The specific conductivity of 0.5N solution is `0.01287 ohm^(-1) cm^(-1)`. What would be its equivalent conductance?

Options

A. 257.4
B. 2.574
C. 25.74
D. `0.2574 ohm^(-1) cm^(2) (g.eq)^(-1)`

Answer

To find the equivalent conductance of a 0.5N solution with a specific conductivity of \(0.01287 \, \text{ohm}^{-1} \text{cm}^{-1}\), we can use the formula for equivalent conductance (\(\Lambda_e\)):

\[ \Lambda_e = \frac{\kappa \times 1000}{N} \]

Where:
- \(\Lambda_e\) = equivalent conductance (in \(\text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1}\))
- \(\kappa\) = specific conductivity (in \(\text{ohm}^{-1} \text{cm}^{-1}\))
- \(N\) = normality of the solution (in eq/L)

### Step-by-step Solution:

1. **Identify the given values**:
- Specific conductivity (\(\kappa\)) = \(0.01287 \, \text{ohm}^{-1} \text{cm}^{-1}\)
- Normality (N) = \(0.5 \, \text{N}\)

2. **Substitute the values into the formula**:
\[ \Lambda_e = \frac{0.01287 \times 1000}{0.5} \]

3. **Calculate the numerator**:
\[ 0.01287 \times 1000 = 12.87 \]

4. **Divide by normality**:
\[ \Lambda_e = \frac{12.87}{0.5} = 25.74 \]

5. **Final result**:
\[ \Lambda_e = 25.74 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \]

### Conclusion:
The equivalent conductance of the 0.5N solution is \(25.74 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1}\).

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