The density of solid argon is $\frac{2}{3}$(amu /${\text{Å}}^3$) at 40K. If the Argon atom is assumed to be sphere to radius $\frac{3}{{\pi }^{1/3}}\stackrel{0}{\mathrm{A}}$, what percentage of solid Argon is apparently without anything?

$\begin{array}{l}\rho =\frac{2}{3}\mathrm{amu}/A^{03}=\frac{2}{3}\times \frac{1.66\times {10}^{-24}}{{10}^{-24}{\mathrm{cm}}^3}\mathrm{g}\left[\begin{array}{c}1\mathrm{amu}=1.66\times {10}^{-24}\mathrm{g}\\ 1A^{\mathrm{\prime }}={10}^{-8}\mathrm{cm}\end{array}\right.\\ \begin{array}{rl} & =1.106\mathrm{g}{\mathrm{cm}}^3\\ \rho & =1.106\mathrm{gm}\end{array}\\ \left|m\right|=1.106\mathrm{g}\end{array}$$\text{ Moles }=\frac{\text{ Given Mess }}{\text{ Mol.mas }}=\frac{1.106}{40}$

Total no of atoms $=\frac{1.106}{40}\times 6.023\times {10}^{23}$

Volume occupied lay total atoms$\to$$\frac{1.106}{40}\times 6.023\times {10}^{23}\times \mathrm{\pi x}\frac{3}{{\pi }^{1/3}}A$

$\begin{array}{l}=\frac{1.106\times 6.023\times 27\times {10}^{-24}\times {10}^{23}{\mathrm{cm}}^3}{40}\\ =6\times {10}^{-1}{\mathrm{cm}}^3\\ \text{vol. ocupid }=0.6\mathrm{ml}\end{array}$$=1-0.6=0.4\times 100$

vol 40%