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The molality of a sulphuric acid in whic...

The molality of a sulphuric acid in which the mole fraction of water is 0.86 is.........

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The correct Answer is:
`rarr 6`
Mole fraction of `H_(2)SO_(4)` in solution
`=1 - 0.86 = 0.14`
`= (n_(2))/(n_(1)+n_(2)) = 0.14`
Molality is `n_(2)` is 1000g of water i.e., `n_(1)`
`= 1000//18 = 55.55` moles.
`(n_(2))/(55.55+n_(2)) = 0.14`
or `n_(2) = 0.14 n_(2) +7.77`
or `0.86 n_(2) = 7.77` or `n_(2) = 9`.
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