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

Molecular weight of $N{H}_3$ = Weight of 1 Nitrogen atom + 3$\times$ Weight of Hydrogen atom

= 14 + 3 $\times$ 1 = 17

17 grams of $N{H}_3$ = 1 mole

4.25 grams of $N{H}_3$ = $\frac{Massofsubstance}{Molecularmassofsubstance}=\frac{4.25}{17}$ = 0.25 mole of $N{H}_3$

No of molecules present in 1 mole of $N{H}_3$ = 6.023 $\times {10}^{23}$

No of molecules present in 0.25 mole of $N{H}_3$ = 6.023 $\times {10}^{23}\times 0.25$

= 1.505 $\times {10}^{23}$

1 mole of $N{H}_3$ = 4 atoms

6.023 $\times {10}^{23}\times 0.25$ molecules of $N{H}_3$ = $0.25\times 4\times 6.023\times {10}^{23}$ atoms

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