A conductivity cell has been calibrated with a `0.01 M 1:1` electrolyte solution (specific conductance, `k=1.25 xx 10^(-3) S cm^(-1)`) in the cell and the measured resistance was 800 ohms at `25^(@)` C. The constant will be-

Options

A. 1.02 cm
B. `0.102 cm^(-1)`
C. `1.00 cm^(-1)`
D. `0.5 cm^(-1)`

Explanation

To find the cell constant (k) of the conductivity cell, we can use the relationship between resistance (R), specific conductance (κ), and the cell constant (k). The formula relating these quantities is:

\[ k = \kappa \times R \]

Where:
- \( k \) is the cell constant (in cm\(^{-1}\)),
- \( \kappa \) is the specific conductance (in S cm\(^{-1}\)),
- \( R \) is the resistance (in ohms).

### Step-by-Step Solution:

1. **Identify the given values**:
- Specific conductance, \( \kappa = 1.25 \times 10^{-3} \, \text{S cm}^{-1} \)
- Resistance, \( R = 800 \, \Omega \)

2. **Substitute the values into the formula**:
\[ k = \kappa \times R \]
\[ k = (1.25 \times 10^{-3} \, \text{S cm}^{-1}) \times (800 \, \Omega) \]

3. **Calculate the cell constant**:
\[ k = 1.25 \times 800 \times 10^{-3} \]
\[ k = 1000 \times 10^{-3} \]
\[ k = 1 \, \text{cm}^{-1} \]

4. **Final answer**:
The cell constant \( k \) is \( 1 \, \text{cm}^{-1} \).