Following Kjeldahl's method, 1g of organic compound released...

{H}_{2}{SO}_{4} + 2{NH}_{3} \to {\left({NH}_{4}\right)}_{2}{SO}_{4}$</p><p>Equivalent of${NH}_{3} =$Equivalent of${H}_{2}{SO}_{4}$</p><p>moles of${NH}_{3}\left({n}_{{NH}_{3}}\right)$\times 1 = \dfrac{2 \times 10 \times 2}{1000}$

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

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

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

Percent of nitrogen$= \dfrac{56 \times {10}^{- 2}}{weightofthecompound} \times 100 = \dfrac{56 \times {10}^{- 2}}{1} \times 100 = 56$