Single Choice

The amount of charge required to liberate 9 gm of aluminium ( atomic weight = 27 and valency = 3 ) in the process of electrolysis is ( Faraday's number = 96500 coulombs/gm equivalent)

A321660 coulombs
B69500 coulombs
C289500 coulombs
D96500 coulombs
Correct Answer

Solution

The atomic mass of $$Al=27 g/mol$$
$$9g $$ of $$Al=\dfrac{9}{27}=0.333 moles$$
1 mole $$Al= 3 $$ moles electrons.
$$0.333$$ moles $$Al=0.333\times 3=1 $$ mole electrons
1 mole electrons =1 faraday of electricity $$=96500 C.$$
Hence, option (D) is the correct answer.

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