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. `(6250Sm^(2)mol^(-1))`
B. `6.25xx10^(-4)sm^(2)mol^(-1))`
C. `(625xx10^(-4) sm^(2) mol^(-1))`
D. `(62.5sm^(2)mol^(-1))`

Explanation

To solve the problem step by step, we will follow the concepts of conductivity and molar conductivity.

### Step 1: Calculate the Cell Constant
We know that the specific conductance (κ) is related to resistance (R) and the cell constant (k) by the formula:
\[ \kappa = \frac{k}{R} \]
For the first solution:
- Specific conductance (κ) = 1.3 S/m
- Resistance (R) = 50 Ω

We can rearrange the formula to find the cell constant (k):
\[ k = \kappa \times R \]
Substituting the values:
\[ k = 1.3 \, \text{S/m} \times 50 \, \Omega = 65 \, \text{m}^{-1} \]

### Step 2: Calculate the Specific Conductance for the Second Solution
For the second solution:
- Resistance (R) = 260 Ω

Using the same formula for specific conductance:
\[ \kappa = \frac{k}{R} \]
Substituting the values:
\[ \kappa = \frac{65 \, \text{m}^{-1}}{260 \, \Omega} = \frac{1}{4} \, \text{S/m} \]

### Step 3: Calculate Molar Conductivity
Molar conductivity (Λ) is given by the formula:
\[ \Lambda = \frac{\kappa \times 1000}{C} \]
Where:
- κ = specific conductance = \(\frac{1}{4}\) S/m
- C = concentration = 0.4 M

Substituting the values:
\[ \Lambda = \frac{\left(\frac{1}{4} \, \text{S/m}\right) \times 1000}{0.4} \]
Calculating this gives:
\[ \Lambda = \frac{250}{0.4} = 625 \, \text{S m}^2/\text{mol} \]

### Final Answer
The molar conductivity of the 0.4 M solution of the electrolyte is:
\[ \Lambda = 6.25 \times 10^{2} \, \text{S m}^2/\text{mol} \]