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Question

To a 30 mL 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.

A
0.374 V
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B
2.544 V
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C
4.244 V
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D
1.12 V
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Solution

The correct option is D 1.12 V
Miliequivalent of 25 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 at NTP.
Equivalent of H2O2/L = Equivalent of O2=V (NTP)Equivalent vol.of O2=x5.6

x5.6=630x=1.12
The volume stregnth of H2O2 is 1.12V

Alternate solution:
Volume strength=Normality×5.6Volume strength=630×5.6=1.12V

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