How many Faraday of electricity are required to deposit 0.20 mole of nickel, if 0.10 Faraday of electricity deposited 2.98g of nickel during electrolysis of its aqueous solution?

(Ni = 58.7, 1F = 96 500 C mol^-)

0.20

0.30

0.40

0.50

Correct answer is C

No of mole → (2.98)/(58.7) ..............0.1

2.0 .................x
∴ x = (0.2)/(2.98) x 58.7 x 0.40

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How many Faraday of electricity are required to deposit 0.20 mole of nickel, if 0.10 Faraday of electricity deposited 2.98g of nickel during electrolysis of its aqueous solution? (Ni = 58.7, 1F = 96 500 C mol-)