The number of chlorine atoms in 20 , mL of ch | vedclass

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(D) At $STP$,the pressure $P = 1 \, bar$ and temperature $T = 273.15 \, K$.Using the ideal gas equation $PV = nRT$,where $n = \frac{N}{N_A}$:$1 	imes

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$1$

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(D) At $STP$,the pressure $P = 1 \, bar$ and temperature $T = 273.15 \, K$.Using the ideal gas equation $PV = nRT$,where $n = \frac{N}{N_A}$:$1 	imes (20 	imes 10^{-3} \, L) = \frac{N}{6.023 	imes 10^{23}} 	imes 0.083 	imes 273.15$$N = \frac{0.020 	imes 6.023 	imes 10^{23}}{0.083 	imes 273.15} \approx 5.31 	imes 10^{20} \, 	ext{molecules of } Cl_2$.Since each $Cl_2$ molecule contains $2$ chlorine atoms:Number of $Cl$ atoms $= 2 	imes 5.31 	imes 10^{20} = 10.62 	imes 10^{20} = 1.062 	imes 10^{21}$.Rounding to the nearest integer,we get $1 	imes 10^{21}$.

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 $STP$,$R = 0.083 \, L \, bar \, mol^{-1} \, K^{-1}$,$N_{A} = 6.023 \times 10^{23}$]

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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 $STP$,$R = 0.083 \, L \, bar \, mol^{-1} \, K^{-1}$,$N_{A} = 6.023 \times 10^{23}$]