The resistance of a `N//10 KCI` solution is 245 ohms. Calculate the specific conductance and the equivalent conductance of the solution if the electrodes in the cell are 4 cm apart and each having an area of 7.0 sq cm.

Explanation

To solve the problem, we will calculate the specific conductance and the equivalent conductance of a \( N/10 \) KCl solution given the resistance, distance between electrodes, and area of the electrodes.

### Step 1: Calculate the Cell Constant
The cell constant \( C \) is given by the formula:
\[ C = \frac{L}{A} \]
where:
- \( L \) = distance between the electrodes (in cm)
- \( A \) = area of the electrodes (in cm²)

Given:
- \( L = 4 \, \text{cm} \)
- \( A = 7.0 \, \text{cm}^2 \)

Substituting the values:
\[ C = \frac{4 \, \text{cm}}{7.0 \, \text{cm}^2} = \frac{4}{7} \, \text{cm}^{-1} \approx 0.5714 \, \text{cm}^{-1} \]

### Step 2: Calculate the Specific Conductance
The specific conductance \( \kappa \) is calculated using the formula:
\[ \kappa = \frac{1}{R} \times C \]
where:
- \( R \) = resistance of the solution (in ohms)

Given:
- \( R = 245 \, \Omega \)

Substituting the values:
\[ \kappa = \frac{1}{245} \times \frac{4}{7} = \frac{4}{1715} \approx 2.33 \times 10^{-3} \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 3: Calculate the Equivalent Conductance
The equivalent conductance \( \Lambda \) is given by the formula:
\[ \Lambda = \kappa \times V \]
where \( V \) is the volume that contains one equivalent of the substance.

For a \( N/10 \) solution:
- Normality = \( \frac{1}{10} \) N, which means 1 equivalent occupies 10,000 cm³ (since 1 N = 1 equivalent per liter).

Substituting the values:
\[ \Lambda = 2.33 \times 10^{-3} \, \text{ohm}^{-1} \text{cm}^{-1} \times 10,000 \, \text{cm}^3 = 23.3 \, \text{ohm}^{-1} \text{cm}^2 \text{equivalent}^{-1} \]

### Final Answers
- Specific Conductance \( \kappa \approx 2.33 \times 10^{-3} \, \text{ohm}^{-1} \text{cm}^{-1} \)
- Equivalent Conductance \( \Lambda \approx 23.3 \, \text{ohm}^{-1} \text{cm}^2 \text{equivalent}^{-1} \)