0 . 8   g of an organic compound was analysed by Kjeldahl's method for the estimation of nitrogen.…

Organic compound : $0.8 \mathrm{gm}$

wt. of $\mathrm{N}=\left(\frac{42}{100}\times 0.8\right)\mathrm{gm}$

mole of $\mathrm{N}=\frac{42\times 0.8}{100\times 14}=\frac{2.4}{100} \mathrm{mol}$

moles of ${\mathrm{NH}}_3=\frac{2.4}{100}$

$\mathrm{Molarity} \times \mathrm{Volume} \left(\mathrm{L}\right) = \mathrm{number} \mathrm{of} \mathrm{moles}$

$\frac{1.2}{100}=1\times \mathrm{V}\left(\mathrm{L}\right)$

$\Rightarrow {\mathrm{V}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}=\frac{1.2}{100}\mathrm{L}=12 \mathrm{mL}$