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To a 25 ml H(2)O(2) solution, excess of ...

To a 25 ml `H_(2)O_(2)` solution, excess of acidified solution of Kl was added. The iodine liberated required 20 ml of 0.3 N `Na_(2)S_(2)O_(3)` solution.
The volme strength of `H_(2)O_(2)` solution is

A

56ml

B

112ml

C

168ml

D

224ml

Text Solution

Verified by Experts

The correct Answer is:
B
Volume of `O_(2)` liberated
= Volume strength `xx` vol. of `H_(2)O_(2)` solution `=20 xx 5.6 = 112ml`
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To a 25 mL of H_(2)O_(2) solution, excess of acidified solution of KI was added. The iodine liberated required 20 mL of 0.3 N Na_(2)S_(2)O_(3) solution. Calculate the volume strength of H_(2)O_2 solution. Strategy : Volume strength of H_(2)O_(2) solution is related to its normality by the following relation Volume strength (V)=5.6xx"Normality" (N) where, Normality =((meq)H_(2)O_(2))/V_(mL) According to the law of equivalence (meq)_(Na_(2)S_(2)O_(3))=(meq)_(I_(2))=(meq)_(H_(2)O_(2))

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