Question

The number of electrons required to reduce $ 4.5 \times 10^{-5}\, $ g of $ Al $ is

• A. $ 1.03\times10^{18} $
• B. $ 3.01\times10^{18} $
• C. $ 4.95\times10^{26} $
• D. $ 7.31\times10^{20} $

Answer 1

Answer: (B)

$\underset{27}{Al^{3+}}+3e^{-} \to \underset{27\,g}{Al}$

$27\, g$ of $Al$ is reduced by $=3 \times 6.023 \times 10^{23} e^{-} s$

$4.5 \times 10^{-5} g$ of $Al$ will be reduced by

$=\frac{3 \times 6.023 \times 10^{23} \times 4.5 \times 10^{-5}}{27}$

$=3.01 \times 10^{18}$ electrons