The conductivity of 0.02 N solution of a cell of KCl at `25^@C" is "2.765 xx 10^(-3) Scm^(-1)`. If the resistance of a cell containing this solution is 4000 ohm, what will be the cell constant?

Options

A. `11.06 cm^(-1)`
B. `1.21 cm^(-1)`
C. `1.19 cm^(-1)`
D. `1.286cm^(-1)`

Explanation

To find the cell constant (L/A) for the given KCl solution, we can use the relationship between conductivity (κ), resistance (R), and the cell constant (L/A). The formula we will use is:

\[ \kappa = \frac{L}{A} \cdot \frac{1}{R} \]

Where:
- κ is the conductivity,
- L is the length of the electrodes,
- A is the cross-sectional area of the electrodes,
- R is the resistance of the solution.

### Step-by-Step Solution:

1. **Identify the given values:**
- Conductivity (κ) = \(2.765 \times 10^{-3} \, \text{S/cm}\)
- Resistance (R) = \(4000 \, \Omega\)

2. **Rearrange the formula to solve for the cell constant (L/A):**
\[ \frac{L}{A} = \kappa \cdot R \]

3. **Substitute the known values into the rearranged formula:**
\[ \frac{L}{A} = (2.765 \times 10^{-3} \, \text{S/cm}) \cdot (4000 \, \Omega) \]

4. **Calculate the cell constant:**
\[ \frac{L}{A} = 2.765 \times 10^{-3} \times 4000 \]
\[ \frac{L}{A} = 11.06 \, \text{cm}^{-1} \]

5. **Final answer:**
The cell constant (L/A) is \(11.06 \, \text{cm}^{-1}\).