Discussion
The mass of Al deposited when 2 ampere current is passed through a solution of Al2(SO4)3 for 1 hour is (F = 96500 C)
Answer: A
The mass of aluminium deposited is 0.67 g.
The charge passed is \(Q = It\), and each Al3+ ion needs 3 electrons, so 3 F deposit one mole (27 g) of Al.
\[ \begin{aligned} Q &= 2 \times 3600 = 7200 \text{ C} \\ m &= \frac{27 \times 7200}{3 \times 96500} \\ &= \frac{194400}{289500} \approx 0.67 \text{ g} \end{aligned} \]
1.34 g is exactly double the correct mass, and 0.067 g and 0.134 g are factor-of-ten slips; remember that each Al3+ ion takes three electrons.
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