A 1. 0 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. 0 mL. An aliquot of 25. 0 mL of this solution requires 17. 0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.

A 1. 0 g sample of Fe_(2)O_(3) solid of 55. 2 % purity is dissolved in

1.	A 1. 0 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. 0 mL. An aliquot of 25. 0 mL of this solution requires 17. 0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.
Answer» Solution :Mass of `Fe_(2) O_(3) - ( 55.2) /( 100) xx 1 = 0.552 G ` moles of`Fe_(2)O_(3)= ( 0.552)/( 160) = 3.45 xx 10^(-3)` Equivalents of the oxidant `- ( n xx 0.0167 xx 17)/( 1000)- 2.84 xx 10^(-4)n ` (n is the 'n' factor of the oxidant ) Since on adding Zinc dust to the `Fe_(2) O_(3)` solution all the `Fe^(+3)` will become `Fe^(+2)` moles of `Fe^(2+)` in 100mL. `= 3.45 xx 10^(-3) xx 2 = 6.9 xx 10^(-3)` `:.` Equivalents of `Fe^(2+)` in the 25 mL , that is reacting with oxidant `= ( 6.9 xx 10^(-3))/(4)= 1.725 xx 10^(-3)` `:.` according to the Law of Equivalents `:. n = ( 1.72 xx 10^(-3))/(2.84 xx 10^(-4)) = 6.07 ~~6`