The no.of atoms of ozone present in 11.2 lit.of ozone at NTP are.........

Volume of 1 mole of gas at NTP is 24.0L.

Or we can say 22.4 lt of a gas at NTP has 6.022 x 10²³ molecules.

 

So, 11.2 lt ozone will have = (6.022 x 10²³  x 11.2 )/ 22.4  molecules of ozone.