The two `Pt` electrodes fitted in a conductance cell are `1.5 cm` apart while the cross`-` sectional area of each electrode is `0.75cm`. What is the cell constant?

Options

A. 1.25cm
B. 0.5 cm
C. `2.0 cm^(-1)`
D. `0.2 cm^(-1)`

Answer

To find the cell constant of a conductance cell with two platinum electrodes, we can use the formula for the cell constant (G*), which is defined as:

\[ G^* = \frac{L}{A} \]

where:
- \( L \) is the distance between the electrodes (in cm),
- \( A \) is the cross-sectional area of the electrodes (in cm²).

### Step-by-Step Solution:

1. **Identify the given values**:
- Distance between the electrodes, \( L = 1.5 \, \text{cm} \)
- Cross-sectional area of each electrode, \( A = 0.75 \, \text{cm}^2 \)

2. **Substitute the values into the formula**:
\[ G^* = \frac{L}{A} = \frac{1.5 \, \text{cm}}{0.75 \, \text{cm}^2} \]

3. **Perform the division**:
\[ G^* = \frac{1.5}{0.75} = 2 \, \text{cm}^{-1} \]

4. **State the final answer**:
The cell constant \( G^* \) is \( 2 \, \text{cm}^{-1} \).

### Final Answer:
The cell constant is \( 2 \, \text{cm}^{-1} \).