Chemistry - Solid State Question with Solution | TestHub

Volume of one atom of Ar$=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^3$

Also, No. of atoms in 1.65g$=\frac{1.65}{40}\times 6.023\times {10}^{23}$

Total volume of all atoms of Ar in solid state.

$=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^3\times \frac{1.65}{40}\times \mathrm{ }6.023\times {10}^{23}$

$=\frac{4}{3}\times \frac{22}{7}\times {\left(1.54\times {10}^{-8}\right)}^3\times \frac{1.65}{40}\times 6.023\times {10}^{23}$

$=0.380\mathrm{ }\mathrm{c}{\mathrm{m}}^3$

Volume of solid argon$=1\mathrm{ }\mathrm{c}{\mathrm{m}}^3$

% Empty space$=\frac{\left[\right[1-0.380]}{1}\times 100=62\mathrm{\%}$