The ammonia evolved from the treatment of $$0.30g$$  of an organic compound for the estimation of nitrogen was passed in $$100ML$$  of $$0.1M$$  sulphuric acid. The excess of acid required $$20mL$$ of $$0.5M$$  sodium hydroxide solution for complete neutralization. The organic compound is

Solution :
$$\eqalign{ & {H_2}S{O_4}\,{\text{is}}\,\,{\text{dibasic}}{\text{.}} \cr & 0.1\,M\,\,{H_2}S{O_4} = 0.2N\,{H_2}S{O_4}\,\,\,\left[ {\because \,M = 2 \times N} \right] \cr & {M_{eq}}\,{\text{of}}\,{H_2}S{O_4}\,{\text{take}}{{\text{n}}}{\text{ = 100}} \times {\text{0}}{\text{.2 = 20}} \cr & {M_{eq}}\,{\text{of}}\,{H_2}S{O_4}\,\,{\text{neutralised}}\,{\text{by}}\,NaOH = 20 \times 0.5 = 10 \cr & {M_{eq}}\,{\text{of}}\,{H_2}S{O_4}\,\,{\text{neutralised}}\,{\text{by}}\,N{H_3} = 20 - 10 = 10 \cr & \% \,of\,{N_2} = \frac{{{\text{1}}{\text{.4}} \times {M_{eq}}\,{\text{of}}\,{\text{acid}}\,{\text{neutrialised}}\,{\text{by}}\,N{H_3}}}{{{\text{wt}}{\text{.}}\,\,{\text{of}}\,{\text{organic}}\,{\text{compound}}}} \cr & = \frac{{1.4 \times 10}}{{0.3}} = 46.6 \cr & \% \,{\text{of}}\,{\text{nitrogen}}\,\,{\text{in}}\,\,{\text{urea}}\, = \frac{{14 \times 2 \times 100}}{{60}} = 46.6\left[ {{\text{Mol}}{\text{.}}\,{\text{wt}}\,{\text{of}}\,{\text{urea = 60}}} \right] \cr & {\text{Similarly}}\,\% \,{\text{of}}\,{\text{Nitrogen}}\,{\text{in}}\,{\text{Benzamide}} \cr & = \frac{{14 \times 100}}{{121}} = 11.5\% \,\left[ {{C_6}{H_5}CON{H_2} = 121} \right] \cr & {\text{Acetamide}} = \frac{{14 \times 1 \times 100}}{{59}} = 23.4\% \,\left[ {C{H_3}CON{H_2} = 59} \right] \cr & {\text{Thiourea}} = \frac{{14 \times 2 \times 100}}{{76}} = 36.8\% \,\left[ {N{H_2}CSN{H_2} = 76} \right] \cr & {\text{Hence}}\,{\text{the}}\,{\text{compound}}\,{\text{must}}\,{\text{be}}\,{\text{urea}}{\text{.}} \cr} $$