`100 mL` of ozone at `STP` was passed through `100 mL` of `10` volume `H_(2)O_(2)` solution. What is the volume strength of `H_(2)O_(2)` after attraction?

Options

A. `9.5`
B. `9.0`
C. `4.75`
D. `4.5`

Answer

A

Explanation

`O_(3)toO_(2)+O....(i)`
`H_(2)O_(2)toH_(2)O+O....(ii)`
`O+OtoO_(2)......(iii)`
`1//2 vol ,1//2 vol ,1vol`
`100` mL of `O_(3)` at `STP` will produce will produce `100 mL` of `O_(2)` as such and `100 mL` of `O_(2)` after reaction with `H_(2)O_(2)`. this new volume of `100 mL` of molecule after oxygen reaction with `H_(2)O_(2)` is contributed equally by `O_(2)` and `H_(2)O_(2)`. thus `50 mL` of oxygen has been contributed `H_(2)O_(2)`. again we know, volume of `H_(2)O_(2)xx`volume strength of `H_(2)O_(2)`
`=` volume of `O_(2) at STP`
`:. 100 mL` of '`10` volume' `H_(2)O_(2)-=1000 mL of O_(2)` at `STP` After unitlisation of `50 mL` of `O_(2)` of `O_(3)` according to Eq(iii) m the balance `(1000-50)=950 mL o fO_(2) at STP`
are still retainable by `100 mL of H_(2)O_(2)`
Hence volume strength of `H_(2)O_(2)` after reaction
`=("Volume of " O_(2) at STP)/("Volume of "H_(2)O_(2))=950/100 =9.5 V`
`:.` Volume strength `=9.5`
Hence the correct option is (a).