For the estimation of nitrogen 1.4g of organic compound was digested by Kjeldahl and the

evolved ammonia was absorbed in 60 mL $\frac{\text{M}}{\text{10}}$ of sulphuric acid. The unreacted acid required 20 ml $\frac{\text{M}}{\text{10}}$of sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is

Molarity of sulphuric acid used in Kjeldha’s method is = 0.1M

            Volume of acid used = 60 Ml

            N = Mx Basicity N = 0.1 × 2

                                           = 0.2 M

            The no of meq of base used to neutralize the acid is

            = N × V

            = 20 × 0.1

            = 2 meq

            The no of meq of Acid (Total) used

            = 60 × 0.2

            = 12 meq

            The acid used for neutralization of ${\text{NH}}_3$ is

            = 12 – 2   = 10 meq = ${10}^{-2}$ equal

            1 equivalent ${\text{NH}}_{\text{3}}$ -  17g

            equivalent - x

            x = 0.17 g

             17 g – 14g “N”

            0.17 g – xg

           $Y=\frac{14\times 0.17}{17}$ 

            = 0.14 g

            $\text{\%\hspace{0.17em}N\hspace{0.17em}=\hspace{0.17em}}\frac{\text{0.14\hspace{0.17em}}}{\text{1.4}}\text{\hspace{0.17em}×\hspace{0.17em}100}$

            = 0.14g