Question

# 10 mL of a solution of H2O2 labelled '10 volume' just decolorises 100 mL of potassium permanganate solution acidified with dilute H2SO4. Calculate the amount of potassium permanganate in the given solution.

A
0.1563 g
B
0.563 g
C
5.63 g
D
0.256 g

Solution

## The correct option is C 0.563 gGiven reaction is: 2H2O2(aq)Δ→2H2O(l)+O2(g) 10 mL (0.01 L) volume H2O2 aqueous solution means on heating 1 mL of the H2O2 aqueous solution will produce 10 mL of O2 gas at STP. Volume strength of H2O2 aqueous solution can be related to the molarity of H2O2 solution as Volume strength of H2O2= Molarity of H2O2×11.210=Molarity×11.2 Molarity of H2O2 solution = 0.0111.2M H2O2(aq)n=2+KMnO4n=5(aq)H+→Mn2+(aq)+O2 n - factor of H2O2 in the above reaction is 2 n - factor of KMnO4 in the above reaction is 5 Equivalents of H2O2=equivalents of KMnO4  n−factor×Molarity×Volume= no. of moles of KMnO4×n−factor 2×0.0111.2×10=x158×5x=0.564 g

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