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

Asked by  | 23rd Feb, 2013, 08:29: PM

Expert Answer:

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.

Answered by  | 24th Feb, 2013, 12:00: PM

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