Question

The density of solid argon is $1.65\, g\, per\, cc$ at - $233^\circ C$. If the argon atom is assumed to be a sphere of radius $1.54 \times 10^{-8}\, cm$, what per cent of solid argon is apparently empty space? $(Ar = 40)$

• A. 0.165
• B. 0.38
• C. 0.5
• D. 0.62

Answer 1

The Correct Option is D

Volume of one molecule
$=\frac{4}{3} \pi r^{3} $
$=\frac{4}{3} \pi\left(1.54 \times 10^{-8}\right)^{3} cm ^{3} $
$= 1.53 \times 10^{-23} cm ^{3}$
Volume of all molecules in $1.65\, g$ Ar
$=\frac{1.65}{40} \times N_{0} \times 1.53 \times 10^{-23}$
$=0.380\, cm ^{3}$
Volume of solid containing $1.65\, g\, Ar =1 \, cm ^{3}$
$\therefore$ Empty space $=1-0.380$
$=0.620$
$\therefore$ Per cent of empty space $=62 \%$