See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(for 32–33) CuSO4.5H2O (s)   CuSO4(s) (A) Blue (B) white 2CuSO4 + 4KI  Cu2I2 + I2 + 2K2SO4 ppt I2 + 2 2 3 2S O   2I + 2 4 6 S O  millimoles of thiosulphate required = 0.25  20 = 5  millimoles of I2 used = 5/2  millimoles of CuSO4 dissolved = 5 2 5 2   AITS-FT-VII-(Paper-2)-PCM (Sol.)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 11  4 ( )CuSO wt = 5  10-3  159.5 = 0.797 gm in 100 mL of solution. Hence in 1 L solution (wt.) of CuSO4 dissolved = 7.97 gm SECTION – C