760=XaPa+XbPb
Xb=1-Xa
solve it u get the ans
35) At 80° C , the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a mixture solution of 'A' and 'B' boils at 80° C and 1 atm pressure, the amount of 'A' in the mixture is (1 atm = 760 mm Hg)
1 ) 34 mol percent
2 ) 48 mol percent
3 ) 50 mol percent
4 ) 52 mol percent