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Question

To a 30 mL of H2O2 solution, excess of acidified solution of KI was added. The iodine liberated required 20 mL of 0.3 N Na2S2O3. Calculate the volume stregnth of H2O2 solution.

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Solution

Miliequivalent of 30 mL H2O2 = Miliequivalent of I2 = Miliequivalent of Na2S2O3 = 20×0.3=6.
Normality of H2O2=630 eq/L
Let, the volume stregnth of H2O2 be x V. Thus, 1 L of H2O2 produces x L of O2.
Volume strength = 5.6 x Normality
= 5.6 (6/30) =1.12 volume
The volume stregnth of H2O2 is 1.12Volume



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