Question

The density of solid argon is $1.65gpercc$ at $-{233}^{o}C$. If the argon atom is assumed to be a sphere of radius $1.5\times {10}^{-8}cm$, what percent of solid argon is apparantly empty space?
Given: ($Ar=40g/mol$)

• A. 16.5%
• B. 38%
• C. 50%
• D. 65%

Answer 1

The correct option is D 65%

(d)
Here the argon atom is assumed to be a sphere so,
Volume of one molecule = $\frac{4}{3}\pi r^3$
= $\frac{4}{3}\pi (1.5\times {10}^{-8}{)}^{3}c{m}^3$
= $1.41\times {10}^{-23}c{m}^3$
Number of molecules in $1.65g$ of Ar is
$=\frac{1.65}{40}$ $\times$ $N_A$
Volume of all molecules in $1.65g$ of Ar is
$=\frac{1.65}{40}\times N_A\times 1.41\times {10}^{-23}$
= $0.35c{m}^3$
$\text{Volume of solid containing}1.65gofAr=1c{m}^3$
$\therefore$
$\text{Empty space = Volume of solid - Volume occupied by molecules}$
$=1-0.350=0.650$
$\therefore$ Percent of empty space = 65%