THE SAME QUANTITY OF ELECTRICAL CHARGE THAT DEPOSITED 0.583 GRAMS OF SILVER WAS PASSED THROUGH A SOLUTION OF GOLD SALT AND 0.355 GRAMS GOLD WAS FORMED .WHAT IS THE OXIDATION STATE OF GOLD IN THIS SALT?

Mass of metal deposited is directltly proportional to its eq. wt.(Faraday''s Law) so, M(Au) / M(Ag) = (197/valency of gold x)/(94/ Valency of silver y) Thus, 355/538 = 197/94 * (y/x) 2/3=2y/x (nearly) Therefore y/x=1/3 Since silver exhibits +1 or +2 states, x=+3 or +6 but gold doesn''t exhibit +6, so Oxidation state of gold is +3 in the compound.