When electricity is passed through a solution of AlCl3, 13.5 g of Al is deposited. The number of Faradays (charge) must be:
A
1
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B
1.5
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C
0.5
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D
2
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Solution
The correct option is D 2 AlCl3→Al3++3Cl−
Using faraday's law of electrolysis w=E.i.tF
where,
w = amount of Al deposited
F= faraday's constant
i = current supplied
t = total time for which current is supplied
E = equivalent weight = Molar mass/ no of electron transfer per mole
put up in equation; 13.5=273i.tF Q=i.t=1.5F