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When electricity is passed through a solution of ${AlCl}_{3}$ , 13.5 g of Al is discharged. The amount of charge passed is (Al = 27)

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1.5 F

0.5 F

1.0 F

2.0 F

answer is A.

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Detailed Solution

$\begin{array}{l} {Al}^{+ 3} + 3 e^{-} \to Al \\ 27 g \\ 1 mole 3 mole \\ 1/2 3/2 = 1.5 F \end{array}$

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