Question

The number of chlorine atoms in 20 mL of chlorine gas at STP is ${10}^{21}$. (Round off to the Nearest Integer).


$[AssumechlorineisanidealgasatSTPR=0.083Lbarmo{l}^{-1}{K}^{-1},N_A=6.023\times {10}^{23}]$

Answer 1

Answer: 1

$n=\frac{PV}{RT}=\frac{1\times 20\times {10}^{-3}}{0.083\times 273}$
Since,

$\text{Number of atoms}=\text{No. of moles}\times 2\times N_A=\frac{1\times 20\times {10}^{-3}}{0.083\times 273}\times 2\times 6.023\times {10}^{23}$

$\text{Number of atoms}=1.06\times {10}^{21}$
Correct answer is 1