`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
C. 4.75
D. 4.5

Answer

A

Explanation

`O_(3)toO_(2)+O` .(i)
`H_(2)O_(2)toH_(2)O+O` ..(ii)
`underset((1)/(2)vol)(O)+underset((1)/(2)vol)(O)tounderset(1 vol)(O_(2))` ..(iii)
From equations (i) and (ii), we inter that 100 " mL of " `O_(3)` at STP will produce 100 " mL of " molecular `O_(2)` as such and 100 " mL of " oxygen molecular 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 have been contributed by `H_(2)O_(2)`.
Again we know
Volume of `H_(2)O_(2)xx` volume stregnth of `H_(2)O_(2)`
`=` Volume of `O_(2)` at STP
`therefore100 " mL of " 10 V H_(2)O_(2)-=1000" mL of " O_(2)` STP
After utilisation of 50 " mL of " `O_(2)`, according to equation (iii), the balance `(1000-50)=950mL` of `O_(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)=9.5V`
`therefore` Volume strength `=9.5`