One mole of an ideal gas at standard temperature and pressure occupies $22.4 \; L$ (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of a hydrogen molecule to be about $1 \; \mathring{A}$). Why is this ratio so large?

(N/A) Radius of hydrogen atom,$r = 0.5 \; \mathring{A} = 0.5 \times 10^{-10} \; m$.
Volume of one hydrogen atom $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.1416 \times (0.5 \times 10^{-10})^3 \approx 0.524 \times 10^{-30} \; m^3$.
Number of atoms in $1$ mole of hydrogen ($H_2$ gas contains $2$ atoms per molecule,but for atomic volume of $1$ mole of atoms,we use Avogadro's number) $= 6.023 \times 10^{23}$.
Volume of $1$ mole of hydrogen atoms,$V_a = 6.023 \times 10^{23} \times 0.524 \times 10^{-30} \approx 3.16 \times 10^{-7} \; m^3$.
Molar volume of $1$ mole of gas at $STP$,$V_m = 22.4 \; L = 22.4 \times 10^{-3} \; m^3$.
Ratio $\frac{V_m}{V_a} = \frac{22.4 \times 10^{-3}}{3.16 \times 10^{-7}} \approx 7.08 \times 10^4$.
This ratio is large because in a gas,the molecules are separated by distances much larger than their own size,meaning the gas is mostly empty space.