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Question

H2O2+2KI40% yieldI2+2KOHH2O2+2KMnO4+3H2SO450% yieldK2SO4+2MnSO4+3O2+4H2O
150 mL of a H2O2 sample was divided into two parts. The first part was treated with KI and the formed KOH required 200mL of M2H2SO4 for complete neutralisation. The other part was treated with KMnO4 yielding 6.72 L of O2 at 1 atm and 273 K. Using the % yields indicated, find the volume strength of the H2O2 sample used.

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Solution

Moles of H2SO4=0.2×12 mol=0.1 mol
Moles of used KOH will be=0.1×2 mol=0.2 mol

H2O2+2KI40% yieldI2+2KOH
Moles of H2O2 used in the first reaction
=0.22×10.4=0.25
According to the question, the moles of O2 produced =6.7222.4=0.3 mol

Given,
H2O2+2KMnO4+3H2SO450% yieldK2SO4+2MnSO4+3O2+4H2O
Moles of H2O2 used in the second reaction
=0.33×0.5=0.2
So, the total moles of H2O2 consumed =(0.25+0.2) mol=0.45 mol
Molarity of H2O2=0.450.15=3 M
Volume strength =11.2×3=33.6

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