InterviewSolution
Saved Bookmarks
InterviewSolution
| 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. 9.5B. 9C. 4.75D. 4.5 |
|
Answer» Correct Answer - A `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` |
|