The specific conductance of a 0.1 N KCI solution at `25^@ C` is 0.015 `ohm^ (-1) cm^(-1)`. The resistances of the cell containing the solution at the same temperature was found to be 60 Omega`. The cell constant `(in cm^(-1))` will be

Options

A. 0.9
B. `4 xx 10^3`
C. `2.4 xx 10^(-4)`
D. `1.11`

Answer

To find the cell constant of the solution, we can follow these steps:

### Step 1: Identify the given values
- Specific conductance (K) of KCl solution = 0.015 ohm⁻¹ cm⁻¹
- Resistance (R) of the cell = 60 ohms

### Step 2: Calculate the conductance (G) of the cell
Conductance (G) is the reciprocal of resistance (R):
\[ G = \frac{1}{R} = \frac{1}{60 \, \text{ohms}} = 0.01667 \, \text{ohm}^{-1} \]

### Step 3: Use the relationship between specific conductance, conductance, and cell constant
The relationship is given by:
\[ K = G \cdot G^* \]
where \( G^* \) is the cell constant. Rearranging the equation to solve for the cell constant:
\[ G^* = \frac{K}{G} \]

### Step 4: Substitute the known values into the equation
Substituting the values of K and G into the equation:
\[ G^* = \frac{0.015 \, \text{ohm}^{-1} \text{cm}^{-1}}{0.01667 \, \text{ohm}^{-1}} \]

### Step 5: Calculate the cell constant
Calculating the above expression:
\[ G^* = \frac{0.015}{0.01667} \approx 0.9 \, \text{cm}^{-1} \]

### Final Answer
The cell constant \( G^* \) is approximately 0.9 cm⁻¹.

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