Question

The density of solid argon is 1.65 g per cc at $-{233}^{o}C$. If the argon atom is assumed to be a sphere of radius $1.54\times {10}^{-8}cm$, the percentage of empty space in solid argon is:

• A. 32%
• B. 54%
• C. 68%
• D. 62%

Answer 1

The correct option is D 62%

Volume of one atom$=\frac{4}{3}\pi r^3$
$=\frac{4}{3}\times \frac{22}{7}\times (1.54\times {10}^{-8}{)}^{3}c{m}^3$
$=1.53\times {10}^{-23}c{m}^3$
Volume of all atoms in 1.65 g $Ar$
$=\frac{1.65}{40}\times 6.02\times {10}^{23}\times 1.53\times {10}^{-23}c{m}^3$
$=0.380c{m}^3$
Volume of solid Ar containing
$1.65g=1c{m}^3$
Empty space $=1-0.380=0.620$
$\therefore$ % = 62%