The density of copper is `"8.94 g mL"^(-1)`. Find the charge needed to plate an area of `10xx10cm^(2)` to a thickness of `10^(-2)cm` using a `CuSO_(4)` solution as electrolyte
(atomic weight of Cu = 63.6 g/mol).

Options

A. `2.7xx10^(4)C`
B. `8.8xx10^(4)C`
C. `18.3xx10^(4)C`
D. `1.7xx10^(4)C`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the Volume of Copper to be Plated
The volume (V) can be calculated using the formula:
\[ V = \text{Area} \times \text{Thickness} \]

Given:
- Area = \( 10 \times 10 \, \text{cm}^2 = 100 \, \text{cm}^2 \)
- Thickness = \( 10^{-2} \, \text{cm} \)

Calculating the volume:
\[ V = 100 \, \text{cm}^2 \times 10^{-2} \, \text{cm} = 100 \times 0.01 = 1 \, \text{cm}^3 \]

### Step 2: Calculate the Mass of Copper
Using the density (d) of copper:
\[ d = 8.94 \, \text{g/mL} \]
Since \( 1 \, \text{cm}^3 = 1 \, \text{mL} \), we can calculate the mass (m) using:
\[ m = d \times V \]

Calculating the mass:
\[ m = 8.94 \, \text{g/mL} \times 1 \, \text{mL} = 8.94 \, \text{g} \]

### Step 3: Use Faraday's Second Law of Electrolysis
According to Faraday's second law:
\[ \text{Weight deposited} = \frac{n \times Q}{F} \]
Where:
- \( n \) = atomic weight of copper = 63.6 g/mol
- \( F \) = Faraday's constant = 96500 C/mol (approximately)

The number of electrons transferred (n-factor) for copper (Cu²⁺ to Cu) is 2.

### Step 4: Rearranging the Equation to Find Charge (Q)
We can rearrange the formula to solve for Q:
\[ Q = \frac{\text{Weight deposited} \times F}{n} \]

Substituting the values:
\[ Q = \frac{8.94 \, \text{g} \times 96500 \, \text{C/mol}}{63.6} \]

### Step 5: Calculate the Charge
Calculating the charge:
\[ Q = \frac{8.94 \times 96500}{63.6} \]
\[ Q = \frac{863910}{63.6} \approx 27129.2453 \, \text{C} \]

### Step 6: Final Result
Thus, the charge needed to plate the copper is approximately:
\[ Q \approx 2.71 \times 10^4 \, \text{C} \]