At `80^@C`, the vapour pressure of pure liquid 'A' is 50 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a solution of 'A and 'B' boils at `80^@C` and 1 atm pressure find the amount of A'A' in the mixture in mol fraction.

Explanation

`P_s=P_A^0.X_A+P_B^0.X_B`
`760=520.X_A+1000(1-X_A)`, since at boiling point vapour pressure of solution is equal to atmospheric pressure.
`X_A=0.5`
Thus mole percent of A = mole fraction of `Axx100=0.5xx100=50`