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Question

To a 25 mL H2O2 solution, excess of an acidified solution of potassium iodide was added. The iodine liberated required 20 mL of 0.3 N sodium thiosulphate solution.

Calculate the volume strength of H2O2 solution (to the nearest integer):

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Solution

Milliequivalents of Na2S2O3=20×0.3=6

Milliequivalents of I2=6

Milliequivalents of KI=6

Milliequivalents of H2O2=6

Normality of H2O2=625

For H2O2, 5.6× Normality = Volume strength

Volume strength=625×5.6=1.344
Hence, the answer is 1 (to the nearest integer).

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