A sample containing KI is dissolved in water and made 100 mL solution. 10 mL of this solution is treated with excess KMnO 4 in weakly basic medium. Oxidised form of Iodine is again treated with exce

Question

A sample containing KI is dissolved in water and made 100 mL solution. 10 mL of this solution is treated with excess KMnO 4 ​ in weakly basic medium. Oxidised form of Iodine is again treated with excess acidified KI. Liberated I 2 ​ required 60 mL of 0.6 M hypo solution for complete titration. How many grams of KI are present in the sample round off to nearest integer (at wt. K=39u, I=127u)

TestHub · Accepted Answer

I − + 2 MnO 4 − ​ OH − ​ 2 MnO 2 ​ + IO 3 − ​ IO 3 − ​ + 5 I − H + ​ 3 I 2 ​ I 2 ​ + 2 S 2 ​ O 3 2 − ​ → S 4 ​ O 6 2 − ​ + 2 I − 18 × 60 × 0.6 = 36 mmol n KI ​ in 10 mL solution = 6 mmol ∴ in 100 mL = 60 mmol m KI ​ = 60 × 1 0 − 3 × 166 = 9.96 ≃ 10 g

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