A) \[{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\]
B) Calcium cyanamide
C) Urea
D) Ammonium nitrate
Correct Answer: C
Solution :
Urea \[(N{{H}_{2}}CON{{H}_{2}})\] has the maximum percentage of nitrogen in it. Molecular weight of\[N{{H}_{2}}CON{{H}_{2}}\] \[=14+2+12+16+14+2=60\] \[\therefore \] % of nitrogen\[=\frac{28}{60}\times 100=46.67%\]
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