Question

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

• A. $6.023\times 10^{23}$
• B. $4\times 6.023\times 10^{23}$
• C. $1.7\times 10^{24}$
• D. $4.5\times 6.023\times 10^{23}$

Answer 1

Answer: (A)

Molar mass $=14+3=17$
No.of moles $=\frac{4.25}{17}=0.25$ moles
1 mole $=4\times 6.022\times 10^{27}$ atoms
$0.25mole=.25\times 4\times 6.022\times 10^{23}=6.022\times 10^{23}$ atoms