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

+3

Approved

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.

Approved