Question

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

• A. $16.5$%
• B. $38$%
• C. $50$%
• D. $62$%

Answer 1

The correct option is D $62$%

Volume of one molecule $=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {(1.54\times {10}^{-8})}^3{cm}^3$
$=1.53\times {10}^{-23}{cm}^3$
Volume of molecules in $1.65g$ $Ar$ $=\frac{1.65}{40}\times N_0\times 1.53\times {10}^{-23}=0.380{cm}^3$
Volume of solid containing $1.65g$ $Ar$ $=1{cm}^3$
$\therefore$ Empty space $=1-0.380=0.620$
$\therefore$ Percent of empty space $=62$%