For the estimation of nitrogen, $1.4 \mathrm{g}$ of an organic compound was digested by the Kjeldahl method and the evolved ammonia was absorbed in $60 \mathrm{ml}$ of $\frac{\text{M}}{10}$ sulphuric acid. The unreacted acid required $20 \mathrm{ml}$ of $\frac{\text{M}}{10}$ sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is

In Kjeldahl method,

$\mathrm{Per} \mathrm{cent} \mathrm{N} = \frac{1.4 \mathrm{N} \mathrm{V}}{\mathrm{W}}$

Where, $\mathrm{NV}$= gram equivalent of ammonia

$\mathrm{W}$= weight of the organic compound 

$\Rightarrow 2{\mathrm{NH}}_3+\underset{\mathrm{excess}}{{\mathrm{H}}_2{\mathrm{SO}}_4}\to {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$

Total of ${\mathrm{H}}_2{\mathrm{SO}}_4$ used $=60\times \frac{1}{10}=\text{6 mm = 6 x 2 =12 gm equivalent}$

$\Rightarrow 2\mathrm{NaOH}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$

$\mathrm{NaOH}$ for back titration $=20\times \frac{1}{10}=\text{2 gm equivalent}$

gm equivalent of remaining ${\mathrm{H}}_2{\mathrm{SO}}_4=12-2=10$

${\mathrm{H}}_2{\mathrm{SO}}_4$used for ${\mathrm{NH}}_3$ neutralisation $=10 \mathrm{gm} \mathrm{equivalent}$

${\mathrm{NH}}_3 \left(\mathrm{NV}\right)=10 \mathrm{gm} \mathrm{equivalent}$
$\text{Per cent of N}=\frac{1.4 x10}{\text{1.4}}=\text{10\%}$