The same quantity of electrical charge deposited 0.583 g of Ag when passed through AgNO3; AuCl3 solution. Calculate the weight of gold formed . ( atomic weight of Au = 197 g mol ) Share with your friends Share 6 Geetha answered this According to Faraday's II law, the amount of substance deposited on each electrode, on passingsame amount of current is equal to their equivanet massWAgWAu = EAgEAu0.583WAu = Molar mass of Ag/nMolar mass of Au/n (where n is the charge of the ion For Ag its 1 and Au its 3)0.583WAu = 108/1197/3WAu = 0.583 × 1973108 = 0.3545 g 21 View Full Answer
