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The number of moles of electrone require...

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)

A

4

B

2

C

3

D

1

Text Solution

Verified by Experts

The correct Answer is:
A
`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
`therefore n = (36 xx 1)/(9) = 4 `moles
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