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  4. When electricity is passed through the solution of AlCl3, 13.5 g of Al are deposited. The number of faradays must be

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Q. When electricity is passed through the solution of $AlCl_{3}, 13.5\, g$ of $Al$ are deposited. The number of faradays must be

Electrochemistry

Solution:

$Eq. of Al =\frac{13.5}{27/3}=1.5 $
Thus 1.5 F is needed.