The specific conductivity of N/50 solution of KCl at 298 K is 0.002765 S `cm^(-1)`. If the resistance of the same solution placed in the cell is 2000 ohms, what is cell constant?

Explanation

To find the cell constant (L/A) for the given solution of KCl, we can use the relationship between specific conductivity (κ), resistance (R), and the cell constant (L/A).

### Step-by-Step Solution:

1. **Understand the Formula**:
The specific conductivity (κ) is related to the resistance (R) and the cell constant (L/A) by the formula:
\[ \kappa = \frac{L}{A} \cdot \frac{1}{R} \]
Rearranging this gives us:
\[ \frac{L}{A} = R \cdot \kappa \]

2. **Identify Given Values**:
- Specific conductivity (κ) = 0.002765 S cm⁻¹
- Resistance (R) = 2000 ohms

3. **Substitute the Values into the Formula**:
Now, substitute the known values into the rearranged formula:
\[ \frac{L}{A} = 2000 \, \text{ohms} \times 0.002765 \, \text{S cm}^{-1} \]

4. **Calculate the Cell Constant**:
Performing the multiplication:
\[ \frac{L}{A} = 2000 \times 0.002765 = 5.53 \, \text{cm}^{-1} \]

5. **Final Answer**:
The cell constant (L/A) is:
\[ \frac{L}{A} = 5.53 \, \text{cm}^{-1} \]