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)`
`93% H_(2)SO_(4)` solution weight by weight by volume indicates that there is `93 g H_(2)SO_(4)` in `100` mL of solution. If we consider `100` mL solution, Weight of solution `=184 g` Weight of `H_(2)O` in `100` mL solution, weight of solution `=184-93=91 g` `implies` Molality `=("Moles of solute")/("Weight of solvent (g)")xx1000` `=93/98xx1000/91=10.42`
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