When 9.65 coulomb of electricity is passed through a solution of silver nitrate (Atomic mass of Ag = 108 g `mol^(-1)`, the amount of silver deposited is :

Options

A. 5.8 mg
B. 10.8 mg
C. 15.8 mg
D. 20.8 mg

Answer

To solve the problem of how much silver is deposited when 9.65 coulombs of electricity is passed through a solution of silver nitrate, we can use Faraday's laws of electrolysis. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between charge, moles, and mass
The amount of substance deposited during electrolysis can be calculated using the formula:
\[ \text{Mass} = \frac{Q \times \text{Molar Mass}}{F} \]
where:
- \( Q \) is the charge in coulombs,
- Molar Mass is the atomic mass of the substance (in grams per mole),
- \( F \) is Faraday's constant, approximately \( 96500 \, C/mol \).

### Step 2: Identify the given values
From the problem, we have:
- \( Q = 9.65 \, C \)
- Molar Mass of silver (Ag) = 108 g/mol
- \( F = 96500 \, C/mol \)

### Step 3: Substitute the values into the formula
Now, substituting the known values into the formula:
\[ \text{Mass} = \frac{9.65 \, C \times 108 \, g/mol}{96500 \, C/mol} \]

### Step 4: Calculate the mass of silver deposited
Now, perform the calculation:
\[ \text{Mass} = \frac{9.65 \times 108}{96500} \]
Calculating the numerator:
\[ 9.65 \times 108 = 1047.4 \]
Now divide by Faraday's constant:
\[ \text{Mass} = \frac{1047.4}{96500} \approx 0.01086 \, g \]

### Step 5: Convert the mass to milligrams
To convert grams to milligrams, multiply by 1000:
\[ 0.01086 \, g \times 1000 = 10.86 \, mg \]

### Final Answer
The amount of silver deposited is approximately **10.86 mg**.