See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

3 2 2 KI O KOH I O     2 2 2 3 2 4 6 I Na S O NaI Na S O    Law of chemical equivalence,       2 2 2 3 3 I Na S O O ne ne ne       2 2 3 3 Na S O O ne ne     3 O 2 1 2 mole       3 mole O 1  Mole of % 3 1 O 100 25% 4   