A quantity of 1.0 g sample of F e 2 O 3 solid of 69.6 per cent purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made up to 100 m

Question

A quantity of 1.0 g sample of F e 2 ​ O 3 ​ solid of 69.6 per cent purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made up to 100 ml. An aliquot of 25 ml of this solution requires 26.05 ml of 0.0167 M solution of an oxidant for titration. The number of electrons taken up by the oxidant in the reaction of the above titration is

TestHub · Accepted Answer

F e 2 ​ O 3 ​ = 0.696 g = 0.696/160 = 4.35 mmol, so F e 2 + = 8.7 mmol in 100 ml; 25 ml aliquot holds 2.175 mmol F e 2 + , releasing 2.175 mmol e − . Oxidant = 0.0167 × 0.02605 = 0.435 mmol. Electrons per oxidant = 2.175/0.435 = 5 (e.g. Mn O 4 − ​ → M n 2 + ). Ans: 5.

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