Question

Calculate the molality of $1$ litre solution of $93\% H_{2}SO_{4}$ (weight/volume). The density of the solution is $1.84 \,g/mL$.

Answer 1

Mass of $H_{2}SO_{4}$ in $100ml$ of $93\% H_{2}SO_{4}$ solution$ =93g$
$\therefore $ Mass of $H_{2}SO_{4}$ in $1000$ ml of the $H_{2}SO_{4}$ solution$ =930g$
Mass of $1000$ ml $H_{2}SO_{4}$ solution$ = 1000\times 1.84 = 1840g$
Mass of water in $1000$ ml of solution $ = 1840 - 930 = 910g$
Moles of $H_{2}SO_{4}=\frac{Wt of H_{2}SO_{4}}{Mol Wt. of H_{2}SO_{4}} = \frac{930}{98}$
$\therefore $ Moles of $H_{2}SO_{4}$ in $1\, kg$ of water
$= \frac{930}{98} \times\frac{1000}{910} = 10.43\, mol\, kg^{-1}$
$\therefore $ Molality of litre solution $= 10.43$