Question

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

Answer 1

Step 1: Analyzing the given data

Given mass of Ammonia$\left({\mathrm{NH}}_3\right)$ $=4.25\text{g}$

Molar mass of Ammonia$\left({\mathrm{NH}}_3\right)$ ${\text{=14.0067 g mol}}^{\text{-1}}\text{+1.0079×3}\text{g}{\mathrm{mol}}^{\text{-1}}=17.0304\text{g}{\mathrm{mol}}^{\text{-1}}$

Step 2: Calculating the number of atoms

The number of molecules of methane can be calculated using the given formula

$\text{Number of molecules=}\frac{\text{Given mass}}{\text{Molar mass}}\text{×Avogadro number}$

Since one Ammonia molecule contains three atoms of hydrogen and one atom of nitrogen. So, the total number of atoms can be calculated as

$$\begin{gathered} \text{Number of atoms=4×}\frac{4.25\text{g}}{17.0304{\text{g mol}}^{-1}}{\text{×6.022×10}}^{23}{\text{mol}}^{\text{-1}} \\ =6.032{\times 10}^{23} \end{gathered}$$

Hence, the number of atoms is $6.032\times {10}^{23}$.