No. of atoms in $4.25g.$ of ${NH}_3$ is approx :

The correct option is D $6\times {10}^{23}$
$1$ $mole$ of $N{H}_3$ has $4$ $moles$ of atoms i.e., $4\times 6.023\times {10}^{23}$ atoms $⟹\left(24.092\right)\times {10}^{23}$

Here, No. of moles of $N{H}_3=\frac{4.25}{17}=\frac{1}{4}$

So, Total number of atoms $=\frac{1}{4}\times \left(24.092\right)\times {10}^{23}$

$⟹$ Total number of atoms $=6.023\times {10}^{23}$ atoms