Question

The density of solid argon is 1.65 g/mL at $-233^{0}C$ . If the argon atom is assumed to be sphere of radius $1.54\times 10^{-8}cm$ , what percentage of solid argon has apparently empty space? (At. wt. of Ar = 40)

• A. 62%
• B. 72%
• C. 52%
• D. 42%

Answer 1

Answer: (A)

Volume of one atom of Ar $=\frac{4}{3}\pi r^3$
Also, No. of atoms in 1.65g $=\frac{1.65}{40}\times 6.023\times 10^{23}$
Total volume of all atoms of Ar in solid state.
$=\frac{4}{3}\pi r^3\times \frac{1.65}{40}\times 6.023\times 10^{23}$
$=\frac{4}{3}\times \frac{22}{7}\times {\left(1.54\times {\left(10\right)}^{-8}\right)}^3\times \frac{1.65}{40}\times 6.023\times {\left(10\right)}^{23}$
$=0.380c{m}^3$
Volume of solid argon $=1c{m}^3$
% Empty space $=\frac{[[1-0.380]}{1}\times 100=62\%$