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The atomic weight of Al is 27. When a current of 5 F is passed through a solution of ${Al}^{3 +}$ ions, the weight of Al deposited is

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27 g

36 g

39 g

45 g

answer is D.

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Detailed Solution

$\begin{array}{l} {Al}^{3 +} + 3 e^{-} \to Al \\ E_{Al} = \frac{27}{3} = 9 \\ W_{Al} = E_{Al} \times Number of faradays = 9 \times 5=45 g \end{array}$

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