At 300 K two pure liquids A and B have 150 mm Hg and 100 mmHg vapour pressures, respectively. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is
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a
0.6
b
0.5
c
0.8
d
0.4
answer is D.
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Detailed Solution
In equimolar liquid mixture xA=0.5,xB=0.5So, p=0.5×150+0.5×100=125Now, let YB be the mole fraction of vapour B then YB=xBpB∘p=0.5×100125=0.4