Test Index

JAMB Chemistry 2010 Paper

© africaexams.com

Question : 24

Total: 49

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

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

© africaexams.com

Go to Question:

Prev Question

Next Question