When a solution of conductanes `1.342 mho m^(-1)` was placed in a conductivity cell with parallel electrodes the resistance was found to be `170.5` ohm. The area of the electrode is `1.86 xx 10^(-4)` sq meter. Calculate the distance between the two electrodes in meter.

Explanation

To solve the problem of calculating the distance between two electrodes in a conductivity cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Conductance (κ) = 1.342 mho/m
- Resistance (R) = 170.5 ohm
- Area of the electrode (A) = 1.86 × 10^(-4) m²

2. **Understand the Relationship**:
The relationship between specific conductance (κ), resistance (R), and the cell constant (G*) is given by:
\[ \kappa = \frac{1}{R} \cdot G^* \]
where the cell constant \( G^* \) is defined as:
\[ G^* = \frac{l}{A} \]
Here, \( l \) is the distance between the electrodes.

3. **Rearranging the Formula**:
We can express the specific conductance in terms of resistance and the area:
\[ \kappa = \frac{1}{R} \cdot \frac{l}{A} \]
Rearranging this gives:
\[ l = \kappa \cdot R \cdot A \]

4. **Substituting the Values**:
Now, substitute the known values into the equation:
\[ l = 1.342 \, \text{mho/m} \times 170.5 \, \text{ohm} \times 1.86 \times 10^{-4} \, \text{m}^2 \]

5. **Calculating**:
- First, calculate \( 1.342 \times 170.5 \):
\[ 1.342 \times 170.5 = 228.821 \]
- Then multiply by the area:
\[ l = 228.821 \times 1.86 \times 10^{-4} \]
- Finally, calculate \( 228.821 \times 1.86 \times 10^{-4} \):
\[ l \approx 4.25 \times 10^{-2} \, \text{m} \]

6. **Final Result**:
The distance between the two electrodes is approximately:
\[ l \approx 0.0425 \, \text{m} \, \text{or} \, 4.25 \times 10^{-2} \, \text{m} \]