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

The correct option is D $6\times {10}^{23}$

1 molecule of ${NH}_3$ has 4 atm 1N+3H$=$4

No. of molecules$=\frac{4.25}{17}\times 6.023\times {10}^{23}$

$=1.51\times {10}^{23}$ molecules

No. of atoms$=4\times 1.51\times {10}^{23}$

$=6.023\times {10}^{23}$