Question

One ampere of current is passed for $9650$ seconds through molten $AlC{l}_3$. What is the weight in grams of $Al$ deposited at the cathode? (Atomic weight of $Al$ $=$ $27$)

• A. $0.9$
• B. $9.0$
• C. $0.09$
• D. $90.0$

Answer 1

The correct option is A $0.9$

Mass of Al deposited $=\frac{I\left(A\right)\times t\left(s\right)}{96500\frac{C}{mol{e}^{-}}}\times \text{mole ratio}\times \text{molar mass}$
Substitute $I\left(A\right)=1A$ and $t\left(s\right)=9650s$ in the above equation.
Hence, mass of Al deposited $=\frac{I\left(A\right)\times t\left(s\right)}{96500\frac{C}{mol{e}^{-}}}\times \text{molar ratio}\times \text{molar mass}$

$=\frac{1A\times 9650s}{96500\frac{C}{mol{e}^{-}}}\times \frac{1molAl}{3mol{e}^{-}}\times 27\frac{g}{mol}=0.9gm$

Option A is correct.