1.

State required quantity of electricity to produce 5.12 kg of Al. (atomic weight of Al=27 gm mol^(-1))

Answer»

`1.83xx10^(6)` COULOMB
`5.49xx10^(5)` Coulomb
`1.83xx10^(7)` Coulomb
`5.49xx10^(7)` coulomb

Solution :`Al^(3+)+3E^(-) to Al`. . . Cathodic reduction
by 3F electricity 1 mol Al=27 GM Al is produced.
If 27 gram Al is produced then `3xx96500` coulomb electricity is REQUIRED.
So, `5.12xx10^(3)` gm Al is obtained by how much coulomb electricity.
* So, coulomb `=(5.12xx10^(3)xx3xx96500)/(27)`
`=5.489xx10^(7)` Coulomb


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