`{:(O_(3)toO_(2)+O),(H_(2)O_(2)toH_(2)O+O),(underset((1//2 vol))(O)+underset((1//2 vol))(O)tounderset((1 vol))(O_(2))):}`
From Eqs. (i) and (ii), we get `50 mL` of `O_(3)` at `STP` will produce `50 mL` of molecular `O_(2)` as such and `50 mL` of oxyten molecule after reaction with `H_(2)O_(2)`.
This new volume of `50 mL` of molecular oxygen after reaction wilth `H_(2)O_(2)` is contributed equally by `O_(3)` and `H_(2)O_(2)`. thus, `25 mL` of oxygen has been contributed by `H_(2)O_(2)`. Volume of `H_(2)O_(2)xx` volume strength of `H_(2)O_(2)`=volume of `O_(2)` at `STP 50 mLxx5V H_(2)O_(2)=250 mL of O_(2)` at `STP`.
After utilisation of `25 mL of O_(2)`, according to Eq. (iii), the balance `(250-25)mL=225 mL of O_(2)` is still available by `50 mL of H_(2)O_(2)`.
Hence, volume strength of `H_(2)O_(2)`, after reaction is
`("Volume of "O_(2) at STP)/("Volume of " H_(2)O_(2))=225/50=4.5`
`:.` Volume strength `4.5`
