The number of atoms in 4.25 g of $N{H}_3$ is approximately

The correct option is D $6\times {10}^{23}$
$\because 17gmN{H}_3$ contains $6\times {10}^{23}$ molecules of $N{H}_3$
$\therefore 4.25gmN{H}_3$ contains $=\frac{6\times {10}^{23}}{17}\times 4.25$
$\therefore$ No. of atoms $=\frac{6\times {10}^{23}\times 4.25}{17}\times 4=6\times {10}^{23}$