A 100 W, 220 V incandescent lamp is connected in series with an electrolytic cell containing copper sulphate solution. What weight of copper will be deposited by 1 A current flowing for 5 hours? (at. Wt. of Cu=63.54).

Explanation

To solve the problem of determining the weight of copper deposited by a 1 A current flowing for 5 hours in an electrolytic cell with copper sulfate solution, we will follow these steps:

### Step 1: Calculate the total charge (Q) passed through the circuit.
The formula to calculate charge is:
\[ Q = I \times T \]
Where:
- \( I \) = current in amperes (A)
- \( T \) = time in seconds (s)

Given:
- \( I = 1 \, A \)
- \( T = 5 \, hours = 5 \times 3600 \, seconds = 18000 \, seconds \)

Now, substituting the values:
\[ Q = 1 \, A \times 18000 \, s = 18000 \, C \]

### Step 2: Use Faraday's law of electrolysis to find the weight of copper deposited.
Faraday's law states that the mass (m) of a substance deposited during electrolysis is given by:
\[ m = \frac{Q \times M}{F} \]
Where:
- \( m \) = mass of the substance deposited (in grams)
- \( Q \) = total charge (in coulombs)
- \( M \) = molar mass of the substance (in grams per mole)
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))

For copper (Cu):
- Molar mass \( M = 63.54 \, g/mol \)
- Faraday's constant \( F = 96500 \, C/mol \)

Substituting the values into the formula:
\[ m = \frac{18000 \, C \times 63.54 \, g/mol}{96500 \, C/mol} \]

### Step 3: Calculate the mass of copper deposited.
Now, performing the calculation:
\[ m = \frac{18000 \times 63.54}{96500} \]
\[ m \approx \frac{1143720}{96500} \]
\[ m \approx 11.85 \, g \]

### Final Answer:
The weight of copper deposited is approximately **11.85 grams**.

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