When a certain conductance cell was finlled with 0.1 mol `L^(-1)` KCI solution, it had a resistance of 85 ohm at 298K. When the same cell was filled with an aqueous solution of 0.052 mol `L^(-1)` of an electrolyte, the resistance was 96 ohm. Calcualte the molar conductance of the electrolyte at this concentration. (Specific conductance of 0.1 mol `L^(-1)` KCI solution is 1.29xx10^(-2)ohm^(-1)cm^(-1))`.

Explanation

To solve the problem, we need to calculate the molar conductance of the electrolyte solution at a concentration of 0.052 mol L^(-1). Here’s the step-by-step solution:

### Step 1: Calculate the Cell Constant (K)
The cell constant (K) can be calculated using the resistance (R) of the KCl solution and its specific conductance (κ).

Given:
- Resistance (R) for 0.1 mol L^(-1) KCl = 85 ohm
- Specific conductance (κ) for 0.1 mol L^(-1) KCl = 1.29 × 10^(-2) ohm^(-1) cm^(-1)

The formula for cell constant is:
\[ K = R \cdot \kappa \]

Substituting the values:
\[ K = 85 \, \text{ohm} \times 1.29 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1} \]
\[ K = 1.0965 \, \text{cm}^{-1} \]

### Step 2: Calculate the Specific Conductance (κ) for the Electrolyte
Now we will calculate the specific conductance of the electrolyte solution using its resistance.

Given:
- Resistance (R) for 0.052 mol L^(-1) electrolyte = 96 ohm

Using the formula:
\[ \kappa = \frac{K}{R} \]

Substituting the values:
\[ \kappa = \frac{1.0965 \, \text{cm}^{-1}}{96 \, \text{ohm}} \]
\[ \kappa = 1.145 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 3: Calculate the Molar Conductance (Λ) of the Electrolyte
The molar conductance (Λ) can be calculated using the specific conductance (κ) and the concentration (C) of the electrolyte.

Given:
- Concentration (C) = 0.052 mol L^(-1)

Using the formula:
\[ \Lambda = \frac{1000 \cdot \kappa}{C} \]

Substituting the values:
\[ \Lambda = \frac{1000 \cdot 1.145 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1}}{0.052} \]
\[ \Lambda = \frac{11.45}{0.052} \]
\[ \Lambda = 220.19 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Final Answer
The molar conductance of the electrolyte at a concentration of 0.052 mol L^(-1) is:
\[ \Lambda = 220.19 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]