Resistance of a decimolar solution between two electrodes `0.02` meter apart and `0.0004m^(2)` in area was fround to be `50 ohm`. Specific conductance `(k)` is `:`

Options

A. `0.1 S-m^(-1)`
B. `1 S-m^(-1)`
C. `10 S-m^(-1)`
D. `4xx10^(-4)S-m^(-1)`

Explanation

To find the specific conductance (k) of the solution, we can use the formula for conductivity in terms of resistance, length, and area:

\[ k = \frac{L}{R \cdot A} \]

Where:
- \( k \) is the specific conductance (conductivity),
- \( L \) is the distance between the electrodes,
- \( R \) is the resistance,
- \( A \) is the area of the electrodes.

### Step-by-step Solution:

1. **Identify the given values**:
- Resistance (\( R \)) = 50 ohms
- Distance between electrodes (\( L \)) = 0.02 m
- Area of the electrodes (\( A \)) = 0.0004 m²

2. **Substitute the values into the formula**:
\[ k = \frac{L}{R \cdot A} \]
\[ k = \frac{0.02}{50 \cdot 0.0004} \]

3. **Calculate the denominator**:
\[ 50 \cdot 0.0004 = 0.02 \]

4. **Substitute back into the equation**:
\[ k = \frac{0.02}{0.02} \]

5. **Calculate \( k \)**:
\[ k = 1 \, \text{S/m} \]

### Final Answer:
The specific conductance \( k \) is \( 1 \, \text{S/m} \).

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