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 $=1.50 \times 10^{-8}\, cm$, what $\%$ of solid $Ar$ is apparently empty space?

• A. $35.64$
• B. $64.36$
• C. $74 \%$
• D. None of these

Answer 1

Answer: (B)

Volume of all atoms in $1.68\, g$

Argon $=\frac{1.68}{40} \times N_{ A } \times \frac{4}{3} \times \pi \times\left(1.5 \times 10^{-8}\right)^{3}=0.3564$

Volume of solid argon $=1\, cm ^{3}$

$\%$ Empty space $=(1-0.3564) \times 100=64.36$