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Number of Faraday's required to deposit ...

Number of Faraday's required to deposit 4 g of `H_(2)` is _________.

Text Solution

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The correct Answer is:
9
`Al^(3+) + 3e^(-) to Al`
For depositing 1 mol or 27 g Al , electricity required = 3 F
For depositing 81 g , electricity required = `(3 xx 81)/(27) = 9 F `
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