Q19 / 120 · VITEEE 2013 · Chemistry
The density of solid argon is $1.65 \mathrm{~g}$ per $\mathrm{cc}$ at $233^{\circ} \mathrm{C}$. If the argon atom is assumed to be a sphere of radius $1.54 \times 10^{-8} \mathrm{~cm}$, what per cent of solid argon is apparently empty space? $(\mathrm{Ar}=40)$
- A$16.5 \%$
- B$38 \%$
- C$50 \%$
- Correct answerD$62 \%$
Solution
Volume of one molecule
$\begin{array}{l}
=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi\left(1.54 \times 10^{-8}\right)^{3} \mathrm{~cm}^{3} \\
=1.53 \times 10^{-23} \mathrm{~cm}^{3}
\end{array}$
Volume of molecules in $1.65 \mathrm{~g} \mathrm{Ar}$
$=\frac{1.65}{40} \times \mathrm{N}_{0} \times 1.53 \times 10^{-23}=0.380 \mathrm{~cm}^{3}$
Volume of solid containing $1.65 \mathrm{~g} \mathrm{Ar}=1 \mathrm{~cm}^{3}$
$\therefore$ Empty space $=1-0.380=0.620$
$\therefore$ Per cent of empty space $=62 \%$
Hence, the correct answer is option: 'D'.