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A 5.0 mL of solution of H(2)O(2) liberat...

A `5.0 mL` of solution of `H_(2)O_(2)` liberates `0.508 g` of iodine from acidified `KI` solution. Calculate the strength of `H_(2)O_(2)` solution in terms of volume strength at `STP`.

A

2.24 V

B

1.12 V

C

4.48 V

D

8.96 V

Text Solution

Verified by Experts

The correct Answer is:
C
meq `H_(2)O_(2)=` meq `I_(2)`
`Nxx5=(0.508xx2)/(254)xx1000` or Normality = 0.8 N
Volume strength = `5.6xxN=5.6xx.8=4.48 V`.
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