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

Weight of `Fe_(2)O_(3) = 0.552 g `
Number of eq. od `Fe_(2)O_(3) = (0.552)/80 {:(,Fe_(2)O_(3)to,2FeO),("+6",," +4"),(eq. wt . of ,Fe_(2)O_(3)= 160/2,=80):}`
Let the number of electrons taken up by the oxidant in the reaction be n ( i..e the change in oxidation number )
` :. ` normality of the oxidant = 0.0167 n N
` :. ` m.e of 25 mL of `Fe^(2+)` solution = 0.0167 n xx 17
` :. ` m.e of 100 mL of `Fe^(2+)` solution = `(68 xx 0.0167 n)/(1000)`
eq . of `Fe_(2)O_(3) ` = eq of FeO
`(0.552)/80 = (68 xx 0.167n)/1000 n = 6`