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

The number of moles present: $\frac{4.25g}{17g/mol}$
Number of moles of atoms: $\frac{4.25}{17}\times 4mol$
Number of atoms:$\frac{4.25}{17}\times 4\times 6.022\times {10}^{23}atoms$
=$6.022\times {10}^{23}atoms$