The specific conductance of a 0.1 N KCI solution at `25^@ C` is 0.015 `ohm^ (-1) cm^(-1)`. The resistances of the cell containing the solution at the same temperature was found to be 60 Omega`. The cell constant `(in cm^(-1))` will be

Options

A. 0.9
B. `4 xx 10^3`
C. `2.4 xx 10^(-4)`
D. `1.11`

Answer

To find the cell constant of the solution, we can use the relationship between specific conductance (κ), resistance (R), and the cell constant (L/A). Here’s the step-by-step solution:

### Step-by-Step Solution:

1. **Understand the Relationship**:
The cell constant (L/A) can be calculated using the formula:
\[ \text{Cell Constant} = \kappa \times R \]
where:
- \( \kappa \) = specific conductance (ohm\(^{-1}\) cm\(^{-1}\))
- \( R \) = resistance (ohms)

2. **Substitute the Given Values**:
From the problem, we know:
- \( \kappa = 0.015 \, \text{ohm}^{-1} \text{cm}^{-1} \)
- \( R = 60 \, \text{ohms} \)

Now, substituting these values into the formula:
\[ \text{Cell Constant} = 0.015 \, \text{ohm}^{-1} \text{cm}^{-1} \times 60 \, \text{ohms} \]

3. **Calculate the Cell Constant**:
\[ \text{Cell Constant} = 0.015 \times 60 = 0.9 \, \text{cm}^{-1} \]

4. **Conclusion**:
The cell constant (L/A) of the solution is \( 0.9 \, \text{cm}^{-1} \).

### Final Answer:
The cell constant is **0.9 cm\(^{-1}\)**.

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