On passing a current through molten KCl, 19.5 g of K is deposited. The amount of Al deposited by the same quantity of electricity if passed through molten AlCl $_{3}$ is :

4.5 g

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9.0 g

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13.5 g

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27 g

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Solution

The correct option is A 4.5 g
Equivalent weight of potassium = $\frac{39}{1} = 39$

Equivalent weight of aluminum = $\frac{27}{3} = 9$

Mass of K deposited = $19.5 g$

Substituting the values in equation:

$\frac{M a s s o f A l d e p o s i t e d}{M a s s o f K d e p o s i t e d} = \frac{E q . W t o f A l}{E q . W t o f K}$

$\frac{M a s s o f A l d e p o s i t e d}{19.5} = \frac{9}{39}$

$M a s s o f A l d e p o s i t e d = \frac{9}{39} \times 19.5 = 4.5 g .$

Hence, option $A$ is correct.

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On passing a current through molten KCl, 19.5 g of K is deposited. The amount of Al deposited by the same quantity of electricity if passed through molten AlCl3 is :

On passing a current through molten KCl, 19.5 g of K is deposited. The amount of Al deposited by the same quantity of electricity if passed through molten AlCl $_{3}$ is :