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 solve the problem of calculating the amount of silver deposited when an electric current is passed through an aqueous silver nitrate solution, we can follow these steps:

### Step 1: Understand the formula
The mass of the substance deposited during electrolysis can be calculated using the formula:
\[ \text{Mass} = Z \times I \times t \]
where:
- \( Z \) is the electrochemical equivalent (in grams per coulomb),
- \( I \) is the current (in amperes),
- \( t \) is the time (in seconds).

### Step 2: Identify the given values
From the question, we have:
- Electrochemical equivalent of silver, \( Z = 0.0011191 \, \text{g/C} \)
- Current, \( I = 0.5 \, \text{A} \)
- Time, \( t = 200 \, \text{s} \)

### Step 3: Calculate the total charge (Q)
The total charge (in coulombs) passed through the solution can be calculated using the formula:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 0.5 \, \text{A} \times 200 \, \text{s} = 100 \, \text{C} \]

### Step 4: Calculate the mass of silver deposited
Now, we can substitute the values into the mass formula:
\[ \text{Mass} = Z \times Q \]
Substituting the known values:
\[ \text{Mass} = 0.0011191 \, \text{g/C} \times 100 \, \text{C} \]
\[ \text{Mass} = 0.11191 \, \text{g} \]

### Step 5: Round the answer
Rounding the answer to an appropriate number of significant figures, we get:
\[ \text{Mass} \approx 0.112 \, \text{g} \]

### Conclusion
The amount of silver deposited is approximately **0.112 grams**.

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