Q.

Vapour pressure of pure liquids A & B are 180 mm and 540 mm respectively. Equimolar mixture of A & B is prepared which behaves like an ideal solution. Molefraction of ‘B’ in vapour phase will be

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

answer is 3.

(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

Ptot = pA0XA+pB0XBGiven XA = XB = 0.5Ptot = 0.5180+0.5540 = 90+270mm = 360 mmIn Vapour phasePB = Ptot×yB; yB = PBPtot = 0.5540360 = 34 = 0.75

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

Connect with our

Expert Counsellors

access to India's best teachers with a record of producing top rankers year on year.

+91

We will send a verification code via OTP.