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In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. How many grams of iron were present in the sample ? |
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Answer» Solution :In this problem `Fe^(2+)` is oxidised to `Fe^(3+)` by `KMnO_(4)` Wt.of IRON = equivalent `xx` eq.wt ...(Eqn.4) ` = 0.0042 xx 56 =0.2352 g` `{Fe^(2+)toFe^(3+):. "of Fe"=("at.wt")/("change in ON")=56/1 = 56}`. |
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