Question

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

• A. 1 X 10²³
• B. 1.5 X 10²³
• C. 2 X 10²³ NH3
• D. 6.02 X 10²³

Answer 1

The correct option is D

6.02 X 10²³

Number of atoms in a $N{H}_3$ molecule is 4.

Atomic weight of Nitrogen is 14.

Atomic weight of hydrogen is 1.

Molecular weight of $N{H}_3$ is

14+3(1)=17

So, no of atoms in the molecule is

$\frac{4.25}{17}\times 4\times 6.023\times {10}^{23}=6.023\times {10}^{23}$