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Earlier discussion
1 older comment kept for context.
Pirisola Ayomikun
First convert the 9g to mole using mole = mass/molar mass
Mole = 9g/27g/mol
Mole = 0.333moles
Write the cathodic equation
Al(s) ====>> Al^3+ + 3e^—
1 mole of Al ==== > 3F (from the electron)
0.333 moles of Al ===> X
X = 0.333mol × 3F/1mol
Mol cancel mol
X = 0.333 × 3F/1
X = 1F
Hence 1F is required to liberated 9g of Aluminium