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`        A 5.0 cm3 solution of H2O2 liberates 0.508 g of iodine from an acidified KI solution. Calculate the strength of H2O2 solution in terms of volume strength at STP.`
5 years ago

Jitender Pal
365 Points
```							Sol. The redox reaction involved is :
H2O2 + 2I- + 2H+ → 2H2O + I2
If M is molarity of H2O2 solution, then
5M  =  (0.508 × 1000)/254 (∵ 1 mole H2O2 ≡ 1 mole I2)
⇒ M = 0.4
Also, n-factor of H2O2 is 2, therefore normality of H2O2 solution is 0.8 N.
⇒ Volume strength = Normality × 5.6
= 0.8 × 5.6 = 4.48 V

```
5 years ago
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