Question

The number of Faraday required to deposit 1g equivalent of aluminium (at. Wt. 27) from a solution of AlCl3 is?

Answer 1

Final Answer : 1 F 

Steps:
1) 1-g Equivalent  = Mass of substance equal to 1 -equivalent  of given chemical species.
Here,
1 - g equivalent of Al has mass 27 / 3 =9 g

2) We have Reaction :
        
        $Al^{3+} (aq.) + 3e^{-} \:\:\: -\ \textgreater \ Al (s) \\ \\ \frac{Q}{F} = 3 =\ \textgreater \ Q=3F$


Therfore, 
By Faradays Law of Electrolysis, 
3 F of charge is required to deposit 1 mole of Al(s).
=> 3F of charge is required to deposit  27 g of Aluminium.
=> 1 F of charge is required to deposit  9 g of Aluminium .
=> 1 F of Charge is required to deposit 1- g equivalent of Aluminium.

Answer 2

Answer:

pls like follow pls and rate