• # question_answer The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is  A) Urea      B) benzamide C) Acetamide D) thiourea

 ${{H}_{2}}S{{O}_{4}}$is dibasic. $0.1M\,{{H}_{2}}S{{O}_{4}}=0.2N\,{{H}_{2}}S{{O}_{4}}$ ${{M}_{eq}}\,of\,{{H}_{2}}S{{O}_{4}}taken\,=100\times 0.2=20$ ${{M}_{eq}}\,of\,{{H}_{2}}S{{O}_{4}}$neutralized by$NaOH$ $=20\times 0.5=10$ ${{M}_{eq}}$ of ${{H}_{2}}S{{O}_{4}}$ neutralized by$N{{H}_{3}}$ $=12-10=10$ $%of N$, $\frac{1.4 \times {{M}_{eq}}of acid neutrialised by N{{H}_{3}}}{Wt. of organic compound}$ $=\frac{1.4\times 10}{0.3}46.6$ % of nitrogen in urea $=\frac{14\times 2\times 100}{60}=46.6$