Ionic conductance of `H^(+) and SO_(4)` are x and yS `cm^(2) mol^(-1)`. Hence equivalent conductivity of `H_(2)SO_(4)` is

Options

A. `2x + (y)/(2)`
B. `x + (y)/(2)`
C. `(x)/(2) + y`
D. `(x)/(2) + (y)/(2)`

Answer

To find the equivalent conductivity of sulfuric acid (H₂SO₄) using the ionic conductance values of H⁺ and SO₄²⁻, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Ionic Conductance**:
- Let the ionic conductance of H⁺ be denoted as \( x \) S cm² mol⁻¹.
- Let the ionic conductance of SO₄²⁻ be denoted as \( y \) S cm² mol⁻¹.

2. **Apply Kohlrausch's Law**:
- According to Kohlrausch's Law, the equivalent conductivity (\( \Lambda \)) of an electrolyte at infinite dilution can be expressed as the sum of the contributions of its constituent ions.
- For H₂SO₄, which dissociates into 2 H⁺ ions and 1 SO₄²⁻ ion, we can write:
\[ \Lambda_{H_2SO_4} = \text{(number of H⁺ ions)} \times \text{conductance of H⁺} + \text{(number of SO₄²⁻ ions)} \times \text{conductance of SO₄²⁻} \]

3. **Substitute the Values**:
- The number of H⁺ ions produced from H₂SO₄ is 2, and the number of SO₄²⁻ ions is 1. Therefore:
\[ \Lambda_{H_2SO_4} = 2 \times x + 1 \times y \]

4. **Calculate the Equivalent Conductivity**:
- Since equivalent conductivity is defined per equivalent of solute, we need to divide the total conductivity by the total number of equivalents. For H₂SO₄, which provides 2 equivalents (2 H⁺ and 1 SO₄²⁻), we can express the equivalent conductivity as:
\[ \Lambda_{eq} = \frac{2x + y}{2} \]

5. **Final Expression**:
- Thus, the equivalent conductivity of H₂SO₄ can be simplified to:
\[ \Lambda_{eq} = \frac{x + y}{2} \]

### Final Answer:
The equivalent conductivity of H₂SO₄ is given by:
\[ \Lambda_{eq} = \frac{x + y}{2} \text{ S cm}^2 \text{ mol}^{-1} \]