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When electricity is passed through a solution of `AlCl_(3)` and `13.5g` of `Al` is deposited, the number of `Faraday of electricity passed must be `………………….F`.

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Correct Answer - `(1.5F)`
`(1.5F)`
`1F=1Eq or Al =(27)/(3)=9g (n ` factor for `Al=3)`
`9g of Alimplies1F`
`13.5 of Al =(1xx13.5)/(9)=1.5F`

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