0.5N solution of a salt placed between two Pt electrodes 2.0 cm apart and having area of cross-section 2.5 `cm^(2)` has resistance of 25 ohms. Calculate the conductance and cell constant.

Explanation

To solve the problem step by step, we will calculate the conductance and cell constant based on the given data.

### Given Data:
- Normality (N) of the salt solution = 0.5 N
- Distance between electrodes (length, L) = 2.0 cm
- Area of cross-section (A) = 2.5 cm²
- Resistance (R) = 25 ohms

### Step 1: Calculate the Cell Constant (k)
The cell constant (k) is defined as the ratio of the distance between the electrodes (L) to the area of cross-section (A).

\[ k = \frac{L}{A} \]

Substituting the given values:

\[ k = \frac{2.0 \, \text{cm}}{2.5 \, \text{cm}^2} \]

Calculating this gives:

\[ k = 0.8 \, \text{cm}^{-1} \]

### Step 2: Calculate the Conductance (G)
Conductance (G) is the reciprocal of resistance (R).

\[ G = \frac{1}{R} \]

Substituting the given resistance:

\[ G = \frac{1}{25 \, \text{ohms}} \]

Calculating this gives:

\[ G = 0.04 \, \text{S} \, (\text{Siemens}) \]

### Final Results:
- Cell Constant (k) = 0.8 cm⁻¹
- Conductance (G) = 0.04 S