A) Urea
B) benzamide
C) Acetamide
D) thiourea
Correct Answer: A
Solution :
\[{{H}_{2}}S{{O}_{4}}\]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 \[=\frac{14\times 2\times 100}{60}=46.6\] |
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