QJEEM19Ch3448: JEE Main 2019 Chemistry - Class 12 JEEM19Ch3448 Solution: The vapour pressures of pure liquids A and B are 400 and 600…

Given

$P_A^0$ = 400 mm of Hg

$P_B^0$ = 600 mm of Hg

Mole fraction of B (x_B) in liquid phase = 0.5

Mole fraction of A (x_A) in liquid phase = 1 - 0.5 = 0.5

Pressure of solution :

P_s = x_A $P_A^0$ + x_B $P_B^0$

= 0.5 $\times$ 400 + 0.5 $\times$ 600

= 500 mm of Hg

From Roult's law we know,

Partial pressure of A (P_A) = x_A $P_A^0$ ......(1)

From Dalton'sRoult's law we know,

Partial pressure of A (P_A) = y_AP_s ........(2)

here y_A = mole fraction of A in vapour phase

From (1) and (2) we can write,

x_A $P_A^0$ = y_AP_s

$\Rightarrow$ y_A = ${{{x_A}P_A^0} \over {{P_s}}}$ = ${{0.5 \times 400} \over {500}}$ = 0.4

Mole fraction of A in vapour phase = 0.4

$\therefore$ mole fraction of B in vapour phase = 1 - 0.4 = 0.6