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.

Explanation

Oxidant `-17xx0.0167=0.2839 m mol`
Let the oxidation state changes by n
20 " mL of " solution `=0.2839n mEq`
100 " mL of " solution `-0.2839nxx4mEq`
`=1.135n m" Eq of "Fe_2O_3`
Weight of `Fe_2O_3=(55.2)/(100)[Ew(Fe_2O_3)=(Mw)/(2)=80]`
" Eq of "`Fe_2O_3=(1)/(80)xx(55.2)/(100)`
`therefore(55.2)/(100xx80)=(1.1356)/(1000)n`
`thereforen=6`