A) Acetamide
B) Benzamide
C) Urea
D) Thiourea
Correct Answer: C
Solution :
\[{{H}_{2}}S{{O}_{4}}\,+2\,NaOH\,\xrightarrow{\,}N{{a}_{2}}S{{O}_{4}}\,+2{{H}_{2}}O\] \[{{M}_{1}}\times n\,-factor\,\times {{V}_{1}}={{M}_{2}}\times n-factor\,\times {{V}_{2}}\] \[({{H}_{2}}S{{O}_{4}})\] (NaOH) \[0.1\times 2\times {{V}_{1}}=0.5\,\times 1\times 20\] \[{{V}_{1}}=50\,ml\] Volume of \[{{H}_{2}}S{{O}_{4}}\] used up to absorb \[N{{H}_{3}}=50\,ml\] \[N%\,=\frac{2.8\times \,(MV)}{W}\,=\frac{2.8\times 0.1\,\times 50}{0.3}\,=46.6\] Thus, the organic compound will be urea, which has 46.6 % \[{{N}_{2}}\]You need to login to perform this action.
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