When electricity is passed through molten AlCl3, 13.5 g. of Al is deposited. The number of Faradays must be

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When electricity is passed through molten $A l C l_{3}$ , 13.5 g. of Al is deposited. The number of Faradays must be

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1.5

1.0

2.0

0.5

answer is C.

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Detailed Solution

${AlCl}_{3} \to {Al}^{3 +} + 3 {Cl}^{-}$

Using electrolysis's first law
$w = \frac{E.i.t}{F}$

putting the values in the equation-
$\begin{array}{l} 13.5 = \frac{\frac{27}{3}}{F} \cdot t \\ = 1.5 F \end{array}$

So, charge passed=1.5F

Therefore, the correct option is C 1.5

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