A) acetamide
B) benzAmide
C) urea
D) thiourea
Correct Answer: C
Solution :
Let unreacted \[0.1\,M(=0.2\,N){{H}_{2}}S{{O}_{4}}=V\,mL\] \[\therefore \] 20 mL of\[0.5M\,NaOH\] \[=VmL\,of\,0.2\,N\,{{H}_{2}}S{{O}_{4}}\] \[\therefore \] \[20\times 0.5=V\times 0.2\] \[\therefore \] \[V=50mL\] Used \[{{H}_{2}}S{{O}_{4}}=100-50=50\text{ }mL\] % Nitrogen\[=\frac{1.4NV}{w}.\] Where \[N=\]normality of \[{{H}_{2}}S{{O}_{4}}\] \[V=\]volume of\[{{H}_{2}}S{{O}_{4}}\]used \[\therefore \] %Nitrogen \[=\frac{1.4\times 0.2\times 50}{0.30}\] \[=46.67%\] % of nitrogen in (a) \[C{{H}_{3}}CON{{H}_{2}}=\frac{14\times 100}{59}=23.73%\] (b)\[{{C}_{6}}{{H}_{5}}CON{{H}_{2}}=\frac{14\times 100}{122}=11.48%\] (c)\[N{{H}_{2}}CON{{H}_{2}}=\frac{28\times 100}{60}=46.67%\] (d)\[N{{H}_{2}}CSN{{H}_{2}}=\frac{28\times 100}{76}=36.84%\]You need to login to perform this action.
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