The resistance of a conductivity cell containing 0.001M KCl solution at 298K is 1500`Omega`. What is the cell constant ( in `mm^(-1)`) if the conductivity of 0.001M KCl solution is `2xx10^(-3)Smm^(-1)`

Answer

To find the cell constant of a conductivity cell containing a 0.001 M KCl solution, we can follow these steps:

### Step 1: Understand the relationship between resistance, conductivity, and cell constant
The relationship between resistance (R), conductivity (K), and the cell constant (L/A) can be expressed using the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) = resistance of the cell
- \( \rho \) = resistivity of the solution
- \( L \) = distance between the electrodes
- \( A \) = surface area of the electrodes

### Step 2: Relate resistivity to conductivity
Resistivity (\( \rho \)) is the reciprocal of conductivity (\( K \)):
\[ \rho = \frac{1}{K} \]
Substituting this into the resistance formula gives:
\[ R = \frac{1}{K} \cdot \frac{L}{A} \]

### Step 3: Rearrange the formula to find the cell constant
We can rearrange the formula to solve for the cell constant (\( \frac{L}{A} \)):
\[ \frac{L}{A} = R \cdot K \]

### Step 4: Substitute the given values
We are given:
- Resistance (\( R \)) = 1500 Ω
- Conductivity (\( K \)) = \( 2 \times 10^{-3} \, S \, mm^{-1} \)

Now, substituting these values into the rearranged formula:
\[ \frac{L}{A} = 1500 \, \Omega \cdot (2 \times 10^{-3} \, S \, mm^{-1}) \]

### Step 5: Calculate the cell constant
Calculating the above expression:
\[ \frac{L}{A} = 1500 \cdot 2 \times 10^{-3} \]
\[ \frac{L}{A} = 3000 \times 10^{-3} \]
\[ \frac{L}{A} = 3 \, mm^{-1} \]

### Final Answer
The cell constant is \( 3 \, mm^{-1} \).

---