What would be the equivalent conductivity of a cell in which 0.5 salt solution offers a resistance of 40 ohm whose electrodes are 2 cm apart and 5 `cm^2` in area?

Options

A. `10 "ohm"^(-1) "cm"^(2) "eq"^(-1)`
B. `20 "ohm"^(-1) "cm"^(2) "eq"^(-1)`
C. `30 "ohm"^(-1) "cm"^(2) "eq"^(-1)`
D. `25 "ohm"^(-1) "cm"^(2) "eq"^(-1)`

Explanation

To find the equivalent conductivity of the cell, we will follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 40 ohms
- Distance between electrodes (l) = 2 cm
- Area of electrodes (A) = 5 cm²
- Normality (N) = 0.5

### Step 2: Calculate the specific conductivity (k)
The formula for specific conductivity (k) is given by:

\[ k = \frac{l}{R \cdot A} \]

Substituting the values into the formula:

\[ k = \frac{2 \, \text{cm}}{40 \, \Omega \cdot 5 \, \text{cm}^2} \]

Calculating the denominator:

\[ 40 \, \Omega \cdot 5 \, \text{cm}^2 = 200 \, \Omega \cdot \text{cm}^2 \]

Now substituting back into the equation:

\[ k = \frac{2}{200} = \frac{1}{100} \, \text{S/cm} \]

### Step 3: Calculate the equivalent conductivity (λ)
The equivalent conductivity (λ) can be calculated using the formula:

\[ \lambda = k \cdot \frac{1000}{N} \]

Substituting the values we have:

\[ \lambda = \left(\frac{1}{100} \, \text{S/cm}\right) \cdot \frac{1000}{0.5} \]

Calculating the right side:

\[ \lambda = \frac{1}{100} \cdot 2000 = 20 \, \text{S cm}^2/\text{equivalent} \]

### Final Answer
The equivalent conductivity of the cell is:

\[ \lambda = 20 \, \text{S cm}^2/\text{equivalent} \]