50 mL of pure and dry oxygen was subjected to silent electric discharge and on cooling to the original temperature, the volume of ozonised oxygen was found to be 47 mL. The gas was then brought in contact with turpentine oil, when after the absorption of ozone, the remaining gas occupied a volume of 41 mL. Find the molecular formula of ozone.

Explanation

Total volume of `O_(2)=50mL`
Volume of `O_(2)` (not converted to `O_(x)`) + ozone produced = 47 mL.
Volume of `O_(2)` (not converted to `O_(x)`) = 41 mL.
(ozone absorbed in turpentine oil)
`therefore` volume of `O_(2)` converted to `O_(x)=50-41=9mL`.
Volume of ozone produced = `47-41=6mL`
Let the formula of ozone be `O_(x)`.
`{:(,O_(2),rarr,O_(x)),(,"9 mL",,"6 mL"),(or,"9 moles",,"6 moles"):}`
Applying POAC for the O atom, we get,
`2xx` moles of `O_(2)=x xx` moles of `O_(x)`
`2xx9=x xx6`
`therefore x=3`,
Hence, the formula of ozone is `O_(3)`.