1.

Reduction of Al_(2)O_(3) is occurred at low potential and high current through electrolysis. If 4.0xx10^(4) ampere current is pass through molten Al_(2)O_(3) solution for 6 hours then how much aluminium is obtained ? (atomic mass of aluminium is 27 gm/mol at 100% efficiency)

Answer»

`8.1xx10^(4)GM`
`2.4xx10^(5)gm`
`1.3xx10^(4)gm`
`9.0xx10^(3)gm`

Solution :Having `Q=ixxt`
`Q=4.0xx10^(4)xx6xx60xx60C`
`=8.64xx10^(8)C`
Now, 96500 C liberate 9 gm AL, So, `8.64xx10^(8)C`
So, `(9)/(96500)xx8.64xx10^(8)` gram Al is liberated
`=8.05xx10^(4)` gram Al.


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