Question

The number of $N$ atoms is $681 \,g $ of $C _{7} H _{5} N _{3} O _{6}$ is $x \times 10^{21}$. The value of $x$ is _______$\left( N _{ A }=6.02 \times\right.$ $10^{23} mol ^{-1}$ ) (Nearest Integer)

Answer 1

M.M. of $C _{7} H _{5} N _{3} O _{6}$ is $84+5+42+96=227$
$n _{ C _{7} H _{5} N _{3} O _{6}}=\frac{681}{227}=3$
$n _{ N }=\frac{681}{227} \times 3=9\, mol$
no. of $N$ atoms $=9 \times 6.02 \times 10^{23}$ $=5418 \times 10^{21}$
$\therefore$ The answer is $5418$ .