Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is `(100Omega)` . The conductivity of this solution is `1.29sm^(-1)`. Resistance of the same cell when filled with 0.2 M of the same solution is `520Omega` . The molar conductivity of (0.02)M solution of the electrolyte will be :

Options

A. ` 124xx10^(-4) sm^(2) mol^(-1)`
B. `1240xx10^(-4) Sm^(2) mol^(-1) `
C. `1.242xx10^(-4)S m ^(2) mol^(-1)`
D. ` 12.4xx10^(-4) sm ^(2) mol^(-1)`

Explanation

To find the molar conductivity of a 0.02 M solution of the electrolyte, we can follow these steps:

### Step 1: Calculate the Cell Constant
The cell constant (k) can be calculated using the formula:
\[ k = \text{Specific Conductivity} \times R \]
Where:
- Specific Conductivity (σ) is given as 1.29 S/m.
- Resistance (R) for the 0.1 M solution is 100 Ω.

Using the formula:
\[ k = 1.29 \, \text{S/m} \times 100 \, \Omega = 129 \, \text{m}^{-1} \]

### Step 2: Calculate Specific Conductivity for 0.2 M Solution
Using the same cell constant, we can find the specific conductivity for the 0.2 M solution:
\[ \sigma = \frac{k}{R} \]
Where:
- Resistance (R) for the 0.2 M solution is 520 Ω.

Substituting the values:
\[ \sigma = \frac{129 \, \text{m}^{-1}}{520 \, \Omega} = 0.247 \, \text{S/m} \]

### Step 3: Calculate Molar Conductivity for 0.02 M Solution
The molar conductivity (Λ_m) can be calculated using the formula:
\[ \Lambda_m = \frac{\sigma}{C} \]
Where:
- σ is the specific conductivity we found for the 0.2 M solution (0.247 S/m).
- C is the concentration of the solution in moles per liter (0.02 M).

Substituting the values:
\[ \Lambda_m = \frac{0.247 \, \text{S/m}}{0.02 \, \text{mol/L}} = 12.35 \, \text{S m}^2/\text{mol} \]

### Step 4: Convert Units to Standard Form
To express the molar conductivity in standard units, we can convert it:
\[ \Lambda_m = 12.35 \, \text{S m}^2/\text{mol} = 1235 \, \text{S cm}^2/\text{mol} \]

### Final Answer
The molar conductivity of the 0.02 M solution of the electrolyte is:
\[ \Lambda_m = 1235 \, \text{S cm}^2/\text{mol} \]

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