First slide

Faraday's Laws

Question

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

Moderate

A

0.5
B

1.0
C

1.5
D

2.0

Solution

$No.of Faradays = \frac{mass deposited}{Eq \cdot wt} = \frac{13.5}{9} = 1.5 F$

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