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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

0.6

0.5

0.8

0.4

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

Watch 3-min video & get full concept clarity