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-

Question

Toppr Admin · Accepted Answer

Correct option is D- 96500 coulombsThe 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 electrons1 mole electrons -1 faraday of electricity -96500 C-Hence- option -D- is the correct answer-

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)

Correct option is D. 96500 coulombsThe 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 electrons1 mole electrons =1 faraday of electricity $$=96500 C.$$Hence, option (D) is the correct answer.