1.

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 of 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.)

Answer»

50 mol per cent
52 mol per cent
34 mol per cent
48 mol per cent

Solution :`p_("total")=p_(A)^(@)x_(A) + p_(B)^(@)x_(B)`
`760=520x_(A)+1000(1-x_(A))`
`760=520 x_(A)+1000-1000x_(A)`
`480x_(A)=240`
`x_(A)=0.5`
`:.` MOLES of A `=50%`


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