The resistance of a 0.5 M solution of an electrolyte was found to be 30 `Omega` enclosed between two platinum electrodes. Calculate the molar conductivity of the solution if the electrodes in the cell are 1.5 cm apart and having an area of cross section 2.0 `cm^(2)`?

Explanation

To calculate the molar conductivity of a 0.5 M solution of an electrolyte with a resistance of 30 Ω, where the electrodes are 1.5 cm apart and have an area of cross-section of 2.0 cm², we can follow these steps:

### Step 1: Calculate the Conductance (G)
The conductance (G) is the reciprocal of resistance (R):
\[ G = \frac{1}{R} \]
Given that \( R = 30 \, \Omega \):
\[ G = \frac{1}{30} \, \text{S} = 0.0333 \, \text{S} \]

### Step 2: Calculate the Cell Constant (x)
The cell constant (x) is calculated using the formula:
\[ x = \frac{L}{A} \]
where:
- \( L = 1.5 \, \text{cm} = 0.015 \, \text{m} \) (convert to meters for standard units)
- \( A = 2.0 \, \text{cm}^2 = 2.0 \times 10^{-4} \, \text{m}^2 \) (convert to square meters)

Now substituting the values:
\[ x = \frac{0.015 \, \text{m}}{2.0 \times 10^{-4} \, \text{m}^2} = 75 \, \text{m}^{-1} \]

### Step 3: Calculate the Conductivity (κ)
The conductivity (κ) can be calculated using the formula:
\[ \kappa = G \times x \]
Substituting the values:
\[ \kappa = 0.0333 \, \text{S} \times 75 \, \text{m}^{-1} = 2.5 \, \text{S/m} \]

### Step 4: Calculate the Molar Conductivity (Λm)
The molar conductivity (Λm) is calculated using the formula:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]
where \( C \) is the concentration in mol/m³. Since the concentration is given in mol/L (0.5 M), we convert it to mol/m³:
\[ C = 0.5 \, \text{mol/L} = 0.5 \times 1000 \, \text{mol/m}^3 = 500 \, \text{mol/m}^3 \]

Now substituting the values:
\[ \Lambda_m = \frac{2.5 \, \text{S/m} \times 1000}{500 \, \text{mol/m}^3} = \frac{2500}{500} = 5 \, \text{S m}^2/\text{mol} \]

### Final Answer
The molar conductivity of the solution is:
\[ \Lambda_m = 5 \, \text{S m}^2/\text{mol} \]