A) Acetamide
B) Benzamide
C) Urea
D) Thiourea
Correct Answer: C
Solution :
[c] Urea Let the volume of acid left unused \[=v\,m\ell \,of\,0.1\,M{{H}_{2}}S{{O}_{4}}\] \[\therefore \,v\times 0.1\times 2=20\times 0.5\times 1\] \[V=50m\,\ell \] \[\therefore Volume\text{ }of\text{ }acid\text{ }used=100-50\] \[=50\,m\ell \,\,of\,0.1\,M\,of\,{{H}_{2}}\,S{{O}_{4}}\] \[%N=\frac{1.4\times 2\times 50\times 0.1}{0.3}=46.6\] \[%\,N\,in\,urea\,(N{{H}_{2}}CON{{H}_{2}})\] \[=\,(28/60)\times 100=46.6%\] In acetamide \[(C{{H}_{3}}CON{{H}_{2}})=(14/59)\times 100\] \[=23.72%\] In benzamide \[({{C}_{6}}{{H}_{5}}CON{{H}_{2}})=(14/212)\times 100\] \[=11.57%\] In thiourea \[({{C}_{6}}{{H}_{5}}CSN{{H}_{2}})=(14/76)\times 100\] Thus, urea [c] is correct.
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