Question

Calculate the molality of $1$ litre solution of $93\%H_{2}S{O}_4$ (weight/volume). The density of the solution is $1.84g/mL$.

Answer 1

Answer: 10.43

Mass of $H_{2}S{O}_4$ in $100ml$ of $93\%H_{2}S{O}_4$ solution$=93g$
$\therefore$ Mass of $H_{2}S{O}_4$ in $1000$ ml of the $H_{2}S{O}_4$ solution$=930g$
Mass of $1000$ ml $H_{2}S{O}_4$ solution$=1000\times 1.84=1840g$
Mass of water in $1000$ ml of solution $=1840-930=910g$
Moles of $H_{2}S{O}_4=\frac{Wtof{H}_{2}S{O}_4}{MolWt.of{H}_{2}S{O}_4}=\frac{930}{98}$
$\therefore$ Moles of $H_{2}S{O}_4$ in $1kg$ of water
$=\frac{930}{98}\times \frac{1000}{910}=10.43molk{g}^{-1}$
$\therefore$ Molality of litre solution $=10.43$