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A bottle of H(2)O(2), is labelled as 10 ...

A bottle of `H_(2)O_(2)`, is labelled as 10 vol `H_(2)O_(2)`. 112 mL of this solution of `H_(2)O_(2)`, is titrated against 0.04 M acidified solution of `KMnO_(4)`, Calculate the volume of `KMnO_(4)`, in terms of litre.

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The correct Answer is:
1
5.6 vol `H_(2) O_(2) =1 N " " " or" " " 10 vol H_(2) O_(2) = (10)/(5.6) ` N
mEq of `H_(2) O_(2) = mEq ` of `MnO_(4)^(-) " " ," " (10)/(5.6) xx 112 = 0.04 xx 5 ` ( n- factor ) `xx v `
V=1000 mL = 1 L
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