When a current of 10 A is passes through molten AlCl 3 for 1 . 608 minutes. The mass of Al deposited will be [Atomic mass of Al = 27 g ]:-

Step-by-step solution

Given data :- Current = 10   A . Time = 1 . 608   min = 96 . 48   s Atomic mass of Al = 27   g / mol Oxidation state of Al = + 3 Mass of aluminium deposited = Atomic   mass   × time × current F × valency   factor Mass of aluminium deposited = 27 × 96 . 48 × 10 96480 × 3 = 0 . 09   g Hence, the mass of Al deposited will be 0 . 09   g .