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Question

An excess of acidified solution of KI is added to a 25 mL H2O2 solution. The iodine liberated requires 20 mL of 0.3 N Na2S2O3 solution. Calculate the volume strength of H2O2:
  1. 1.144
  2. 1.244
  3. 1.344
  4. 1.444

A
1.344
B
1.444
C
1.244
D
1.144
Solution
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I2+2Na2S2O3Na2S4O6+2NaI
20 ml of 0.3M(=0.3N) Na2S2O3 will react with (20)(0.3)(1000)(2) moles of I2 which is 0.003 moles or 0.006 equivalents. Thus there are 0.006 equivalents in 25 ml of H2O2, hence, its normality is 0.24 N and thus, the volume strength is 0.24×5.6=1.344.

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