In a conductivity cell the two platinum electrodes each of area 10 sq. cm are fixed 1.5 cm apart. The cell. Contained `0.05N` solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, find equivalent conductance of the salt solution.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the cell constant (L/A)
The cell constant (k) is given by the formula:
\[ k = \frac{L}{A} \]
Where:
- \( L \) = distance between the electrodes = 1.5 cm
- \( A \) = area of the electrodes (since they are half dipped, we take half of 10 cm²) = \( \frac{10}{2} = 5 \) cm²

Now substituting the values:
\[ k = \frac{1.5 \, \text{cm}}{5 \, \text{cm}^2} = 0.3 \, \text{cm}^{-1} \]

### Step 2: Calculate the conductance (G)
Conductance (G) is the reciprocal of resistance (R):
\[ G = \frac{1}{R} \]
Given that the resistance \( R = 50 \, \Omega \):
\[ G = \frac{1}{50} = 0.02 \, \text{S} \, (\text{Siemens}) \]

### Step 3: Calculate the conductivity (κ)
The conductivity (κ) can be calculated using the formula:
\[ \kappa = G \cdot k \]
Substituting the values we found:
\[ \kappa = 0.02 \, \text{S} \cdot 0.3 \, \text{cm}^{-1} = 0.006 \, \text{S/cm} \]

### Step 4: Calculate the equivalent conductance (λ)
The equivalent conductance (λ) can be calculated using the formula:
\[ \lambda = \frac{\kappa \cdot 1000}{N} \]
Where:
- \( N \) = normality of the solution = 0.05 N

Substituting the values:
\[ \lambda = \frac{0.006 \, \text{S/cm} \cdot 1000}{0.05} \]
\[ \lambda = \frac{6}{0.05} = 120 \, \text{S cm}^{-1} \text{equivalent}^{-1} \]

### Final Answer
The equivalent conductance of the salt solution is:
\[ \lambda = 120 \, \text{S cm}^{-1} \text{equivalent}^{-1} \]

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