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On passing current through molten KCl, 19.5 g of ‘K’ is deposited. The amount of Al deposited by the same quantity of electricity when passed through molten AlCl₃ is ? (At mass of ‘K’ = 39 ‘Al’ = 27)

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4.5 g

9 g

13.5 g

2.7 g

answer is A.

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Detailed Solution

$\frac{w_{k}}{w_{A l}} = \frac{E_{k}}{E_{A l}} \frac{19.5}{w_{A l}} = \frac{39}{9} w_{A l} = 4.5 g$

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