The number of chlorine atoms in 20 mL of chlorine gas at STP is $x\times {10}^{21}$. The value of x (Rounded off to the Nearest integer) is
.
[Assume chlorine is an ideal gas at STP,
$N_A=6.023\times {10}^{23}$]

Moles of $C{l}_2$ in $20mL$ of it at STP $=\frac{20}{22400}=8.9\times {10}^{-4}mol$
Thus, no. of $Cl$ atoms $=moles\times N_A\times atomicity$ ..... eq. (i)
Atomicity is the number of atoms per molecule of a species.
$\therefore$ Atomicity of $C{l}_2=2$
Putting values in eq(i), we get;
no. of $Cl$ atoms $=8.9\times {10}^{-4}\times 6.022\times {10}^{23}\times 2\approx 1\times {10}^{21}$
Thus value of $x=1$.