The number of Faraday required to deposit 1g equivalent of aluminium (at. Wt. 27) from a solution of AlCl3 is?

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  • -19
1 gm, equivalent of aluminium is equal to 27/3 =9gm
Al+3 +3e- ------------>Al
for 27 gm or 1 mole of aluminium 1F is required
for the decomposition of 9gm faradays of electricty required =F/3
  • -67
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