Question

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

• A. $1.00\times {10}^{23}$
• B. $1.50\times {10}^{23}$
• C. $2.00\times {10}^{23}$
• D. $6.02\times {10}^{23}$

Answer 1

The correct option is D $6.02\times {10}^{23}$

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