The specific conductivity of 0.1 N KCl solution is ` 0.0129Omega^(-1)cm ^(-1)` . The cell constant of the cell is `0.01 cm^(-1)` then conductance will be:

Options

A. `(1.10)`
B. `(1.29)`
C. `(0.56)`
D. `(2.80)`

Explanation

To find the conductance of the given solution, we can use the relationship between specific conductivity (κ), cell constant (G), and conductance (C). The formula is:

\[ \kappa = G \times C \]

Where:
- \( \kappa \) is the specific conductivity,
- \( G \) is the cell constant,
- \( C \) is the conductance.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Specific conductivity (\( \kappa \)) = 0.0129 Ω⁻¹ cm⁻¹
- Cell constant (\( G \)) = 0.01 cm⁻¹

2. **Rearrange the Formula**:
To find conductance (\( C \)), we can rearrange the formula:
\[ C = \frac{\kappa}{G} \]

3. **Substitute the Values**:
Substitute the known values into the rearranged formula:
\[ C = \frac{0.0129 \, \text{Ω}^{-1} \text{cm}^{-1}}{0.01 \, \text{cm}^{-1}} \]

4. **Calculate Conductance**:
Perform the division:
\[ C = 0.0129 \, \text{Ω}^{-1} \text{cm}^{-1} \div 0.01 \, \text{cm}^{-1} = 1.29 \, \text{Ω}^{-1} \]

5. **Final Result**:
Thus, the conductance of the given solution is:
\[ C = 1.29 \, \text{Ω}^{-1} \]