The electrochemical equivalent of silver is `0.0011191` g. When an electric current of `0.5` ampere is passed through an aquesus silver nitrate solution of 200 sec, the amount fo silver deposited is:

Options

A. `1.1191` g
B. `0.11191` g
C. `5.590` g
D. `0.5590` g

Answer

To find the amount of silver deposited when an electric current is passed through an aqueous silver nitrate solution, we can use the formula related to electrochemical equivalent (Z):

### Step-by-Step Solution:

1. **Understand the Formula**: The amount of substance deposited (m) can be calculated using the formula:
\[ m = Z \times I \times t \]
where:
- \( m \) = mass of the substance deposited (in grams)
- \( Z \) = electrochemical equivalent (in grams per coulomb)
- \( I \) = current (in amperes)
- \( t \) = time (in seconds)

2. **Identify Given Values**:
- Electrochemical equivalent of silver, \( Z = 0.0011191 \, \text{g/C} \)
- Current, \( I = 0.5 \, \text{A} \)
- Time, \( t = 200 \, \text{s} \)

3. **Calculate the Total Charge (Q)**:
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 0.5 \, \text{A} \times 200 \, \text{s} = 100 \, \text{C} \]

4. **Calculate the Mass of Silver Deposited**:
Now, substitute the values into the formula for mass:
\[ m = Z \times Q \]
Substituting the values:
\[ m = 0.0011191 \, \text{g/C} \times 100 \, \text{C} = 0.11191 \, \text{g} \]

5. **Final Answer**:
The amount of silver deposited is approximately:
\[ m \approx 0.11191 \, \text{g} \]

### Conclusion:
The amount of silver deposited when a current of 0.5 amperes is passed for 200 seconds is approximately **0.11191 grams**.