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A 5-0cm^(3) solutions of H(2)O(2) libera...

A `5-0cm^(3)` solutions of `H_(2)O_(2)` liberates of 0.508g of iodine from acidified KI solution. Calculate the volume strength of `H_(2)O_(2)` at N.T.P.

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`"meq of "H_(2)O_(2)="meq of "I_(2)." That means, "((w)/(17))xx1000=[(0.508)/(127)]xx1000 therefore w=0.068g`
This means that `N_(H_(2)O_(2))=(0.068xx1000)/(17xx5)=0.8M" Volume Strength"=Nxx5.6=4.48" vol"`
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