The ionization constant of a weak electrolyte is `2.5 xx 10^(-5)`, while of the equivalent conductance of its `0.1 M` solution is `19.6 s cm^(2) eq^(-1)`. The equivalent conductance of the electrolyte at infinite dilution is :

Options

A. `250`
B. `196`
C. `392`
D. `384`

Explanation

To find the equivalent conductance of the electrolyte at infinite dilution, we can follow these steps:

### Step 1: Understand the relationship between ionization constant, concentration, and degree of ionization.
The ionization constant \( K_i \) for a weak electrolyte is given by the formula:
\[ K_i = C \alpha^2 \]
where:
- \( K_i \) = ionization constant
- \( C \) = concentration of the solution
- \( \alpha \) = degree of ionization

### Step 2: Calculate the degree of ionization (\( \alpha \)).
We can rearrange the formula to find \( \alpha \):
\[ \alpha = \sqrt{\frac{K_i}{C}} \]
Substituting the given values:
- \( K_i = 2.5 \times 10^{-5} \)
- \( C = 0.1 \)

\[ \alpha = \sqrt{\frac{2.5 \times 10^{-5}}{0.1}} = \sqrt{2.5 \times 10^{-4}} \approx 5 \times 10^{-3} \]

### Step 3: Use the relationship between equivalent conductance and degree of ionization.
The equivalent conductance \( \Lambda \) at a given concentration is related to the equivalent conductance at infinite dilution \( \Lambda^0 \) and the degree of ionization \( \alpha \) by the formula:
\[ \alpha = \frac{\Lambda}{\Lambda^0} \]
Rearranging gives us:
\[ \Lambda^0 = \frac{\Lambda}{\alpha} \]

### Step 4: Substitute the known values to find \( \Lambda^0 \).
We know:
- \( \Lambda = 19.6 \, \text{s cm}^2 \, \text{eq}^{-1} \)
- \( \alpha \approx 5 \times 10^{-3} \)

Substituting these values:
\[ \Lambda^0 = \frac{19.6}{5 \times 10^{-3}} = 3920 \, \text{s cm}^2 \, \text{eq}^{-1} \]

### Final Answer:
The equivalent conductance of the electrolyte at infinite dilution is approximately:
\[ \Lambda^0 \approx 3920 \, \text{s cm}^2 \, \text{eq}^{-1} \]

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