Given Weight of H2O2 solution = 1.0 g % of H2O2 = X% Volume of KMnO4 used = X mL Equivalent weight of H2O2 = 17 We have to find the normality of KMnO4 solution Now, Milliequivalent = Normality × Volume Or Milliequivalent =
Weight
Equivalentweight
× Volume Milli eq of KMnO4 = NKMnO4×X Milli eq. of H2O2 =
X×1000
100×17
For complete neutralization Milli eq. of KMnO4 = Meq of H2O2 ... (i) Substituting all the values in equation (i), we get NKMnO4×X =
X×1000
100×17
NKMnO4 = 0.58823 ~ 0.588 Hence option 'D' is correct.