Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2 MH 2 SO 4…

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

Equivalent of ${\mathrm{NH}}_3=$Equivalent of ${\mathrm{H}}_2{\mathrm{SO}}_4$

moles of ${\mathrm{NH}}_3\left({\mathrm{n}}_{{\mathrm{NH}}_3}\right)$$\times 1=\frac{2\times 10\times 2}{1000}$

                                            $=4\times {10}^{-2 }\mathrm{moles}$

Moles of nitrogen$=4\times {10}^{-2} \mathrm{moles}$

Weight of Nitrogen$$\begin{gathered} =4\times {10}^{-2}\times 14 \\ =56\times {10}^{-2} \mathrm{gm} \end{gathered}$$

Percent of nitrogen$$\begin{gathered} =\frac{56\times {10}^{-2}}{\mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{compound}}\times 100 \\ =\frac{56\times {10}^{-2}}{1}\times 100=56\% \end{gathered}$$