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.

30 mL of a H_(2)O_(2) solution after acidification required 30 mL of N

1.	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.
Answer» Solution :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 determinethe 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