Question

$49$% (w/s) $H_2{SO}_4$ stock solution has a density of $1.2g{ml}^{-1}$. The volume of stock solution read to prepare $100ml$ solution of $2{NH}_2{SO}_4$.

Answer 1

Normality $=$ Molarity $\times$ n-factor

For $H_{2}S{O}_4;$ n-factor $=2$

$\therefore$ Molarity $=\frac{2}{2}=1M$

Moles of $H_{2}S{O}_4$ in $100$ $mL$ solution $=100\times 1=100$ $mmoles=0.1$ $moles$

Mass of pure $H_{2}S{O}_4=$ Moles $\times$ Molecular weight $=0.1\times 98=9.8$ $grams$

$\therefore$ Mass of $49\%$ by weight solution of $H_{2}S{O}_4$ which contains $9.8$ $grams$ pure $H_{2}S{O}_4$ is $=\frac{9.8}{0.49}=20$ $grams$

Volume of $49\%$ solution of $H_{2}S{O}_4=\frac{Mass}{\rho }=\frac{20g}{1.2g/ml}=16.67$ $mL$