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Calculate the molality of 1 litre soluti...

Calculate the molality of 1 litre solution of 93% `H_(2)SO_(4) ("weight"//"volume")`. The density of solution is 1.84 g `mL^(-1)`.

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The correct Answer is:
10.43 m

`93% H_(2)SO_(4)(w//v)=93 g H_(2)SO_(4)" in 100 cm"^(3)" of the solution = 93 g in 184 g of the solution"`
`therefore" Solvent (water) "=184-93 = 91 g = 0.091 kg," Molality "=("93/98 mol")/("0.091 kg")=10.43 m.`
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