Q: The vapour pressures of ethanol and methanol are 44.5mm and 88.7mm Hg respectively at the same temperature. An ideal solution is formed by mixing 60gm of ethanol and 40gm of methanol. The mole fraction of methanol in the vapour phase is.

(A) 0.66

(B) 0.55

(C) 0.11

(D) 0.33

Solution: Mole of CH_{3}CH = 40/32 = 1.25

Mole of CH_{3}CH_{2}OH = 60/46 = 1.304

Let CH_{3}CH → A, C_{2}H_{5}OH → B

$\large X_A = \frac{n_A}{n_A + n_B} = \frac{1.25}{1.25 + 1.304} $ XA

= 0.49

∴ X_{B} = 0.51

$\large P_T = P^o_A X_A + P^o_B X_B $

= 88.7 × 0.49 + 44.5 × 0.51

P_{T} = 43.480 + 22.695 = 66.175

P_{A} = 43.48

X_{A} = P_{A} /P_{T} = 43.48/66.175

= 0.66

Answer is (A)