Vapour pressure due to vapours above solution

$P_T = X_AP_A^0 + X_BP_B^0$

$P_T = 300X_A + 100 X_B$

It is given that in vapour phase each of A and B are one mole each hence each of them have mole fraction 0.5 in vapour phase

After condensation of vapours

In condensate (1) $X′_A = 0.5, X′_B = 0.5 $

$P′_T = 0.5 \times 300 + 0.5 \times 100 = 150 + 50 = 200 mm$

Mole fraction of A and B in vapour phase of condensate

$Y'_A=\large\frac{P_A}{P'_T}$$=P_A^0=0.75$

$Y'_B=1-0.75=0.25$

When the vapours of the condensate (1) will again be vapourised in condensate (2) liquid L.

$X''_A=0.75$

$P''_T=300\times 0.75+100\times 0.25=225+25=250mm$

mole fraction of A in vapour phase of the condensate (2) is given by

$Y_A''=\large\frac{225}{250}$$=0.9$

Hence (A) is the correct answer.