Explanation

(c)
In $Ca\left( NO_{3} \right)_{2};\ \ \ \%\ of\ N = \frac{20}{164} \times 100 = 17.07\%$
In $\left( NH_{4} \right)_{2}SO_{4};\ \ \%\ of\ N = \frac{28}{132} \times 100 = 21.21\%$
In $NH_{2}CONH_{2};\ \ \%\ of\ N = \frac{28}{60} \times 100 = 46.66\%$
In $NH_{4}NO_{3};\ \ \%\ of\ N = \frac{28}{80} \times 100 = 35.00\%$