96.5 amperes current is passed through the molten AlCl ₃ for 100 seconds. The mass of aluminium deposited at the cathode is (atomic weight of Al =27 u )

Step-by-step solution

Given, I=96.5 ~A t=100 ~s m=? Atomic weight of Al =27 u Q=I t=96.5 100=9650 C aligned & AlCl ₃ Al ³⁺+3 Cl ⁻ & Al ³⁺+3 e⁻ Al & 96500 3 C of electricity deposits, 1 mole Al . aligned 9650 C will deposit Al = 9650 96500 3 =0.0333 ~mol . Weight of Al deposited =n molar mass aligned & =0.0333 27 & 0.90 ~g aligned