The specific conductivity of N/10 KCl solution at `20^(@)C` is `0.0212 ohm^(-1) cm^(-1)` and the resistance of the cell containing this solution at `20^(@)C` is 55 ohm. The cell constant is :

Options

A. `4.616 cm^(-1)`
B. `1.166 cm^(-1)`
C. `2.173 cm^(-1)`
D. `3.324 cm^(-1)`

Answer

To find the cell constant for the given N/10 KCl solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Specific conductivity (κ) = 0.0212 ohm⁻¹ cm⁻¹
- Resistance (R) = 55 ohm

2. **Understand the Relationship:**
- The relationship between specific conductivity (κ), cell constant (x), and conductance (G) is given by the formula:
\[ \kappa = x \cdot G \]
- Conductance (G) is the reciprocal of resistance (R):
\[ G = \frac{1}{R} \]

3. **Calculate Conductance:**
- Substitute the value of resistance into the conductance formula:
\[ G = \frac{1}{55} \text{ ohm}^{-1} \]

4. **Substitute into the Conductivity Formula:**
- Now substitute G into the equation for specific conductivity:
\[ \kappa = x \cdot \left(\frac{1}{55}\right) \]
- Rearranging gives:
\[ x = \kappa \cdot R \]

5. **Calculate the Cell Constant:**
- Substitute the values of κ and R:
\[ x = 0.0212 \cdot 55 \]
- Performing the multiplication:
\[ x = 1.166 \text{ cm}^{-1} \]

6. **Conclusion:**
- The cell constant is:
\[ \text{Cell constant} = 1.166 \text{ cm}^{-1} \]

### Final Answer:
The cell constant is **1.166 cm⁻¹**.

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