At 80oC, the vapour pressure of pure liquid 'A' is 520 mmHg and that of pure liquid ‘B’ is 1000 mmHg. If a mixture solution of 'A' and ' B' boils at 80oC and 1 atm pressure, the amount of 'A' in the mixture is: (1 atm = 760 mmHg):

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

52 mol percent

34 mol percent

48 mol percent

50 mol percent

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

Pm=PAoXA+PBoXB760=520XA+1000(1-XA)760 = 520XA+1000-1000XA1000-760 = 1000XA-520XA240 = 480XAXA=240480∴XA=0.5=50%

Watch 3-min video & get full concept clarity