Question

The density of solid argon is $1.65g/mL$ at $-{233}^{o}C$. If the argon atom is assumed to be sphere of radius $1.54\times {10}^{-8}cm$, what percentage of solid argon is apparently the empty space?
$\text{Atomic weight of}Ar=40g/mol$
Take:
${1.54}^3=3.64$

• A. $32$%
• B. $52$%
• C. $62$%
• D. $72$%

Answer 1

The correct option is C $62$%

$Ar$ atom is assumed to be sphere,
$\text{Volume of one atom of}Ar=\frac{4}{3}\pi r^3$
$\text{Density of solid}Ar=1.65g/mL$
$\text{Number of atoms in}1.65g\text{in}1mL\text{is},$

$=\frac{1.65}{40}\times 6.023\times {10}^{23}$

So,
$\text{Total volume}=\frac{4}{3}\pi r^3\times \frac{1.65}{40}\times 6.023\times {10}^{23}$

$V=\frac{4}{3}\times \frac{22}{7}\times {(1.54\times {10}^{-8})}^3\times \frac{1.65}{40}\times 6.023\times {10}^{23}$

$V=0.380c{m}^3$

Hence,
$\%\text{of empty space}=\frac{(1-0.380)}{1}\times 100=62\%$