The density of solid argon is 1.65 g/mL at –233ºC. If the argon atom is assumed to be sphere of radius 1.54 × 10^–8 cm, what percentage of solid argon has apparently empty space? (atomic weight of Ar = 40).

Volume of one atom of Ar$=\frac{4}{3}{\mathrm{\pi r}}^3$
Also, number of atoms in 1.65
$\mathrm{g}=\frac{165}{40}\times 6.023\times {10}^{21}$
$\therefore$total volume of all atoms of Ar in solid state
$\begin{array}{l}=\frac{4}{3}\pi r^3\times \frac{1.65}{40}\times 6.023\times {10}^{23}\\ =\frac{4}{3}\times \frac{22}{7}\times {\left(1.54\times {10}^{-8}\right)}^3\times \frac{165}{40}\times 6.023\times {10}^{23}\\ =0.380{\mathrm{cm}}^3\end{array}$
Volume of solid argon = 1 cm³
$\therefore \mathrm{\%}\text{ empty space }=\frac{[1-0.380]}{1}\times 100=62\mathrm{\%}$