A mixture of ethyl alcohol and propyl alcohol has a vapour
pressure of 290 mm Hg at 300 K. The vapour pressure of
propyl alcohol is 200 mm Hg. If the mole fraction of ethyl
alcohol is 0.6, its vapour pressure (in mm Hg) at the same
temperature will be
(a) 360 (b) 350 (c) 300 (d) 700
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The vapour pressure will be pA^0 = 350
option (B) is correct.
Explanation:
Rault's Law:
The total vapour pressure of binary mixture of miscible liquids be having ideally is given by
PT = PA^0 xA + PB^0 xB
Where xA and xB are mole fraction of A and B in liquid phase.
wherein
PA^0 and PB^0 are vapour pressures of pure liquids.
wherein
PA^0 and PB^0 are vapour pressures of pure liquids.
290 = pA^0 x 0.6 + 200 x (1 - 0.6)
290 = 0.6 x pA^0 + 80
pA^0 = 350
Thus the vapour pressure will be pA^0 = 350
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