The volume of oxygen obtained when $9.60 g$ ozone undergoes decomposition at STP is

The reaction of oxygen formation from ozone is

               $2{\mathrm{O}}_3⟶3{\mathrm{O}}_2$

weight of O₃ =  $9.6 \mathrm{g}$, 

 no. of moles of O₃ $=9.6/48=0.2$ moles

$2$ moles  O₃ of produce $3$ moles of O₂

$0.2$ moles of  O₃ produce $0.3$ moles of  O₂

Volume of any gas occupies $22.4$lit at STP

Then volume of O₂ $=0.3\times 22.4=6.72$ lit