The number of atoms present in $142 g$ of Chlorine is

Chlorine atomic weight is $35.5 g$, so for ${Cl}_2$ it is $2\times 35.5=71 g$

$71 g$ of ${Cl}_2$ contains Avogadro number of molecules i.e., $6.023\times {10}^{23}$

Then for $142 g$ of ${Cl}_2$ contains $=142\times 6.023\times {10}^{23}/71=1.2\times {10}^{24}$

There are $2$chlorine atoms in ${Cl}_2=1.2\times {10}^{24}\times 2=2.4\times {10}^{24}$