Answer 1

AlCl₃  Al⁺³ + 3Cl^-

Al⁺³ + 3e^-  Al

1 mole of Al requires passage of 3 moles of electrons. The charge on 3 moles of e^- is 3 Faraday.

Moles of Al produced = `"mass of Al"/"molar mass of Al"`

= `13/27`

= 0.48 moles

As 3F of electricity produces 1 mole of Al

∴ No. of faradays of electricity required to produce 0.48 mole of Al

= 0.48 x 3

= 1.44 Faraday