# Liquids A and B form an ideal solution. The vapour pressure of A and B at 100$^{\large\circ}$C are 300 and 100 mm Hg respectively. Suppose that vapour above solution is composed of 1 mole of A and 1 mole of B is collected and condensed. This condensate is then heated at 100$^{\large\circ}$C and vapour is again condensed to form a liquid L. What is the mole fraction of A in the vapours of L?

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.
answered Jun 9, 2014