Question:
The number of chlorine atoms in $20 \mathrm{~mL}$ of chlorine gas at STP is $10^{21}$. (Round off to the Nearest Integer).
[Assume chlorine is an ideal gas at STP
$\mathrm{R}=0.083 \mathrm{~L}$ bar $\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}, \mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23}\right]$
Solution:
$\mathrm{PV}=\mathrm{nRT}$
$1.0 \times \frac{20}{1000}=\frac{\mathrm{N}}{6.023 \times 10^{23}} \times 0.083 \times 273$
$\therefore$ Number of $\mathrm{Cl}_{2}$ molecules, $\mathrm{N}=5.3 \times 10^{20}$
Hence, Number of Cl-atoms $=1.06 \times 10^{21}$
$\approx 1 \times 10^{21}$