Question

Mole fraction of A in vapours phase above the ideal liquid solution of A and B $(X_A=0.4)$ will be: $(P_A^O=100mm{P}_B^O=200mm)$ :

• A. 0.4
• B. 0.8
• C. 0.25
• D. None of these

Answer 1

The correct option is C 0.25

The option marked is incorrect. The correct option is C

Given that,

Mole fraction of A, $X_A$ = 0.4

So, the mole fraction of B, $X_A$ = 1 - 0.4 = 0.6

P =$P_A$ + $P_B$

P = $(P_A^O=(X_A)+P_B^O=(X_B\left)\right)$

P = 100(0.4) + 200(0.6)

P = 160 mm

Now, let the mole fraction of A in vapour phase be $Y_A$

$Y_A$ = $\frac{P_A}{P}$

$Y_A$ = 40/160

$Y_A$= 0.25

Therefore the mole fraction of A in vapour phase is 0.25