The ionization constant of a weak electrolyte is `25 xx 10^(-6)` while the equivalent conductance of its 0.01 M solution is 19.6 s `cm^(2) eq^(-1)`. The equivalent conductance of the electrolyte at infinite dilution (in S `cm^(2)eq^(-1)`) will be

Options

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

Answer

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

### Step 1: Understand the relationship between degree of dissociation (α), equivalent conductance at concentration (Λ), and equivalent conductance at infinite dilution (Λ°).

The degree of dissociation (α) can be expressed as:
\[ \alpha = \frac{\Lambda}{\Lambda^\circ} \]

Where:
- \( \Lambda \) = equivalent conductance at concentration (given as 19.6 S cm² eq⁻¹)
- \( \Lambda^\circ \) = equivalent conductance at infinite dilution (what we need to find)

### Step 2: Use the relationship between ionization constant (K) and degree of dissociation (α).

The degree of dissociation can also be expressed in terms of the ionization constant (K) and concentration (C):
\[ \alpha = \sqrt{\frac{K}{C}} \]

Where:
- \( K \) = ionization constant (given as \( 25 \times 10^{-6} \))
- \( C \) = concentration (given as 0.01 M)

### Step 3: Calculate the degree of dissociation (α).

Substituting the values into the equation:
\[ \alpha = \sqrt{\frac{25 \times 10^{-6}}{0.01}} \]

Calculating the right side:
\[ \alpha = \sqrt{2.5 \times 10^{-4}} = 0.01581 \]

### Step 4: Substitute α back into the first equation to find Λ°.

Now we can substitute α into the first equation:
\[ 0.01581 = \frac{19.6}{\Lambda^\circ} \]

Rearranging to solve for \( \Lambda^\circ \):
\[ \Lambda^\circ = \frac{19.6}{0.01581} \]

Calculating this gives:
\[ \Lambda^\circ \approx 1240.76 \, \text{S cm}^2 \text{eq}^{-1} \]

### Step 5: Final answer.

Thus, the equivalent conductance of the electrolyte at infinite dilution is approximately:
\[ \Lambda^\circ \approx 1240.76 \, \text{S cm}^2 \text{eq}^{-1} \]