1.

The number of moles of electrone required to deposits 36 g of Al from an aqueous solution of Al(NO_(3))_(3) is (atomic mass of Al = 27)

Answer»

4
2
3
1

Solution :`Al^(3+) + 3e^(-) to Al` .
`1 F-= 96500 C -= 1 g ` EQ. of Al `= (27)/(3) = 9g`
`therefore` 1 MOLE of electrons i.e. 36 g of Al = n
`thereforen = (36 xx 1)/(9) = 4 `moles


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