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`        8.0 mL solution of H2O2 liberates 0.508 g of iodine from excess of acidified KI solution. Calculate volume strength of H2O2 solution. (I = 127)`
one year ago

Arun
18664 Points
```							The redox reaction involved is : H2O2 + 2I- + 2H+ → 2H2O + I2 If M is molarity of H2O2 solution, then 8M = (0.508 × 1000)/254 (∵ 1 mole H2O2 ≡ 1 mole I2) ⇒ M = 0.25Also, n-factor of H2O2 is 2, therefore normality of H2O2 solution is 0.5 N. ⇒ Volume strength = Normality × 5.6 = 0.5 × 5.6 = 2.80 V  RegardsArun(askIITians forum expert)
```
one year ago
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