The number of chlorine atoms in 20 mL of chlorine gas at STP is____ 1021. (Round off to the Nearest Integer)(Assume chlorine is an ideal gas at STPR=0.083 L bar mol-1 K-1, NA=-6.023×1023 )

$P V = n R T$

Where $P =$ Pressure, $V =$ Volume, $n =$ Number of moles, $R =$ constant and $T =$ Temperature

Compute the number of moles:

-Substitute the known values in the formula $n = \frac{P V}{R T}$ , $n = \frac{1 \times 20 \times 10^{- 3}}{0.083 \times 273}$

$n = 0 \cdot 882 \times 10^{- 3}$

Determine the number of atoms:

The number of atoms $= 0 \cdot 882 \times 10^{- 3} \times 2 \times 6 \cdot 023 \times 10^{23}$ $= 1 \cdot 06 \times 10^{21}$ atoms.

Hence the correct answer is 1.06.

The number of chlorine atoms in $20 m L$ of chlorine gas at STP is____ $10^{21}$ . (Round off to the Nearest Integer)
(Assume chlorine is an ideal gas at STP
$R = 0.083 L$ bar ${m o l}^{- 1} K^{- 1}, N_{A} = - 6.023 \times 10^{23}$ )