The specific conductance of a 0.12 N solution of an electrolyte is `2.4xx10^(-2) S cm^(-1)`. Calculate its equivalent conductance.

Explanation

To calculate the equivalent conductance of the given electrolyte solution, we can use the following formula:

\[ \Lambda = \frac{K \times 1000}{N} \]

Where:
- \(\Lambda\) = Equivalent conductance (in S cm² eq⁻¹)
- \(K\) = Specific conductance (in S cm⁻¹)
- \(N\) = Normality of the solution (in eq L⁻¹)

### Step-by-Step Solution:

1. **Identify the given values**:
- Specific conductance \(K = 2.4 \times 10^{-2} \, \text{S cm}^{-1}\)
- Normality \(N = 0.12 \, \text{N}\)

2. **Substitute the values into the formula**:
\[ \Lambda = \frac{K \times 1000}{N} \]
\[ \Lambda = \frac{(2.4 \times 10^{-2} \, \text{S cm}^{-1}) \times 1000}{0.12} \]

3. **Calculate the numerator**:
\[ (2.4 \times 10^{-2} \, \text{S cm}^{-1}) \times 1000 = 24 \, \text{S cm}^{-1} \]

4. **Now divide by the normality**:
\[ \Lambda = \frac{24 \, \text{S cm}^{-1}}{0.12} \]

5. **Perform the division**:
\[ \Lambda = 200 \, \text{S cm}^{-2} \text{eq}^{-1} \]

### Final Answer:
The equivalent conductance of the solution is \(200 \, \text{S cm}^{-2} \text{eq}^{-1}\).

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