How much time is required to deposit `1xx10^(-3)` cm thick layer of silver (density is 1.05 g `cm^(-3)` ) on a surface of area 100 `cm^2` by passing a current of 5 A through `AgNO_3` solution?

Options

A. 125 s
B. 115 s
C. 18.7 s
D. 27.25 s

Explanation

To solve the problem of how much time is required to deposit a silver layer of thickness \(1 \times 10^{-3}\) cm on a surface area of \(100 \, \text{cm}^2\) by passing a current of \(5 \, \text{A}\) through an \(AgNO_3\) solution, we can follow these steps:

### Step 1: Calculate the volume of silver to be deposited
The volume \(V\) of the silver layer can be calculated using the formula:
\[ V = \text{thickness} \times \text{area} \]
Given:
- Thickness = \(1 \times 10^{-3} \, \text{cm}\)
- Area = \(100 \, \text{cm}^2\)

Substituting the values:
\[ V = (1 \times 10^{-3} \, \text{cm}) \times (100 \, \text{cm}^2) = 0.1 \, \text{cm}^3 \]

### Step 2: Calculate the mass of silver using its density
The mass \(m\) of the silver can be calculated using the formula:
\[ m = \text{density} \times \text{volume} \]
Given:
- Density of silver = \(1.05 \, \text{g/cm}^3\)

Substituting the values:
\[ m = 1.05 \, \text{g/cm}^3 \times 0.1 \, \text{cm}^3 = 0.105 \, \text{g} \]

### Step 3: Calculate the equivalent weight of silver
The equivalent weight \(E\) of silver can be calculated using its molar mass. The molar mass of silver (Ag) is approximately \(108 \, \text{g/mol}\). Since silver has a valency of 1, the equivalent weight is:
\[ E = \frac{\text{Molar mass}}{\text{Valency}} = \frac{108 \, \text{g/mol}}{1} = 108 \, \text{g/equiv} \]

### Step 4: Calculate the time required using Faraday's laws of electrolysis
Using the formula:
\[ t = \frac{m \times 96500}{I \times E} \]
Where:
- \(m = 0.105 \, \text{g}\)
- \(I = 5 \, \text{A}\)
- \(E = 108 \, \text{g/equiv}\)

Substituting the values:
\[ t = \frac{0.105 \, \text{g} \times 96500 \, \text{C/equiv}}{5 \, \text{A} \times 108 \, \text{g/equiv}} \]

Calculating the numerator:
\[ 0.105 \times 96500 = 10193.5 \, \text{C} \]

Calculating the denominator:
\[ 5 \times 108 = 540 \, \text{A} \]

Now, substituting these values into the time equation:
\[ t = \frac{10193.5}{540} \approx 18.87 \, \text{s} \]

### Final Answer
Thus, the time required to deposit the silver layer is approximately \(18.87 \, \text{s}\).

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