At 80^(@)C, the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 nm Hg. If a mxiture of solution of 'A' and 'B' boils at 80^(@)C and I atm pressurc the amountof 'A' in the mixture is (1 atm =760 mm Hg)

At 80^(@)C, the vapour pressure of pure liquid 'A' is 520 mm Hg and th

1.	At 80^(@)C, the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 nm Hg. If a mxiture of solution of 'A' and 'B' boils at 80^(@)C and I atm pressurc the amountof 'A' in the mixture is (1 atm =760 mm Hg)
Answer» 52 mol percelnt 34 mol percent 48 mol percent 50 mol percent Solution :At 1 atmospheric pressure the boiling point of mixture is `80^(@)C.` At boiling point the vapour pressure of mixture, `P _(tau) =1` atmosphere = 760 nm Hg. Using the RELATION, `P _(I) =P_(A)^(@)X_(A) + P_(B)^(@) X_(B), ` we GET ` P _(T) =520 X_(A) +1000 (1-X_(A))` `{P_(A)^(@)=520` mm Hg. `p _(B)^(@) =1000` mm Hg, `X_(A) + X_(B) =1}` or `760 =520 X_(A) +1000-1000 X_(A) or 480 X_(A) =240` or ` X_(A) =(240)/(480) =1/2 ` or mol percent i.e, The correct answer is