Question

The density of solid argon (Ar = 40 g / mol) is 1.68 g/mL at 40 K. If the argon atom is assumed to be a sphere of radius 1.50 x ${10}^{-8}$ cm, what % of solid Ar is apparently empty space? (use $N_A=6\times {10}^{23}$)

• A. 35.64
• B. 64.36
• C. 74%
• D. none of these

Answer 1

Answer: (D)

none of these

Volume of one atom of $Ar=\frac{4}{3}\pi r^3$

Also, number of atoms in 1.65 g/mL $=\frac{1.65}{40}\times 6.023\times {10}^{23}$

∴ Total volume of all atoms of Ar in solid state $=\frac{4}{3}\pi r^3\times \frac{1.65}{40}\times 6.023\times {10}^{23}$

$=\frac{4}{3}\times 3.14\times (1.54\times {10}^{-8}{)}^3\times \frac{1.65}{40}\times 6.023\times {10}^{23}=0.380c{m}^3$

Volume of solid argon $=1c{m}^3$

∴% Empty space$=\frac{1-0.380}{1}\times 100=62$%