The specific conductance of a 0.1 N KCl solution at `23^(@)C` is 0.012 `"ohm"^(-1)"cm"^(-1)`. The resistance of the cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be

Options

A. `0.142 cm^(-1)`
B. `0.66 cm^(-1)`
C. `0.918 cm^(-1)`
D. `1.12 cm^(-1)`

Explanation

To find the cell constant of the solution, we can use the relationship between specific conductance (K), resistance (R), and cell constant (G). The formula is given by:

\[ K = \frac{G}{R} \]

Where:
- \( K \) is the specific conductance,
- \( G \) is the cell constant,
- \( R \) is the resistance of the cell.

We can rearrange this formula to solve for the cell constant \( G \):

\[ G = K \times R \]

Now, let's substitute the values given in the question:

1. Specific conductance \( K = 0.012 \, \text{ohm}^{-1} \text{cm}^{-1} \)
2. Resistance \( R = 55 \, \text{ohm} \)

Now we can calculate the cell constant:

\[ G = 0.012 \, \text{ohm}^{-1} \text{cm}^{-1} \times 55 \, \text{ohm} \]

Calculating this gives:

\[ G = 0.012 \times 55 = 0.66 \, \text{cm}^{-1} \]

Thus, the cell constant \( G \) is:

\[ G = 0.66 \, \text{cm}^{-1} \]

### Final Answer:
The cell constant is \( 0.66 \, \text{cm}^{-1} \).