Plz give the solution step by step. Share with your friends Share 0 Akanksha C. answered this Dear Student,Two Cu+ replaced by 1 Cu2+Total loss of Cu =0.2 moleCu2S = Cu+S2- but composition is found Cu1.8S, so loss of Cu = 2-1.8 = 0.2 mole So, mole of Cu2+ present in Cu1.8S = 0.2 Therefore the percentage of Cu2+ in total copper content =0.21.8×100 =11.11 %Regards 1 View Full Answer