A) Urea
B) Ammonium sulphate
C) Ammonium nitrate
D) Calcium nitrate
Correct Answer: A
Solution :
\[N{{H}_{2}}CON{{H}_{2}}\] % of N =\[\frac{\text{Mass of }N}{\text{Mass of compound}}\times 100\] = \[\frac{28}{60}\times 100\]= 46%.
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