Question

The density of solid argon is $1.65$ g mL${}^{-1}$ at $-{233}^{\circ }C$. If the argon atom is assumed to be the sphere of radius $1.54\times {10}^{-8}$ cm, what percentage of solid argon is apparently empty space?

[Atomic mass of Ar is $40$ amu.]

• A. $62\%$
• B. $76\%$
• C. $68\%$
• D. $52\%$

Answer 1

The correct option is A $62\%$

Volume of one atom of $Ar=\frac{4}{3}\pi r^3$
Also, number of atoms in $1.65$ g or in $1$ mL $=\frac{1.65}{40}\times 6.023\times {10}^{23}$
$\therefore$ Total volume of all atoms of $Ar$ in solid state $=\frac{4}{3}\pi r^3\times \frac{1.65}{40}\times 6.023\times {10}^{23}$ $=\frac{4}{3}\times \frac{22}{7}\times (1.54\times {10}^{-8}{)}^3\times \frac{1.65}{40}\times 6.023\times {10}^{23}$ $=0.380$ cm${}^3$

Volume of solid argon $=1$ cm${}^3$
$\therefore \%$ Empty space $=\frac{[1-0.380]}{1}\times 100=62\%$