Question

When $0.1$ Faraday of electricity is passed in aqueous solution of $AlCl_3$, the amount of $Al$ deposited on cathode is

• A. 27 g
• B. 9 g
• C. 0.27 g
• D. 0.9 g

Answer 1

Answer: (D)

0.9 g

Answer 2

In $AlCl_3$ the e wt. of $Al = \frac{27}{3} =9$ $1\,F = 9\,g$ of $Al$ $0.1 F \equiv 9\times 0.1 = 0.9\,g \,Al$