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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