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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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Answer» `(meq)_(Na_(2)S_(2)O_(3))=V_(mL)xxN` `=(20)(0.3)=6` `=(meq)_(H_(2)O_(2))` Thus, normality of `H_(2)O_(2)=(meq)_(H_(2)O_(2))//V_(mL)` `6/25eq L^(-1)` Volume strength `(V)=5.6xx "Normality"` `=(5.6)(6//25)` `=1.344` |
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