JAMB - Chemistry (2010 - No. 24)

What quantity of aluminum is deposited when a current of 10A is passed through a solution of an aluminum salt for 1930s?
[Al = 27, F = 96500 C mol -1]

0.2 g

1.8 g

5.4 g

14.2 g

Explanation

Q = It = 10 x 1930 ( given)

27g → 3F( since aluminum is trivalent)

27g → 3 x 96500

x → 19300

x = \(\frac{27 \times 19300}{3 \times 96500}\)

x = 1.8g

JAMB - Chemistry (2010 - No. 24)

Explanation

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