At `25°C` the specific conductance of a 1N solution of KCl is` 0.002765` mho. The resistance of cell is `400 Omega`. The cell constant is:

Options

A. `0.815 cm^(–1)`
B. `1.016 cm^(–1)`
C. `1.106 cm^(–1)`
D. `2.016 cm^(–1)`

Answer

To find the cell constant of a conductivity cell, we can use the relationship between specific conductance (κ), resistance (R), and the cell constant (L/A). The cell constant is defined as the ratio of the length of the cell (L) to the cross-sectional area (A) of the electrodes.

### Step-by-step Solution:

1. **Identify the given values:**
- Specific conductance (κ) = 0.002765 mho/cm (or ohm⁻¹ cm⁻¹)
- Resistance (R) = 400 Ω

2. **Understand the relationship:**
The cell constant (L/A) can be calculated using the formula:
\[ \text{Cell Constant} (L/A) = \kappa \times R \]
where κ is the specific conductance and R is the resistance.

3. **Substitute the values into the formula:**
\[ L/A = 0.002765 \, \text{ohm}^{-1} \, \text{cm}^{-1} \times 400 \, \text{Ω} \]

4. **Calculate the cell constant:**
\[ L/A = 0.002765 \times 400 \]
\[ L/A = 1.106 \, \text{cm}^{-1} \]

5. **Final Answer:**
The cell constant is \(1.106 \, \text{cm}^{-1}\).