For the oxidation of 0 . 2   M   FeSO 4 solution 0 . 965 amperes current is passed through it for 1 hour. The volume of the solution that is oxidised in mL is

Step-by-step solution

Given that, 0 . 2   M   FeSO 4 and current = 0 . 965   amp - hr = 0 . 965 × 3600   coulomb Dissociation of FeSO 4 = Fe 2 + + SO 4 2 - n - factor = 2 We know that, Normality = Molarity × n - factor ⇒ Normality = 2 × 0 . 2 = 0 . 4   N Here, 96500   coulomb = 1   eq .   Fe 2 + . ⇒ 0 . 965 × 3600 coulomb = 0 . 965 × 3600 96500   eq .   of   Fe 2 + = 0 . 036   eq . Also, Normality = No .   of   gram   eq . Vol