Resistance of 0.2 M solution of an electrolyte is `50 Omega`. The specific conductance of the solution of 0.5 M solution of same electrolyte is `1.4 S m^(-1)` and resistance of same solution of the same electrolyte is `280 Omega`. The molar conductivity of 0.5 M solutions of the electrolyte is `5 m^(2) mol^(-1)` is

Options

A. `5 xx 10^(-4)`
B. `5 xx 10^(-3)`
C. `5 xx 10^(3)`
D. `5 xx 10^(2)`

Answer

To find the molar conductivity of the 0.5 M solution of the electrolyte, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Resistance of 0.2 M solution, \( R_1 = 50 \, \Omega \)
- Specific conductance of 0.5 M solution, \( \kappa = 1.4 \, \text{S m}^{-1} \)
- Resistance of 0.5 M solution, \( R_2 = 280 \, \Omega \)
- Molar conductivity of 0.5 M solution, \( \Lambda_m = ? \)

2. **Calculate the Conductance of the 0.2 M Solution:**
- Conductance \( G_1 = \frac{1}{R_1} = \frac{1}{50} = 0.02 \, \text{S} \)

3. **Calculate the Specific Conductance of the 0.2 M Solution:**
- The specific conductance \( \kappa_1 \) can be calculated using the formula:
\[ \kappa_1 = G_1 \cdot \text{cell constant} \]
- We need to find the cell constant \( k \) first. Since we have the specific conductance of the 0.5 M solution, we can use that to find the cell constant.

4. **Calculate the Conductance of the 0.5 M Solution:**
- Conductance \( G_2 = \frac{1}{R_2} = \frac{1}{280} \approx 0.00357 \, \text{S} \)

5. **Use the Specific Conductance to Find Cell Constant:**
- From the specific conductance of the 0.5 M solution:
\[ \kappa = G_2 \cdot \text{cell constant} \]
- Rearranging gives us:
\[ \text{cell constant} = \frac{\kappa}{G_2} = \frac{1.4}{0.00357} \approx 392.4 \, \text{m}^{-1} \]

6. **Calculate Molar Conductivity:**
- Molar conductivity \( \Lambda_m \) is given by:
\[ \Lambda_m = \kappa \cdot \frac{1000}{C} \]
- Where \( C \) is the concentration in mol/m³. For 0.5 M:
\[ C = 0.5 \, \text{mol/L} = 0.5 \times 10^{-3} \, \text{mol/m}^3 \]
- Substituting the values:
\[ \Lambda_m = 1.4 \cdot \frac{1000}{0.5} = 1.4 \cdot 2000 = 2800 \, \text{S m}^2/\text{mol} \]

### Final Answer:
The molar conductivity of the 0.5 M solution of the electrolyte is \( 2800 \, \text{S m}^2/\text{mol} \).