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

`100 mL` of ozone at `STP` was passed through `100 mL` of `10` volume

1.	`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?A. `9.5`B. `9.0`C. `4.75`D. `4.5`
Answer» Correct Answer - A `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 molecular oxygen after reaction with `H_(2)O_(2)` is contributed equally by `O_(3)` and `H_(2)O_(2)`. thus `50 mL` of oxygen has been contributed by `H_(2)O_(2)`. Again, we know that Volume of `H_(2)O_(2)xx` vol stre ngth of `H_(2)O_(2)` =vol of `O_(2)` at `STP` After utilisation of `50 mL` of `O_(2)`, according to Eq. (iii), the balance `(1000-50)=950 mL` of `O_(2)` at `STP` are still retainable by `100 mL of H_(2)O_(2)`. hence vol 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`

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

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