Question

The density of solid argon is $1.65\text{g/mL}\text{at}-{233}^{\circ }C$. If the argon atom is assumed to be sphere of radius $1.54\times {10}^{-8}\text{cm}$, what percentage of solid argon is apparently empty space ?
Give your answer in the nearest integer value.
$(\text{Atomic weight of Ar}=40\text{u},\pi =\frac{22}{7},N_A=6\times {10}^{23})$

Answer 1

Volume of one atom of $Ar=\frac{4}{3}\pi r^3$
Also, number of atoms of $Ar$ in $1.65\text{g}$ or in one mL
$=\frac{1.65}{40}\times 6\times {10}^{23}$
$\therefore$ Total volume of all atoms of Ar in solid state
$=\frac{4}{3}\pi r^3\times \frac{1.65}{40}\times 6\times {10}^{23}$
$=\frac{4}{3}\times \frac{22}{7}\times (1.54\times {10}^{-8}{)}^3\times \frac{1.65}{40}\times 6\times {10}^{23}$
$\approx 0.379{\text{cm}}^3$
Volume of solid argon = $1c{m}^3$
$\therefore$ empty space $=\frac{[1-0.379]}{1}\times 100=62.1\%\approx 62\%$