See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Q = It 190.5 g Cu = 3 mole ½ mole Cu deposited by 1 F 3 mole Cu deposited = 6 F = 6  96500 = 579000 coulomb 579000 = I  60  4 or 579000 I 2412.5 240   amp AITS-FT-IV (Paper-2)-PCM(Sol.)-JEE(Advanced)/20 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 9 Mathematics PART – III SECTION – A