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Balancing Redox Reaction

50 ml of H2O2...

Question

50 ml of $H_{2} O_{2}$ solution when titrated with 0.1 M $K M n O_{4}$ solution in acidic medium, volume required is 40 ml. Volume strength of $H_{2} O_{2}$ solution is

11.2

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22.4

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1.12

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2.24

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Solution

The correct option is D 2.24
Meqv of $K M n O_{4}$ solution $= 0.1 \times 5 \times 40 m e q v = 20 m e q v$
So, Meqv of $H_{2} O_{2}$ solution = 20 meqv $∴ M \times v o l \times n - f a c = M \times 50 \times 2 = 20$
So, Molarity of $H_{2} O_{2} s o l u t i o n = M = 0.2 M$
So, Volume strength of $H_{2} O_{2}$ solution $= 11.2 \times M = 11.2 \times 0.2 = 2.24 v o l u m e$

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50 ml of H2O2 solution when titrated with 0.1 M KMnO4 solution in acidic medium, volume required is 40 ml. Volume strength of H2O2 solution is

The correct option is D 2.24Meqv of KMnO4 solution =0.1×5×40 meqv=20 meqv So, Meqv of H2O2 solution = 20 meqv ∴M×vol×n−fac=M×50×2=20 So, Molarity of H2O2 solution=M=0.2 M So, Volume strength of H2O2 solution =11.2×M=11.2×0.2=2.24 volume

50 ml of $H_{2} O_{2}$ solution when titrated with 0.1 M $K M n O_{4}$ solution in acidic medium, volume required is 40 ml. Volume strength of $H_{2} O_{2}$ solution is