Calculate the molality of $1$ litre solution of $98 %$ $H_{2} S O_{4}$ (weight/volume). The density of the solution is $1.88 g / m L$ .

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Solution

$98 % (w / V)$ $H_{2} S O_{4}$ means $98 g$ $H_{2} S O_{4}$ in $100 m L$ of the solution.
Again, density of the solution is $1.88 g / m L$
So mass of $1000 m L$ solution = $1880 g$
Mass of $H_{2} S O_{4}$ in $1000 m L$ solution = $980 g$
So mass of the solvent = $1880 - 980 = 900 g = 0.90 k g$
Hence molality = $\frac{\frac{980}{98}}{0.90} = 11.11 m$

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