Select the equivalent conductivity of `1.0 M H_(2)SO_(4)`, if its conductivity is `0.26 ohm^(-1) cm^(-1)`:

Options

A. `1.3 xx 10^(2) S cm^(2) eq^(-1)`
B. `1.6 xx 10^(2) S cm^(-1)`
C. `13S cm^(2) mol^(-1)`
D. `1.3 xx 10^(3) S cm^(2) mol^(-1)`

Answer

To find the equivalent conductivity of 1.0 M \( H_2SO_4 \) given its conductivity, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Conductivity (\( \kappa \)) of \( H_2SO_4 \) = 0.26 \( \Omega^{-1} \, \text{cm}^{-1} \)
- Molarity (C) of \( H_2SO_4 \) = 1.0 M

2. **Determine the n-factor of \( H_2SO_4 \):**
- \( H_2SO_4 \) is a diprotic acid, meaning it can donate two protons (H\(^+\)). Therefore, the n-factor = 2.

3. **Calculate Normality (N):**
- Normality (N) is calculated as:
\[ N = n \times C \]
- Here, \( n = 2 \) and \( C = 1.0 \, \text{M} \):
\[ N = 2 \times 1.0 = 2.0 \, \text{N} \]

4. **Use the Formula for Equivalent Conductivity (\( \Lambda \)):**
- The formula relating conductivity to equivalent conductivity is:
\[ \Lambda = \frac{\kappa \times 1000}{N} \]
- Where \( \kappa \) is in \( \Omega^{-1} \, \text{cm}^{-1} \) and \( N \) is in equivalents per liter. The factor of 1000 is used to convert liters to milliliters.

5. **Substitute the Values:**
- Substitute \( \kappa = 0.26 \, \Omega^{-1} \, \text{cm}^{-1} \) and \( N = 2.0 \):
\[ \Lambda = \frac{0.26 \times 1000}{2.0} \]

6. **Calculate Equivalent Conductivity:**
- Performing the calculation:
\[ \Lambda = \frac{260}{2} = 130 \, \text{S cm}^2 \, \text{equivalent}^{-1} \]
- This can be expressed in scientific notation as:
\[ \Lambda = 1.3 \times 10^2 \, \text{S cm}^2 \, \text{equivalent}^{-1} \]

7. **Final Answer:**
- The equivalent conductivity of \( 1.0 \, M \, H_2SO_4 \) is \( 1.3 \times 10^2 \, \text{S cm}^2 \, \text{equivalent}^{-1} \).

### Conclusion:
The correct option is \( 1.3 \times 10^2 \, \text{S cm}^2 \, \text{equivalent}^{-1} \).