UPSEE Solved Model Paper 7

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.