JEE Main 2019 — Solutions Question with Solution

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 = $$\frac{x_A{P}_A^0}{P_s}$$ = $$\frac{0.5\times 400}{500}$$ = 0.4

Mole fraction of A in vapour phase = 0.4

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