The resistance of `0.5M` solution of an electrolyte in a cell was found to be `50 Omega`. If the electrodes in the cell are `2.2 cm` apart and have an area of `4.4 cm^(2)` then the molar conductivity (in `S m^(2) mol^(-1))` of the solution is

Options

A. `0.2`
B. `0.02`
C. `0.002`
D. none of these

Explanation

To find the molar conductivity of the given electrolyte solution, we will follow these steps:

### Step 1: Calculate Conductivity (κ)

The formula for conductivity (κ) is given by:

\[ \kappa = \frac{1}{R} \cdot \frac{A}{L} \]

Where:
- \( R \) = resistance of the solution = 50 Ω
- \( A \) = area of the electrodes = 4.4 cm² = \( 4.4 \times 10^{-4} \, m^2 \) (since \( 1 \, cm^2 = 10^{-4} \, m^2 \))
- \( L \) = distance between electrodes = 2.2 cm = \( 0.022 \, m \)

Now substituting the values:

\[ \kappa = \frac{1}{50} \cdot \frac{4.4 \times 10^{-4}}{0.022} \]

Calculating the fraction:

\[ \kappa = \frac{1}{50} \cdot 0.02 = \frac{0.02}{50} = 0.0004 \, S/m \]

### Step 2: Convert Conductivity to S/cm

Since \( 1 \, S/m = 100 \, S/cm \):

\[ \kappa = 0.0004 \, S/m \times 100 = 0.04 \, S/cm \]

### Step 3: Calculate Molar Conductivity (Λm)

The formula for molar conductivity (Λm) is:

\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]

Where:
- \( C \) = concentration of the solution = 0.5 M

Substituting the values:

\[ \Lambda_m = \frac{0.04 \times 1000}{0.5} \]

Calculating this gives:

\[ \Lambda_m = \frac{40}{0.5} = 80 \, S \, cm^2 \, mol^{-1} \]

### Step 4: Convert Molar Conductivity to S m² mol⁻¹

To convert from \( S \, cm^2 \, mol^{-1} \) to \( S \, m^2 \, mol^{-1} \):

\[ \Lambda_m = 80 \, S \, cm^2 \, mol^{-1} \times 10^{-4} = 0.008 \, S \, m^2 \, mol^{-1} \]

### Final Answer

Thus, the molar conductivity of the solution is:

\[ \Lambda_m = 0.008 \, S \, m^2 \, mol^{-1} \]