What is the charge required to deposit $40.5 g$ of Al (Atomic mass= $27.0 g / m o l e$ ) from the fused
$A l_{2}$ ( $S O_{4})_{3}$ ?

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Solution

$w = Z q$
$w = \frac{M}{96500 \times n} \times q$ [where $n = n$ factor for $A l = 3$ ]

$∴ q = \frac{n \times w \times 96500}{M}$

$= \frac{3 \times 40.5 \times 96500}{27} = 4.34 \times 10^{5} C$

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