Question

The number of atoms in $4.25\text{g}$ of $N{H}_3$ is approximately:

• A. $4\times {10}^{23}$
• B. $2\times {10}^{23}$
• C. $1\times {10}^{23}$
• D. $6\times {10}^{23}$

Answer 1

The correct option is D $6\times {10}^{23}$

$17\text{g of}N{H}_3=4N_A\text{atoms}$
$4.25\text{g of}N{H}_3=\frac{4N_A}{17}\times 4.25\text{atoms}$
$=N_A\text{atoms}=6\times {10}^{23}$