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The charge required to deposit 40.5 ...

The charge required to deposit 40.5 g of Al (atomic mass 27 g ) from fused ` Al _ 2 ( SO _ 4 ) _ 3 ` is

A

`4 34xx10^(5)C`

B

`43 4 xx10^(5)C`

C

`044xx10^(5)C`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
Gram equivalents of AI
` = (40.5)/(9 ) = 4.5 `
` because ` To deposit 1 g equivalent
1 Faraday or 96500 C electricity is required.
So, the charge required
` = 4.5 xx 96500 C = 4.34 xx 10 ^ 5 C `
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Knowledge Check

  • The change required to deposite 9 g of Al from Al^(3+) solution is (at. Wt. of Al = 27.0) :

    A
    3216.3 C
    B
    96500 C
    C
    9650 C
    D
    32163 C

  • The number of electrons required to reduced 4.5 xx 10^(-5)g of Al is

    A
    ` 1.03xx10^(18)`
    B
    `3.01 xx 10^(18)`
    C
    `4.95 xx 10^(26)`
    D
    `7.31 xx 10^(20)`

  • The equivalent mass of Al in Al_(2)O_(3) is (atomic mass of Al = 27)

    A
    27
    B
    13.5
    C
    9
    D
    52

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    The number of electrons required to reduce 4.5 xx 10^(–5) g of Al^(+3) is :

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