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Volume strength of H(2)O(2) is 5.6, its ...

Volume strength of `H_(2)O_(2)` is 5.6, its normality will be :

Text Solution

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The correct Answer is:
1
`V=Nxx5.6`
`5.6xxNxx5.6`
`therefore N=1`
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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))