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Find the molality fo `H_(2)SO_(4)` solution whose specific gravity is `1.98 g mL^(-1)` and 98% (Weight/volume) `H_(2)SO_(4)`.

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Correct Answer - A::B::D
`H_(2) SO_(4)` is by volume,
weight of `H_(2) SO_(4) = 95 g`
Volume of solution `= 100 mL`
Weight of solution `= 100 xx 1.98 = 198 g`
Weight of `H_(2) O = 198 - 95 = 103 g`
`m = (W_(2) xx 1000)/(Mw xx W_(1)) = (95 xx 1000)/(98 xx 103) = 9.412`

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