A `1 g` sample of `Fe_(2)O_(3)` solid of `55.2%` purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made upto `100 mL`. An aliquot of `25 mL` of this solution requires `17 mL` of `0.0167M` solution of an oxidant for titration. Calculate no.of electrons taken up by oxidant in the above titration.

Answer

`(1.04 xx 10^(4))`

Explanation

Mass of `Fe_(2)O_(3) = 0.552g`
millimol of `Fe_(2)O_(3) = (0.552)/(160) xx 100 = 3.45`
During treatment with `Zn`-dust, all `Fe^(3+)` is reduced to `Fe^(2+)`, hence
millimol of `Fe^(2+)` is oxidised to `Fe^(3+)`, liberating one electron per `Fe^(2+)` ion. Therefore, total electrons taken up by oxidant.
`=1.725 xx 10^(-3) xx 6.023 xx 10^(23) = 1.04 xx 10^(21)`