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Question

Number of coulombs required to deposit 90 g of Al when the electrode reaction Al3++3eAl is

A
9.65 ×106
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B
8.685 ×104
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C
9.65 ×104
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D
6.955
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Solution

The correct option is A 9.65 ×106
According to equation 1 mole of Al i.e 27 grams of Al is deposited by 3 Faraday electricity
So 90 grams of Al is deposited by
90×3×9650027=9650000
which can be written as 9.65×106

1210122_1378000_ans_b161dfbfe0bf4e82832026a635486d4d.png

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