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
Correct answer is B
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