Question

What is the number of atoms present in 4.25 grams of $N{H}_3$ is approximately??

Answer 1

Molecular mass of ammonia ($N{H}_3$)= 14+3=17g

In 1 molecule ($N{H}_3$) there are 1+3=4 atoms

1 mole (17 g) of $N{H}_3$ = $6.022\times {10}^{23}$atoms
1 g of $N{H}_3$ = $\frac{6.022\times {10}^{23}}{17}$
4.25 g of $N{H}_3$ = $\frac{6.022\times {10}^{23}\times 4.25}{17}=1.505\times {10}^{23}$
One molecule contains a total 4 atoms.
Thus, total no. of atoms in ammonia = $4\times 1.505\times {10}^{23}=6.022\times {10}^{23}$