The resistance of 0.5 M solution of and electrolyte enclosed enclosed between two platinuim electrodes 1.56 cm apart and having an are of `2.0cm^(3)` was found to be 30 cm. Calculate the molar conductivity of the electrolyte/.

Explanation

To calculate the molar conductivity of the electrolyte, we will follow these steps:

### Step 1: Calculate the Cell Constant (K)
The cell constant (K) can be calculated using the formula:

\[ K = \frac{L}{A} \]

Where:
- \( L \) = distance between the electrodes (in cm)
- \( A \) = area of the electrodes (in cm²)

Given:
- \( L = 1.56 \, \text{cm} \)
- \( A = 2.0 \, \text{cm}^2 \)

Substituting the values:

\[ K = \frac{1.56 \, \text{cm}}{2.0 \, \text{cm}^2} = 0.78 \, \text{cm}^{-1} \]

### Step 2: Calculate the Specific Conductance (κ)
The specific conductance (κ) can be calculated using the formula:

\[ \kappa = \frac{1}{R} \times K \]

Where:
- \( R \) = resistance of the solution (in ohms)

Given:
- \( R = 30 \, \Omega \)

Substituting the values:

\[ \kappa = \frac{1}{30 \, \Omega} \times 0.78 \, \text{cm}^{-1} = \frac{0.78}{30} = 0.026 \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 3: Calculate the Molar Conductivity (Λ)
The molar conductivity (Λ) can be calculated using the formula:

\[ \Lambda = \frac{1000 \times \kappa}{C} \]

Where:
- \( C \) = molarity of the solution (in mol/L)

Given:
- \( C = 0.5 \, \text{mol/L} \)

Substituting the values:

\[ \Lambda = \frac{1000 \times 0.026 \, \text{ohm}^{-1} \text{cm}^{-1}}{0.5 \, \text{mol/L}} = \frac{26}{0.5} = 52 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Final Answer
The molar conductivity of the electrolyte is:

\[ \Lambda = 52 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

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