Resistance of `0.2 M` solution of an electrolyte is `50 Omega`. The specific conductance of the solution is ` 1.3 S m^(-1)`. If resistance of the `0.4 M` solution of the same electrolyte is `260 Omega`, its molar conductivity is .

Options

A. ` 6250 S m^2 "mol"^(-1)`
B. ` 6.25 xx 10^(-4) Sm^2 "mol"^(-1)`
C. `625 xx 10^(-4) S m^2 "mol"^(-1)`
D. ` 62.5 S m^2 "mol"^(-1)`

Explanation

To solve the problem step by step, we will first find the cell constant (L/A) using the data from the 0.2 M solution, and then use that to find the molar conductivity of the 0.4 M solution.

### Step 1: Calculate the Cell Constant (L/A)
We know from the formula for resistance:
\[ R = \frac{\rho L}{A} \]
Where:
- \( R \) = resistance
- \( \rho \) = resistivity
- \( L/A \) = cell constant

Given:
- Resistance \( R = 50 \, \Omega \)
- Specific conductance \( K = 1.3 \, S/m \) (which is the same as conductivity)

Resistivity \( \rho \) can be calculated as:
\[ \rho = \frac{1}{K} = \frac{1}{1.3} \, \Omega \cdot m \]

Now substituting the values into the resistance formula:
\[ 50 = \frac{\frac{1}{1.3} \cdot L}{A} \]
Rearranging gives us:
\[ L/A = 50 \cdot 1.3 \]
\[ L/A = 65 \, m^{-1} \]

### Step 2: Use the Cell Constant to Find Conductivity (K) for 0.4 M Solution
For the 0.4 M solution, we use the same cell constant \( L/A \):
\[ R = \frac{1}{K} \cdot \frac{L}{A} \]
Given:
- Resistance \( R = 260 \, \Omega \)

Substituting the values:
\[ 260 = \frac{1}{K} \cdot 65 \]
Rearranging gives us:
\[ K = \frac{65}{260} \]
\[ K = 0.25 \, S/m \]

### Step 3: Calculate Molar Conductivity (Λm)
Molar conductivity \( \Lambda_m \) is given by the formula:
\[ \Lambda_m = K \cdot \frac{1000}{C} \]
Where:
- \( C \) = concentration in mol/m³ (for 0.4 M solution, \( C = 0.4 \, mol/L = 0.4 \, mol/m^3 \))

Substituting the values:
\[ \Lambda_m = 0.25 \cdot \frac{1000}{0.4} \]
\[ \Lambda_m = 0.25 \cdot 2500 \]
\[ \Lambda_m = 625 \, S \cdot m^2 \cdot mol^{-1} \]

### Step 4: Convert to cm²
Since we often express molar conductivity in \( S \cdot cm^2 \cdot mol^{-1} \), we convert:
\[ 625 \, S \cdot m^2 \cdot mol^{-1} = 625 \cdot 10^4 \, S \cdot cm^2 \cdot mol^{-1} \]

### Final Answer
The molar conductivity \( \Lambda_m \) of the 0.4 M solution is:
\[ \Lambda_m = 6.25 \times 10^2 \, S \cdot cm^2 \cdot mol^{-1} \]