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)`)

Options

A. `4.6 ohm^(-1) cm^(2)" equiv"^(-1)`
B. `5.6 ohm^(-1) cm^(2)" equive"^(-1)`
C. `6.6 ohm^(-1) cm^(2)" equiv"^(-1)`
D. `7.6 ohm^(-1) cm^(2)" equiv"^(-1)`

Explanation

To solve the problem of finding 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. **Understand the Formula for Equivalent Conductance**:
The equivalent conductance (\( \Lambda \)) is given by the formula:
\[ \Lambda = \frac{\kappa \times 1000}{C} \]
where \( \kappa \) is the conductivity and \( C \) is the concentration in normality.

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

3. **Calculate Conductivity**:
The conductivity (\( \kappa \)) can be calculated using the cell constant (\( k \)) and conductance (\( G \)):
\[ \kappa = k \times G \]
Given \( k = 1.15 \, \text{cm}^{-1} \):
\[ \kappa = 1.15 \times 0.0004 = 0.00046 \, \text{S/cm} \]

4. **Substitute Values into the Equivalent Conductance Formula**:
Now, substitute the values of conductivity and concentration into the equivalent conductance formula:
\[ \Lambda = \frac{0.00046 \times 1000}{0.1} \]
\[ \Lambda = \frac{0.46}{0.1} = 4.6 \, \text{S cm}^{-1} \text{equivalent}^{-1} \]

5. **Final Answer**:
The equivalent conductance of the solution is:
\[ \Lambda = 4.6 \, \text{S cm}^{-1} \text{equivalent}^{-1} \]

### Summary:
The equivalent conductance of the 0.1 N solution of salt is \( 4.6 \, \text{S cm}^{-1} \text{equivalent}^{-1} \).