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Question
On passing 1.8 l of oxygen through ozonizer gives 1.6 l of a mixture of ozone and oxygen. What is the mole fraction of ozone in resultant mixture?
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Solution
The reaction occurring is
3 O2 ---> 2 O3
Initially there is 1.8 L of O2 gas.
Let us assume that V volume of O3 gas is formed , so volume of O2 is reduced, considering pressure and temperature remains unchanged.(Since at constant pressure and temperature number of moles is directly proportional to their volume.)
Hence, final volume of mixture = 1.8 - +V = 1.6 => V = 0.4 L of O3 gas. and 1.2 L of O2 gas left.
Mole fraction of O3 in resultant mixture = = 0.25
The reaction occurring is
3 O2 ---> 2 O3
Initially there is 1.8 L of O2 gas.
Let us assume that V volume of O3 gas is formed , so volume of O2 is reduced, considering pressure and temperature remains unchanged.(Since at constant pressure and temperature number of moles is directly proportional to their volume.)
Hence, final volume of mixture = 1.8 - +V = 1.6 => V = 0.4 L of O3 gas. and 1.2 L of O2 gas left.
Mole fraction of O3 in resultant mixture = = 0.25
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