See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

  5 1 0 2 3 f f f K I O K I excess I 5 n 5 n 1 n 3        Moles of I2 = moles of KIO3 × 3 = 10 × 3 = 30 Milliequivalents of I2 = moles of nf × 1000 5 30 1000 3    50 1000   2 2 2 3 2 4 6 f f 1 1 2 2 I 2Na S O Na S O 2NaI n 2 n 1 N V N V       1 1 2 2 2 2 3 2 N V of I N of Na S O V  50000 6 5 20 1000     = 3 Strength of 2 2 3 Na S O N E   3 158 474 g / 1     hypo S 10 x 2   2370 