Question

The density of solid argon is 1.4 g/mL at $-{233}^{\circ }C$ If the argon atom is assumed to be sphere of radius $2.0\times {10}^{-8}$ cm, What percentage of solid argon is apparently empty space? (atomic wt. Of Ar = 40)

(take $N_A=6\times {10}^{23}$

Answer 1

The number of Ar atoms in 40 g (1 mole) $=6\times {10}^{23}$
$\Rightarrow$ Number of atom in $1.4g=\frac{1.4}{40}\times 6\times {10}^{23}$
As density = 1.4 g/mL or 1.4 g/$c{m}^3$
Number of atoms in 1 $c{m}^3$ = No. of atoms in $1.4g=\frac{1.4}{40}\times 6\times {10}^{23}$ atoms.
Since the volume of each atom is $\frac{4}{3}\pi r^3$ where r = $2.0\times {10}^{-8}$cm, we can say volume occupied by atoms
$=\frac{4}{3}\pi r^3\times \frac{1.4}{40}\times 6\times {10}^{23}=\frac{4}{3}\times \frac{22}{7}\times (2.0\times {10}^{-8}{)}^3\times \frac{1.4}{40}\times 6\times {10}^{23}=0.704c{m}^3$
$\therefore$ per 1 $c{m}^3$ of solid 0.704 $c{m}^3$ volume is occupied by the atoms.
Hence % of empty space $=\frac{[1-0.704]}{1}\times 100=29.6$