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]
A
0.2 g
B
1.8 g
C
5.4 g
D
14.2 g
Verified Correct Answer: B.
Step-by-Step Mathematical Solution
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
Source: WAEC / JAMB / NECO Classroom Archive (Record #4486)