Chemistry
JAMB 2024
The amount of Faraday required to discharge 4.5 moles of Al\(^{3+}\) is
-
A.
13.5
-
B.
4.5
-
C.
7.5
-
D.
1.5
Correct Answer: Option A
Explanation
Firstly, Al\(^{3+}\) + 3e\(^-\) → Al
1 mole of Al = 1F X 3e\(^-\)
1mole = 3F
4.5moles = X
X = 4.5 X 3 = 13.5 F
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