The resistance of 0.01 M `AgNO_(3)` solution dipped in a conductivity cell at `25^(@) C` was 1412 ohms . If the molar conductivity of this solution `132.6 ohm^(-1) cm^(2) mol^(-1)` , what is the cell constant of the conductivity cell ?

Explanation

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

### Step 1: Understand the relationship between cell constant, conductivity, and resistance.
The cell constant (K) is defined as:
\[ K = \frac{\kappa}{G} \]
where:
- \( \kappa \) is the specific conductivity (conductivity) of the solution.
- \( G \) is the conductance of the cell, which is the reciprocal of resistance (R):
\[ G = \frac{1}{R} \]

### Step 2: Calculate the conductance (G).
Given that the resistance (R) of the solution is 1412 ohms:
\[ G = \frac{1}{R} = \frac{1}{1412 \, \Omega} \]
Calculating this gives:
\[ G \approx 0.000707 \, S \, (Siemens) \]

### Step 3: Use the molar conductivity to find the specific conductivity (κ).
The molar conductivity (\( \Lambda_m \)) is related to the specific conductivity (\( \kappa \)) by the formula:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]
where \( C \) is the concentration in mol/L. Given:
- \( \Lambda_m = 132.6 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
- \( C = 0.01 \, mol/L \)

Rearranging the formula to find \( \kappa \):
\[ \kappa = \frac{\Lambda_m \times C}{1000} \]
Substituting the values:
\[ \kappa = \frac{132.6 \, \Omega^{-1} \, cm^2 \, mol^{-1} \times 0.01 \, mol/L}{1000} \]
Calculating this gives:
\[ \kappa = 1.326 \times 10^{-3} \, \Omega^{-1} \, cm^{-1} \]

### Step 4: Calculate the cell constant (K).
Now substituting the values of \( \kappa \) and \( G \) into the cell constant formula:
\[ K = \kappa \times R \]
\[ K = (1.326 \times 10^{-3} \, \Omega^{-1} \, cm^{-1}) \times (1412 \, \Omega) \]
Calculating this gives:
\[ K \approx 1.87 \, cm^{-1} \]

### Final Answer:
The cell constant of the conductivity cell is approximately \( 1.87 \, cm^{-1} \).

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