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.
$=\frac{4}{3}\pi r^{3}\times \frac{1.65}{40}\times \, 6.023\times 10^{23}$
$=\frac{4}{3}\times \frac{22}{7}\times \left(1.54 \times \left(10\right)^{- 8}\right)^{3}\times \frac{1.65}{40}\times 6.023\times \left(10\right)^{23}$
$=0.380 \, cm^{3}$
Volume of solid argon $=1 \, cm^{3}$
% Empty space $=\frac{\left[\right. \left[\right. 1 - 0.380 \left]\right.}{1}\times 100=62\%$