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Calculate the molality of 1 L solution o...

Calculate the molality of `1 L` solution of `93% H_(2)SO_(4)` (Weight/volume) The density of the solution is `1.84 g`.

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The correct Answer is:
10.42
Given: 93% `H_(2)SO_(4)` solution (w/v) , 93g `H_(2)SO_(4)` is present in `rarr 100` ml of solution
100 ml solution `rArr` 184 gm solution (d = 1.84 gm/cc)
`:.` 184 – 93 = 91 gm solvent has 93 gm `H_(2)SO_(4) = (93)/(98)` moles `H_(2)SO_(4)`
Hence, 1000 gm solvent has `= (93)/(98) xx (1000)/(91) = 10.42` molal
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