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Question

When electricity is passed through a solution of $A l C l_{3}$ , 13.5 g of Al are deposited. The number of Faradays of electricity passed is: ( write your answer to nearest integer)

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Solution

$Q = n \times n_{f} \times F n = \frac{13.5}{27} n_{f} = 3 (f o r {A l}^{+ 3}) Q = \frac{13.5}{27} \times 3 \times F Q = 1.5 F$

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