The number of Nitrogen atoms in 681 g of C7H5 N3O6 is x.yz × 1024. The value of x.yz is_____________ NA=6.02×1023 mol-1) (Nearest Integer)

The number of Nitrogen atoms in 681 g of $C_{7} H_{5} N_{3} O_{6}$ is $x . yz \times 10^{24}$ . The value of x.yz is_____________ $(N_{A} = 6.02 \times$ $10^{23} {mol}^{- 1})$ (Nearest Integer)

Solution:

$M.wt. of C_{7} H_{5} N_{3} O_{6} is 84 + 5 + 42 + 96 = 227 \begin{aligned} n_{C_{7} H_{5} N_{3} O_{6}} = \frac{681}{227} = 3 \\ n_{N} = \frac{681}{227} \times 3 = 9 mol \\ no. of N atoms = 9 \times 6.02 \times 10^{23} \\ = 5418 \times 10^{21} \end{aligned} ∴ The answer is 5418 .$