A current of strength ` 2.5 A` was passed through ` CuSO_4` solution for 6 minute `264` seconds. The amount of copper deposited is (At. Of ` Cu = 63. 5, 1 F = 96500 C`) .

Options

A. ` 0. 3175 g`
B. ` 1. 02 8 g`
C. ` 0. 51 g`
D. ` 6. 35 g`

Explanation

To solve the problem of calculating the amount of copper deposited when a current of 2.5 A is passed through a CuSO₄ solution for a total time of 6 minutes and 264 seconds, we can follow these steps:

### Step 1: Convert Time to Seconds
First, we need to convert the total time from minutes and seconds into seconds.

- **6 minutes** = 6 × 60 = 360 seconds
- **Total time** = 360 seconds + 264 seconds = 624 seconds

### Step 2: Calculate Total Charge (Q)
Using the formula for electric charge, \( Q = I \times t \), where:
- \( I \) = current in amperes (2.5 A)
- \( t \) = time in seconds (624 s)

So,
\[ Q = 2.5 \, \text{A} \times 624 \, \text{s} = 1560 \, \text{C} \]

### Step 3: Determine the Number of Moles of Electrons
Using Faraday's law, we know that 1 mole of electrons corresponds to 96500 C (1 Faraday). Thus, the number of moles of electrons (n) can be calculated as:
\[ n = \frac{Q}{F} \]
Where:
- \( F \) = Faraday's constant (96500 C)

Substituting the values:
\[ n = \frac{1560 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.0162 \, \text{mol} \]

### Step 4: Calculate the Amount of Copper Deposited
The amount of copper deposited can be calculated using the formula:
\[ w = \frac{(n \times \text{Atomic mass of Cu})}{\text{n}} \]
Where:
- Atomic mass of Cu = 63.5 g/mol
- Since copper (Cu) has a valency of 2 in CuSO₄, we use n = 2.

Thus, the formula becomes:
\[ w = \frac{(n \times \text{Atomic mass of Cu})}{2} \]
Substituting the values:
\[ w = \frac{(0.0162 \, \text{mol} \times 63.5 \, \text{g/mol})}{2} \]
\[ w = \frac{1.0307 \, \text{g}}{2} \approx 0.5153 \, \text{g} \]

### Final Answer
The amount of copper deposited is approximately **0.51 grams**.

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