Question

Calculate the number of electrons required to deposit 40.5g of Al.

Answer 1

$A{l}^{3+}+3e^{-}⟶Al$
1mole of Al = 27g deposited by 3F of electricity
therefore, 40.5gm of Al is deposited by = $\frac{3}{27}$ X 40.5F
=$\frac{3}{27}\times 40.5\times 96500$
= 434250 Coulombs
1 C = $6.24\times {10}^{18}$ electrons
434250 C = $6.24\times {10}^{18}\times 434250=2709720\times {10}^{18}=2.709720\times {10}^{12}$ electrons.