Chemistry: Molecular Approach (4th Edition)

$\underline{36.9\times {{10}^{-10}}}$
Radius of neon atom is $\text{69 pm}$. $\text{1 pm}=\text{1}{{\text{0}}^{-12}}\text{ m}$ Thus, radius of neon atom in decimeter is written as follows: $r=69\times {{10}^{-12}}\text{ m}$ Assuming a spherical shape: $V=\frac{4}{3}\pi {{r}^{3}}$ Thus, \begin{align} & V=\frac{4}{3}\pi {{\left( 69\times {{10}^{-12}}\text{ m} \right)}^{3}} \\ & =1.37\times {{10}^{-30}}\text{ }{{\text{m}}^{3}} \end{align} Now, $\text{1 L}=\text{1}{{\text{0}}^{-3}}\text{ }{{\text{m}}^{3}}$ Thus, \begin{align} & V=1.37\times {{10}^{-30}}\,\times \,{{10}^{-3}}\text{ L} \\ & \,\,\,\,\text{=}\,1.37\times {{10}^{-33}}\,\,\text{L} \\ \end{align} There are $2.69\,\times \,{{10}^{22}}\,Ne\,atoms\,$ present in $1\,L$of gaseous sample. Thus, the volume occupied by $2.69\,\times \,{{10}^{22}}\,Ne\,atoms\,$is : \begin{align} & 2.69\,\times \,{{10}^{22}}\,\,\,\times \,\,1.37\times {{10}^{-33}}\,\,\text{L}\, \\ & =\,3.68\times {{10}^{-11}}\,\,L\, \\ \end{align} The total volume of the gaseous sample is $1\,L$. The fraction of space occupied by the atoms is the ratio of volume occupied by the atoms to total volume of the gaseous sample. Thus, \begin{align} & \text{Fraqction of space}=\frac{3.69\times {{10}^{-11}}\,\,\text{L}}{1\text{ L}} \\ & =3.69\times {{10}^{-11}}\,\approx \,\,36.9\times {{10}^{-10}} \end{align} The fraction of the space occupied by the atoms is $\underline{\,\,36.9\times {{10}^{-10}}}$.