The ionization constant `(K_(a))` of a weak electrolyte is `2.5xx10^(-7)`, while `wedge_(eq)` of its `0.01M` solution is `19.65S cm^(2) eq^(-1)wedge^(@)._(eq)` is `……………….` .

Explanation

To solve the problem, we need to find the equivalent conductance at infinite dilution (Λ₀) of a weak electrolyte given its ionization constant (Kₐ) and the equivalent conductance of its 0.01 M solution (Λ).

### Step-by-Step Solution:

1. **Identify the given values**:
- Ionization constant, Kₐ = 2.5 × 10^(-7)
- Concentration, c = 0.01 M
- Equivalent conductance at 0.01 M, Λ = 19.65 S cm² eq^(-1)

2. **Use the relation from Ostwald's dilution law**:
- For a weak electrolyte, the ionization constant Kₐ can be expressed as:
\[ K_a = c \cdot \alpha^2 \]
where α is the degree of ionization.

3. **Rearranging the formula to find α**:
- We can rearrange the equation to solve for α:
\[ \alpha = \sqrt{\frac{K_a}{c}} \]

4. **Substituting the values**:
- Substitute Kₐ and c into the equation:
\[ \alpha = \sqrt{\frac{2.5 \times 10^{-7}}{0.01}} = \sqrt{2.5 \times 10^{-5}} = \sqrt{25 \times 10^{-6}} = 5 \times 10^{-3} \]

5. **Relate α to equivalent conductance**:
- The degree of ionization α is also related to the equivalent conductance at a given concentration (Λ) and the equivalent conductance at infinite dilution (Λ₀):
\[ \alpha = \frac{\Lambda}{\Lambda_0} \]

6. **Rearranging to find Λ₀**:
- Rearranging the equation gives:
\[ \Lambda_0 = \frac{\Lambda}{\alpha} \]

7. **Substituting the known values**:
- Substitute Λ and α into the equation:
\[ \Lambda_0 = \frac{19.65}{5 \times 10^{-3}} = 19.65 \times 10^3 = 3930 \, \text{S cm}^2 \text{eq}^{-1} \]

8. **Final result**:
- Therefore, the equivalent conductance at infinite dilution (Λ₀) is:
\[ \Lambda_0 = 3930 \, \text{S cm}^2 \text{eq}^{-1} \]