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)`)

Options

A. 135 gm
B. 121.5 gm
C. 108 gm
D. 94.5 gm

Answer

C

Explanation

`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.