Calculate the total number of electrons involved in the deposition of 24 g of copper from a copper (II) solution.

Copper (II) ions gain electrons to form solid copper.

$$\text{Cu}^{2+}(aq) + 2e^- \to \text{Cu}(s)$$

This reaction shows that 2 moles of electrons are required to deposit 1 mole of copper.

Given mass of copper = 24 g
Molar mass of copper (Cu) = 64 g/mol

$$\text{Moles of Cu} = \frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}$$ $$\text{Moles of Cu} = \frac{24 \text{ g}}{64 \text{ g/mol}}$$ $$\text{Moles of Cu} = 0.375 \text{ mol}$$

Avogadro's number ($N_A$) = $6.022 \times 10^{23}$ atoms/mol

$$\text{Number of Cu atoms} = \text{Moles of Cu} \times N_A$$ $$\text{Number of Cu atoms} = 0.375 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol}$$ $$\text{Number of Cu atoms} = 2.25825 \times 10^{23} \text{ atoms}$$

From Step 1, 1 atom of Cu requires 2 electrons.

$$\text{Total number of electrons} = \text{Number of Cu atoms} \times 2$$ $$\text{Total number of electrons} = 2.25825 \times 10^{23} \times 2$$ $$\text{Total number of electrons} = 4.5165 \times 10^{23}$$

The number of electrons present is $\text{4.5165} \times 10^{23}$.