When exposed to UV radiation pure oxygen converts to Ozone in a steel cylinder. if the initial pressure of oxygen was 750 mm and after conversion to ozone was 720 mm, find the molar composition of the final mixture and the percentage of oxygen converted to ozone.

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Solution

Dear Student,

The computation is expressed below,

$f r o m s t a t e m e n t w e g e t, \frac{P V}{P V} = \frac{n R T}{n R T} \frac{750}{720} = \frac{n_{o x y g e n}}{n_{o z o n e}} \frac{n_{o x y g e n}}{n_{o z o n e}} = 1.042 % o f o x y g e n c o n v e r t e d t o o z o n e = \frac{1}{1.042} \times 100 % = 95.96 %$
Regards.

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