A conductance cell was filled with a 0.02 M KCl solution which has a specific conductance of `2.768xx10^(-3)ohm^(-1)cm^(-1)` . If its resistance is 82.4 ohm at `25^(@)` C the cell constant is:

Options

A. `0.2182cm^(-1)`
B. `0.2281cm^(-1)`
C. `0.2821cm^(-1)`
D. `0.2381cm^(-1)`

Answer

To find the cell constant of a conductance cell filled with a 0.02 M KCl solution, we can follow these steps:

### Step 1: Understand the relationship between resistance, specific conductance, and cell constant
The relationship can be expressed as:
\[ R = \frac{\rho L}{A} \]
Where:
- \( R \) = resistance of the solution (in ohms)
- \( \rho \) = resistivity (which is the inverse of specific conductance, \( k \))
- \( L \) = distance between the electrodes
- \( A \) = area of the electrodes

### Step 2: Identify the specific conductance
The specific conductance \( k \) is given as:
\[ k = 2.768 \times 10^{-3} \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 3: Calculate the resistivity
Resistivity \( \rho \) can be calculated as the inverse of specific conductance:
\[ \rho = \frac{1}{k} = \frac{1}{2.768 \times 10^{-3}} \]

### Step 4: Calculate the resistivity value
Calculating the resistivity:
\[ \rho = \frac{1}{2.768 \times 10^{-3}} \approx 362.5 \, \text{ohm cm} \]

### Step 5: Use the formula to find the cell constant
The cell constant \( \frac{L}{A} \) can be derived from the rearranged formula:
\[ \frac{L}{A} = R \cdot k \]

### Step 6: Substitute the values into the equation
Substituting the known values:
\[ \frac{L}{A} = 82.4 \, \text{ohm} \times 2.768 \times 10^{-3} \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 7: Calculate the cell constant
Calculating the cell constant:
\[ \frac{L}{A} = 82.4 \times 2.768 \times 10^{-3} \approx 0.2281 \, \text{cm}^{-1} \]

### Final Answer
The cell constant is approximately:
\[ \frac{L}{A} \approx 0.2281 \, \text{cm}^{-1} \]

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