When during electrolusis of a solution of `AgNO_3, 9650` coulmbs of charge pass through the electroplationg bath, the mass of silver deposited on the cathode will be:

Options

A. 108 g
B. 10.8 g
C. 1.08 g
D. 216 g

Explanation

To find the mass of silver deposited on the cathode during the electrolysis of a solution of silver nitrate when 9650 coulombs of charge pass through, we can use Faraday's laws of electrolysis. Here's the step-by-step solution:

### Step 1: Understand the Formula
According to Faraday's first law of electrolysis, the mass (W) of a substance deposited at an electrode is directly proportional to the quantity of electricity (Q) that passes through the electrolyte. The formula is given by:
\[ W = Z \cdot Q \]
where \( Z \) is the electrochemical equivalent of the substance.

### Step 2: Determine the Electrochemical Equivalent (Z)
From Faraday's second law, we know:
\[ Z = \frac{E}{F} \]
where:
- \( E \) is the equivalent weight of the substance (for silver, \( E = 108 \, \text{g} \)),
- \( F \) is Faraday's constant, which is approximately \( 96500 \, \text{C/mol} \).

### Step 3: Substitute Values to Find Z
Now we can substitute the values into the equation for \( Z \):
\[ Z = \frac{108 \, \text{g}}{96500 \, \text{C/mol}} \]

### Step 4: Calculate Z
Calculating \( Z \):
\[ Z = \frac{108}{96500} \approx 0.00112 \, \text{g/C} \]

### Step 5: Use the Charge to Find Mass (W)
Now, we can substitute \( Z \) back into the first law of electrolysis:
\[ W = Z \cdot Q \]
Substituting the values:
\[ W = 0.00112 \, \text{g/C} \cdot 9650 \, \text{C} \]

### Step 6: Calculate W
Calculating \( W \):
\[ W = 0.00112 \cdot 9650 \approx 10.8 \, \text{g} \]

### Conclusion
The mass of silver deposited on the cathode will be **10.8 g**.