Hint: Let's convert the volume of a hydrogen atom from angstroms to meters using the given conversion factor, and then multiply that volume by the number of atoms in a mole. Then use the formula for the volume of a sphere $V = \dfrac{4}{3}\pi r^3$ to calculate the volume of a hydrogen atom.

**Solution:**

**Equation**

$V = \dfrac{4}{3}\pi r^3$

where:

V is the volume of the hydrogen atom

r is the radius of the hydrogen atom

To calculate the total atomic volume of a mole of hydrogen atoms, we first need to determine the volume of a single hydrogen atom and then multiply it by Avogadro's number.

The volume of a single hydrogen atom can be calculated using the formula for the volume of a sphere:

$V = \dfrac{4}{3}\pi r^3$

Substituting the given value of

r (0.5 Å), we get:

$V = \dfrac{4}{3}\pi (0.5 \times 10^{-10} \text{ m})^3$

$V = 1.25 \times 10^{-23} \text{ m}^3$

To calculate the total atomic volume of a mole of hydrogen atoms, we multiply the volume of a single atom by Avogadro's number (6.022 × 10²³ atoms/mole):

$\text{Total Volume} = V \times N$

$\text{Total Volume} = 1.25 \times 10^{-23} \text{ m}^3/\text{atom} \times 6.022 \times 10^{23} \text{ atoms}/\text{mole}$

$\text{Total Volume} = 3.157 \times 10^{-7} \text{ m}^3/\text{mole}$

**Answer**

The total atomic volume of a mole of hydrogen atoms is approximately 3.157 × 10⁻⁷ m³. Rounding off the value of 3.157 × 10⁻⁷ m³ to two decimal places gives us 3.16 × 10⁻⁷ m³.

**Note**

This is just a simplified solution to the problem. In reality, the calculation of atomic volume is more complex, as it takes into account the packing arrangement of atoms in a solid.