The number of atoms present in $4.25$ grams of $N{H}_3$ is approximately:

The correct option is A $6.023\times {10}^{23}$

No. of moles = Given mass is divided by molar mass.

No. of moles of ammonia = 4.25/17 = $0.25$ mol
No. of molecules of ammonia = $6.023\times {10}^{23}\times 0.25$

One molecule contains a total 4 atoms.
Thus, total no. of atoms in ammonia $=6.023\times {10}^{23}\times 0.25\times 4=6.023\times {10}^{23}$ atoms