The number of chlorine atoms in 20 mL of chlorine gas at STP is ________ $$\times 10^{21}$$. (Round off to the Nearest Integer).
[Assume chlorine is an ideal gas at STP. R = 0.083 L bar mol$$^{-1}$$ K$$^{-1}$$, $$N_A = 6.023 \times 10^{23}$$]

Given:

Volume ($V$) = $20\text{ mL}=0.02\text{ L}$

Pressure at STP ($P$) = $1\text{ bar}$ (indicated by the units of the given R value)

Temperature at STP ($T$) = $273\text{ K}$

Universal Gas Constant ($R$) = $0.083\text{ L bar mol}^{-1}\text{ K}^{-1}$

Avogadro's Number ($N_A$) = $6.023\times 10^{23}$

Using the Ideal Gas Law:

$$n=\frac{PV}{RT}$$
$$n=\frac{1\times 0.02}{0.083\times 273}$$
$$n=\frac{0.02}{22.659}$$
$$n\approx 8.826\times 10^{-4}\text{ moles}$$

Multiply the number of moles by Avogadro's number:

$$\text{Molecules of }Cl_2=n\times N_A$$
$$\text{Molecules of }Cl_2=(8.826\times 10^{-4})\times (6.023\times 10^{23})$$
$$\text{Molecules of }Cl_2\approx 5.316\times 10^{20}$$

Chlorine gas is diatomic ($Cl_2$), meaning there are 2 atoms per molecule:

$$\text{Atoms of Cl}=2\times 5.316\times 10^{20}$$$$\text{Atoms of Cl}=10.632\times 10^{20}$$$$\text{Atoms of Cl}=1.0632\times 10^{21}$$

Therefore, the closest integer is 1.