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
The correct answer is: B
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
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