20 mL of a `H_(2)O_(2)` solution after acidification required 20 mL of N/10 KMn`O_(4)` solution for complete oxidation. Calculate the percentage and volume strength of `H_(2)O_(2)` solution.

Explanation

From the given data, for `H_(2)O_(2)` solution, `V_(1)` = 20 mL and for KMn`O_(4)` solution `V_(2)` = 20 mL , `N_(2) = (N)/(10)` .
Applying normality equation, `N_(1)V_(1) = N_(2) V_(2)`
or, `20 xx N_(1) = 20 xx (1)/(10) " " therefore N_(1) = 0.1(N)`
Thus, normality of the `H_(2)O_(2)` solution = 0.1(N).
Now, amount of `H_(2)O_(2)` in 1 L solution = 0.1 `xx 17 = 1.7` g
`therefore` The amount of `H_(2)O_(2)` in 100 mL in of the solution
` = (1.7 xx 100)/(1000) = 0.17` g
`therefore` the percentage strength of the solution = 0.17 %
Now, 68 g of `H_(2) O_(2)` produces 22400 mL of `O_(2)` at STP
`therefore` 1.7g of `H_(2) O_(2)` produces `(22400)/(68) xx 1.7` = 560 mL `O_(2)` at STP.
This 1.7 g of `H_(2) O_(2)` is present in 1000 mL of `H_(2) O_(2)` solution.
Hence, 1000 mL of `H_(2)O_(2)` solution gives 560 mL of `O_(2)` at STP.
`therefore " "1 "mL" H_(2)O_(2)` solution gives `(560)/(1000)` = 0.56 mL of `O_(2)` at STP`.