Question

The density of solid argon (atomic mass $=40$ amu) 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, then the percentage of solid Ar which is space will be:

[use $N_A=6\times {10}^{23}$]

• A. $35.64$
• B. $64.36$
• C. $73$
• D. None of the above

Answer 1

The correct option is B $64.36$

As we know, density $=1.68=\frac{nM}{V{N}_A}=\frac{n\times 40}{V\times 6\times {10}^{23}}$
$⟹V=39.53\times {10}^{-24}n$
The radius of argon is $1.5\times {10}^{-8}$ cm and so, the empty space is $(1-\frac{4}{3})\times n\times \pi r^3\times \frac{100}{V}=64.3\%$