Aluminium oxide may be electrolysed at $${1000^ \circ }C$$  to furnish aluminium metal ( $$At.\,Mass = 27\,amu;$$     1 Faraday = 96,500 Coulombs ). The cathode reaction is $$ - A{l^{3 + }} + 3{e^ - } \to Al$$
To prepare $$5.12\,kg$$  of aluminium metal by this method we require electricity of

Solution :
$$\eqalign{ & 1\,mole\,\,{\text{of}}\,{e^ - } = 1F = 96500\,C \cr & 27\,g\,\,{\text{of}}\,Al\,\,{\text{is deposited by}}\,{\text{3}} \times 96500\,C \cr & 5120\,g\,\,{\text{of}}\,Al\,\,{\text{will be deposited by}} \cr & = \frac{{3 \times 96500 \times 5120}}{{27}} \cr & = 5.49 \times {10^7}C \cr} $$