Question

The number of moles of electrons required to deposit $36 \,g$ of $Al$ from an aqueous solution of $Al \left( NO _{3}\right)_{3}$ is (At. wt. of $Al =27$ )

• A. 4
• B. 2
• C. 3
• D. 1

Answer 1

Answer: (A)

$Al ^{3+}+\underset{\text{ 3 mol}}{O3 e^{-}} \ce{->}\underset{ 1 \,mol =27 g}{Al}$

$\because 27 \,g$ of $Al$ is deposited by $3$ moles of electrons

$\therefore 36\, g Al$ will be deposited by electrons

$=\frac{3}{27} \times 36=4 \,mol$