Under similar conditions of temperature and pressure,2 volumes of ozone `(O_(3))` on decomposition give 3 volumes oxygen `(O_(2))`. 20 mL of the gaseous mixture when heated and then cooled expands to 21 mL. Find the percentage volume ozone in the gaseous mixture.

Explanation

Let the volume of ozone in the gaseous mixture `= x mL`
`:.` Volume of oxygen in the gaseous mixture `= (20-x)mL`
Now, 2L of ozone will form oxygen = 3 mL
x mL of ozone will form oxygen `= 3//2x mL`
Total volume of oxygen after the reaction `=(20 -x+3//2 x)mL`
But the volume of oxygen which is actually formed = 21 mL
`:. 20 - x + 3//2x = 21 or x = 2`
`:.` Volume of ozone in the mixture `= 2mL`
Percentage of ozone in the mixture `= ((2 mL))/((20 mL)) xx 100 = 10 %`.