0.124 g of iron wire was dissolved in dilute `H_(2)SO_(4)` in oxygen free atmosphere and the resultant solution was titrated against 0.09672 N solution of `KMnO_(4)`. The titre value was 22.90 mL. Calculate the percentage purity of iron wire.
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`22.90 mL 0.09672 N KMnO_(4)-=22.90 mL 0.09672 N FeSO_(4)` Amount of `FeSO_(4)` in the solution `=(0.09672xx152xx22.90)/(1000)` =0.3366 g Amount of iron in 0.3366 g of `FeSO_(4)=(56)/(152)xx0.3366` =0.124 g Thus, percentage `=(0.124)/(0.124)xx100=100` The iron wire is 100% pure.
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