The amount of silver deposited by passing `241. 25 C` of current through silver nitrated solution is .

Options

A. 2.7 g
B. 2.7 mg
C. 0.27 g
D. 0.54 g

Answer

To find the amount of silver deposited by passing 241.25 C of current through a silver nitrate solution, we can use Faraday's first law of electrolysis. Here’s the step-by-step solution:

### Step 1: Understand the Formula
According to Faraday's first law, the weight (W) of a substance deposited during electrolysis is given by the formula:
\[ W = Z \times Q \]
Where:
- \( W \) = weight of the substance deposited (in grams)
- \( Z \) = electrochemical equivalent (in grams per coulomb)
- \( Q \) = total electric charge (in coulombs)

### Step 2: Calculate the Electrochemical Equivalent (Z)
The electrochemical equivalent \( Z \) can be calculated using the formula:
\[ Z = \frac{M}{n \times F} \]
Where:
- \( M \) = molar mass of the substance (for silver, \( M = 108 \, \text{g/mol} \))
- \( n \) = number of electrons exchanged (for silver, \( n = 1 \) since Ag\(^+\) gains 1 electron to become Ag)
- \( F \) = Faraday's constant (\( F = 96500 \, \text{C/mol} \))

Substituting the values:
\[ Z = \frac{108}{1 \times 96500} \]

### Step 3: Calculate Z
Now, calculate \( Z \):
\[ Z = \frac{108}{96500} \approx 0.00112 \, \text{g/C} \]

### Step 4: Calculate the Weight Deposited (W)
Now, we can use the charge \( Q = 241.25 \, \text{C} \) to find the weight deposited:
\[ W = Z \times Q \]
Substituting the values:
\[ W = 0.00112 \, \text{g/C} \times 241.25 \, \text{C} \]

### Step 5: Calculate W
Now, calculate \( W \):
\[ W \approx 0.27 \, \text{g} \]

### Conclusion
The amount of silver deposited is approximately **0.27 grams**.

### Final Answer
The correct option is **0.27 grams**.

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