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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 ]:-

Options

  1. A0 .09 g
  2. B0 .81 g
  3. C1 .35 g
  4. D0 .27 g

Correct answer

A. 0 .09 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 .

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