1.

If 15 Faraday quantity of electricity is passed through Al_((l))^(3+) solution then how many gram of Al metal will be obtained ? (cell effeciency is 80%)(at wt. Al = 27 gm"mol"^(-1))

Answer»

135 gm
121.5 gm
108 gm
94.5 gm

Solution :`Al^(+3) + 3e^(-) to Al THEREFORE CE=("At. Wt.")/(V)=(27)/(3)=9`
`therefore ` 1 Faraday of electricity DEPOSITES 9 G of Al
`therefore ` 15 Faraday will PRODUCE 135 g of Al.
If cell efficiency is 80% then `(80xx135)/(100)` =108 g of Al.


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