vapour pressure 70 torr (1 torr=1 mm) at ${25}^{o}C$ for a certain mole fraction X of A. For the same mole fraction X for B in the mixture, the vapour pressure of mixture is 90 torr at ${25}^{o}C$. If $P_A^0-P_B^0=40$ torr, calculate $P_A^0,P_B^0$ and X :

• A. $\text{100 torr, 60 torr, 0.25}$
• B. $\text{120 torr, 80 torr, 0.25}$
• C. $\text{200 torr, 160 torr, 0.25}$
• D. $\text{220 torr, 180 torr, 0.25}$

Answer 1

The correct option is A $\text{100 torr, 60 torr, 0.25}$

$70=P_A^{0}X+(1-X)P_B^0$......(1)

$90=P_A^0(1-X)+P_B^{0}X$......(2)

Subtracting equation (1) from equation (2):

$90-70=P_A^0-2P_A^{0}X+2P_B^{0}X-P_B^0$

$20=P_A^0-P_B^0-2X(P_A^0-P_B^0)$

But $P_A^0-P_B^0=40$

$20=40-2X\left(40\right)$

$80X=20$

$X=0.25$

Combining equation (1) and (2):

$70+90=P_A^0+P_B^0$......(3)

Also, $P_A^0=40+P_B^0$......(4)

Substitute equation (4) in equation (3)

$160=40+P_B^0+P_B^0$

$120=2P_B^0$

$P^{0}B=60torr$

$P_A^0=160-P_B^0=160-60=100torr$

Hence, option A is correct.