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 solve the problem, we need to find the molar conductivity of a 0.5 M solution of an electrolyte given the resistance, distance between electrodes, and area of the electrodes. Here’s the step-by-step solution:

### Step 1: Understand the relationship between resistance, cell constant, and specific conductance
The specific conductance (κ) can be calculated using the formula:
\[ \kappa = \frac{1}{R} \cdot \frac{L}{A} \]
where:
- \( R \) = Resistance of the solution (in ohms)
- \( L \) = Distance between the electrodes (in cm)
- \( A \) = Area of the electrodes (in cm²)

### Step 2: Substitute the given values into the formula
Given:
- \( R = 50 \, \Omega \)
- \( L = 2.2 \, \text{cm} \)
- \( A = 4.4 \, \text{cm}^2 \)

Substituting these values into the formula:
\[ \kappa = \frac{1}{50} \cdot \frac{2.2}{4.4} \]

### Step 3: Calculate specific conductance (κ)
Calculating the fraction:
\[ \frac{2.2}{4.4} = 0.5 \]
Now substituting this back into the equation:
\[ \kappa = \frac{1}{50} \cdot 0.5 = \frac{0.5}{50} = 0.01 \, \Omega^{-1} \text{cm}^{-1} \]

### Step 4: Use the specific conductance to find molar conductivity (Λm)
The molar conductivity (Λm) is given by the formula:
\[ \Lambda_m = \frac{1000 \cdot \kappa}{C} \]
where:
- \( C \) = Concentration of the solution (in mol/L)

Given:
- \( C = 0.5 \, \text{mol/L} \)

Substituting the values:
\[ \Lambda_m = \frac{1000 \cdot 0.01}{0.5} \]

### Step 5: Calculate molar conductivity (Λm)
Calculating:
\[ \Lambda_m = \frac{10}{0.5} = 20 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Step 6: Convert units from cm² to m²
To convert from cm² to m², we divide by \( 10000 \):
\[ \Lambda_m = \frac{20}{10000} = 0.002 \, \Omega^{-1} \text{m}^2 \text{mol}^{-1} \]

### Final Answer
The molar conductivity of the solution is:
\[ \Lambda_m = 0.002 \, \text{S m}^{2} \text{mol}^{-1} \]