Question:
The number of chlorine atoms in $20 \mathrm{~mL}$ of chlorine gas at $S T P$ is____________ $10^{21}$. (Round off to the Nearest Integer). [Assume chlorine is an ideal gas at STP $\mathrm{R}=0.083 \mathrm{~L}$ bar $\mathrm{mol}^{-1} \mathrm{~K}^{-1}, \mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23}$ ]
Solution:
(1)
$\mathrm{PV}=\mathrm{nRT}$
$1.0 \times \frac{20}{1000}=\frac{\mathrm{N}}{6.023 \times 10^{23}} \times 0.083 \times 273$
$\therefore$ Number of $\mathrm{Cl}_{2}$ molecules, $\mathrm{N}=5.3 \times 10^{20}$
Hence, Number of Cl-atoms $=1.06 \times 10^{21}$
$\approx 1 \times 10^{21}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.