Chemistry - Solid State Question with Solution | TestHub

Volume of one molecule $\begin{array}{l}=\frac{4}{3}\pi {\mathrm{r}}^3=\frac{4}{3}\pi {\left(1.54\times 10^{-8}\right)}^3{\mathrm{cm}}^3\\ =1.53\times 10^{-23}{\mathrm{cm}}^3\end{array}$ Volume of molecules in $1.65\mathrm{g}\mathrm{Ar}$ $=\frac{1.65}{40}\times {\mathrm{N}}_0\times 1.53\times 10^{-23}=0.380{\mathrm{cm}}^3$ Volume of solid containing $1.65\mathrm{g}\mathrm{Ar}=1{\mathrm{cm}}^3$ $\therefore$ Empty space $=1-0.380=0.620$ $\therefore$ Per cent of empty space $=62\%$