A 1.0 g sample of Fe2O3 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 up to 100.0 mL. An aliquot of 25.0 mL of this solution requires for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.

Sol. Mass of Fe2O3 = 0.552 g Millimol of Fe2O3 = 0.552/160 × 1000 = 3.45 During treatment with Zn-dust, all Fe3+ is reduced to Fe2+, hence millimol of Fe2+ (in 100 mL) = 3.45 × 2 = 6.90 ⇒In 25 mL aliquot, 6.90/4 = 1.725 millimol Fe2+ ion. Finally Fe2+ is oxidation to Fe3+, liberating one electron per Fe2+ ion. Therefore, total electrons taken up by oxidant. = 1.725 × 10-3 × 6.023 × 1023 = 1.04 × 10^21

Mass of Fe2O3 in the sample = 55.2/100 * 1 = 0.552 g Number of moles of Fe2O3 = 0.552/159.8 = 3.454 * 10-3Number of moles of Fe3+ ions = 2 * 3.454 * 10-3 = 6.9 * 10-3mol = 6.90 mmol Since its only 1 electron is exchanged in the conversion of Fe3+ to Fe2+, the molecular mass is the same as equivalent mass. ׵ Amount of Fe2+ ion in 100 ml. of sol. = 6.90 meq Volume of oxidant used for 100 ml of Fe2+ sol = 17 * 4 = 68 ml. Amount of oxidation used = 68 * 0.0167 mmol = 1.1356 mmol Let the number of electrons taken by the oxidant = n׵ No. of meq.of oxidant used = ͳ.ͳ͵ͷ6 * nThus 1.1356 * n = 6.90 N = 6.90/1.1356 = 6