Question

The atomic weight of $\Delta_{m}1.5$ is 27. When a current of 5 Faradays is passed through a solution of $\Delta_{m}20$ions, the weight of $\rightarrow$ deposited is.

• A. 27 gm
• B. 36 gm
• C. 45 gm
• D. 39 gm

Answer 1

Answer: (C)

45 gm

Answer 2

$Al^{3 +} + 3e^{-} \rightarrow Al$

$E_{Al} = \frac{27}{3} = 9$

$W_{Al} = E_{Al} \times \text{No.} ⥂ \text{of } ⥂ \text{faradays} = 9 \times 5 = 45gm$.