At 298 K the resistance of a 0.1M KCl solution is found to be
39.0ohm. If the conductivity (k) of this solution is
1.29`xx10^(-2)ohm^(-1)cm^(-1)` at 298K, what is cell constant

Options

A. `5.03xx10^(-1)cm^(-1)`
B. `10.06xx10^(-1)cm^(-1)`
C. `15.09xx10^(-1)cm^(-1)`
D. `2.51xx10^(-1)cm^(-1)`

Explanation

To find the cell constant (G*) of the KCl solution, we can use the relationship between conductivity (k), conductance (G), and resistance (R). The formula we will use is:

\[ k = G \times G^* \]

Where:
- \( k \) is the conductivity,
- \( G \) is the conductance,
- \( G^* \) is the cell constant.

Since conductance (G) is the reciprocal of resistance (R), we can express it as:

\[ G = \frac{1}{R} \]

Thus, we can rewrite the conductivity equation as:

\[ k = \frac{1}{R} \times G^* \]

From this, we can solve for the cell constant (G*):

\[ G^* = k \times R \]

Now, we can substitute the values given in the problem:

1. **Given Values**:
- Resistance (R) = 39.0 ohms
- Conductivity (k) = \( 1.29 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1} \)

2. **Substituting Values**:
\[ G^* = (1.29 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1}) \times (39.0 \, \text{ohms}) \]

3. **Calculating G***:
\[ G^* = 1.29 \times 10^{-2} \times 39.0 = 0.5031 \, \text{cm}^{-1} \]

4. **Final Result**:
\[ G^* \approx 5.03 \times 10^{-1} \, \text{cm}^{-1} \]

Thus, the cell constant is approximately \( 5.03 \, \text{cm}^{-1} \).