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 0.1 N KCl solution, we can use the relationship between specific conductivity (κ), cell constant (G), and conductance (K). The formula we will use is:

\[ G = \frac{\kappa}{\text{cell constant}} \]

Where:
- \( G \) is the conductance,
- \( \kappa \) is the specific conductivity,
- The cell constant is given as \( 0.01 \, \text{cm}^{-1} \).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Specific conductivity \( \kappa = 0.0129 \, \Omega^{-1} \text{cm}^{-1} \)
- Cell constant \( = 0.01 \, \text{cm}^{-1} \)

2. **Use the Formula for Conductance**:
\[ G = \frac{\kappa}{\text{cell constant}} \]

3. **Substitute the Values into the Formula**:
\[ G = \frac{0.0129 \, \Omega^{-1} \text{cm}^{-1}}{0.01 \, \text{cm}^{-1}} \]

4. **Perform the Calculation**:
\[ G = 0.0129 \div 0.01 = 1.29 \, \Omega^{-1} \]

5. **Final Result**:
The conductance \( G \) of the solution is \( 1.29 \, \Omega^{-1} \).