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)

Options

A. 52 mol percelnt
B. 34 mol percent
C. 48 mol percent
D. 50 mol percent

Answer

D

Explanation

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) [phone] X_(A) or 480 X_(A) =240`
or ` X_(A) =(240)/(480) =1/2 ` or mol percent
i.e, The correct answer is