The resistance of 0.1 N solution of a salt is found to be `2.5xx10^(3) Omega`. The equivalent conductance of the solution is (Cell constant `=1.15 cm^(-1)`)

Answer

To find the equivalent conductance of a 0.1 N solution of a salt with a resistance of \( 2.5 \times 10^3 \, \Omega \) and a cell constant of \( 1.15 \, \text{cm}^{-1} \), we can follow these steps:

### Step-by-step Solution:

1. **Identify Given Values:**
- Resistance (\( R \)) = \( 2.5 \times 10^3 \, \Omega \)
- Cell constant (\( g^* \)) = \( 1.15 \, \text{cm}^{-1} \)
- Normality (\( N \)) = \( 0.1 \, N \)

2. **Calculate Conductance (\( G \)):**
Conductance is the reciprocal of resistance.
\[ G = \frac{1}{R} = \frac{1}{2.5 \times 10^3} \, \text{S} \]
\[ G = 0.0004 \, \text{S} = 4 \times 10^{-4} \, \text{S} \]

3. **Calculate Conductivity (\( k \)):**
The conductivity can be calculated using the cell constant and conductance.
\[ k = g^* \times G \]
\[ k = 1.15 \, \text{cm}^{-1} \times 4 \times 10^{-4} \, \text{S} \]
\[ k = 4.6 \times 10^{-4} \, \text{S/cm} \]

4. **Calculate Equivalent Conductance (\( \Lambda \)):**
The equivalent conductance is given by the formula:
\[ \Lambda = \frac{k \times 1000}{N} \]
Substituting the values:
\[ \Lambda = \frac{4.6 \times 10^{-4} \, \text{S/cm} \times 1000}{0.1} \]
\[ \Lambda = \frac{4.6 \times 10^{-1} \, \text{S/cm}}{0.1} \]
\[ \Lambda = 4.6 \, \text{S cm}^2/\text{equivalent} \]

### Final Answer:
The equivalent conductance of the solution is \( 4.6 \, \text{S cm}^2/\text{equivalent} \).