A current of `5.00` A flowing for `30.00 ` min deposits `3.048` g of zinc at the cathode. Hence, equilvalent mass of zinc is:

Options

A. `65.4` g
B. `130.8` g
C. `32.7` g
D. `16.85` g

Answer

To find the equivalent mass of zinc deposited at the cathode, we will use Faraday's laws of electrolysis. Here are the steps to solve the problem:

### Step 1: Calculate the total charge (Q) using the formula:
\[ Q = I \times t \]
Where:
- \( I \) = current in amperes (A)
- \( t \) = time in seconds (s)

Given:
- \( I = 5.00 \, \text{A} \)
- \( t = 30.00 \, \text{min} = 30 \times 60 \, \text{s} = 1800 \, \text{s} \)

Now, substituting the values:
\[ Q = 5.00 \, \text{A} \times 1800 \, \text{s} = 9000 \, \text{C} \]

### Step 2: Use Faraday's law to find the equivalent mass (E) of zinc:
The formula for equivalent mass is given by:
\[ E = \frac{m}{Q/F} \]
Where:
- \( m \) = mass of zinc deposited (in grams)
- \( Q \) = total charge (in coulombs)
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))

Given:
- \( m = 3.048 \, \text{g} \)

Now, substituting the values:
\[ E = \frac{3.048 \, \text{g}}{9000 \, \text{C} / 96500 \, \text{C/mol}} \]

### Step 3: Calculate the equivalent mass:
First, calculate \( Q/F \):
\[ \frac{9000 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.0932 \, \text{mol} \]

Now, substituting this back into the equivalent mass formula:
\[ E = \frac{3.048 \, \text{g}}{0.0932 \, \text{mol}} \approx 32.7 \, \text{g/mol} \]

### Conclusion:
The equivalent mass of zinc is approximately \( 32.7 \, \text{g/mol} \).

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