30 mL of a `H_(2)O_(2)` solution after acidification required 30 mL of N/10 `KMnO_(4)` solution for complete oxidation . Calculate the percentage and volume strength of `H_(2)O_(2)` solution.

Explanation

To determine the normality of `H_(2)O_(2)` solution . From the given data,
for `H_(2)O_(2), V_(1)=30 mL, N_(1)=?`
For `KMnO_(4), V_(2)=30 mL, N_(2)=N//10`
Applying normallity equation , `N_(1)V_(1)=N_(2)V_(2) , i.e., 30xxN_(1)=30xx1//10 therefore N_(1)=0.1 N`
Thus, the normality of `H_(2)O_(2)` solution =0.1 N
Step 2. To determine the percentage strength of `H_(2)O_(2)` solution,
We know that, `H_(2)O_(2) to 2H^(+) + O_(2) + 2e^(-)" "therefore "` Eq. wt. of `H_(2)O_(2) =34//2=17`
Hence, strength of `H_(2)O_(2)=(1.7xx100)/(1000)=0.17%`
Step 3. To determine the volume strength of `H_(2)O_(2)` solution.
Consider the chemical equation,
`underset(68 g)(2H_(2)O_(2)) to 2H_(2)O + underset(22400 mL at N.T.P.)O_(2)`
Now 68 g of `H_(2)O_(2)` give `O_(2)` at N.T.P. =22400 mL
`therefore 1.7 g ` of `H_(2)O_(2)` will give `O_(2)=(22400)/(68)xx1.7=560` mL
But 1.7 g of `H_(2)O_(2)` are 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 N.T.P.
`therefore ` 1 mL of `H_(2)O_(2)` solution will give `=(560)/(1000)=0.56 ` mL of `O_(2)` at N.T.P.
`therefore` Volume strength of `H_(2)O_(2)` solution =0.56