To find the final molarity of the CuSO₄ solution, we need to calculate how much copper sulfate is reduced during the electrolysis process. We'll follow these steps:

Step 1: Calculate the total charge passed through the solution

Given:

Current, $$I = 3.0 \, \text{A}$$
Time, $$t = 2 \, \text{hours} = 2 \times 3600 \, \text{s} = 7200 \, \text{s}$$

The total charge $$Q$$ can be calculated using the formula:

$$ Q = I \times t $$

$$ Q = 3.0 \, \text{A} \times 7200 \, \text{s} = 21600 \, \text{C} $$

Step 2: Calculate the amount of copper deposited

The electrochemical equivalent (z) of copper (Cu) can be derived from Faraday's laws, where the valency $$n$$ of copper is 2, and Faraday's constant $$F = 96485 \, \text{C/mol}$$.

For copper,

$$M = 63.5 \, \text{g/mol}$$

(molar mass of Cu).

The number of moles of copper deposited can be found using the formula:

$$ \text{n(Cu)} = \frac{Q \times \text{current efficiency}}{n \times F} $$

$$ \text{n(Cu)} = \frac{21600 \times 0.90}{2 \times 96485} $$

$$ \text{n(Cu)} = \frac{19440}{192970} $$

$$ \text{n(Cu)} = 0.1008 \, \text{moles} $$

Step 3: Calculate the change in the moles of CuSO₄

One mole of Cu is deposited from one mole of CuSO₄ solution:

$$ \text{n(CuSO}_4\text{) depleted} = \text{n(Cu)} = 0.1008 \, \text{moles} $$

Step 4: Calculate the initial moles of CuSO₄

Initial molarity of CuSO₄ solution is 0.45 M and the volume is 0.750 L:

$$ \text{n(CuSO}_4\text{) initial} = \text{molarity} \times \text{volume} $$

$$ \text{n(CuSO}_4\text{) initial} = 0.45 \times 0.750 $$

$$ \text{n(CuSO}_4\text{) initial} = 0.3375 \, \text{moles} $$

Step 5: Calculate the final moles of CuSO₄

$$ \text{n(CuSO}_4\text{) final} = \text{n(CuSO}_4\text{) initial} - \text{n(CuSO}_4\text{) depleted} $$

$$ \text{n(CuSO}_4\text{) final} = 0.3375 - 0.1008 $$

$$ \text{n(CuSO}_4\text{) final} = 0.2367 \, \text{moles} $$

Step 6: Calculate the final molarity of the CuSO₄ solution

The final molarity is given by the number of moles divided by the volume:

$$ \text{Molarity}_{\text{final}} = \frac{\text{moles of CuSO}_4\text{ remaining}}{\text{volume}} $$

$$ \text{Molarity}_{\text{final}} = \frac{0.2367}{0.750} $$

$$ \text{Molarity}_{\text{final}} = 0.3156 \, \text{M} $$

Rounding to three significant figures, the answer is:

Option D: 0.316 M

Electrochemistry

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