The specific conductance of a0.5 N solution of an electrolyte at `25^@C` is 0.00045 `Scm^(-1)` . The equivalent conductance of this electrolyte at infinite dilution is 300 S `cm^2eq^(-1)` . The degree of dissociation of the electrolyte is

Options

A. 0.66
B. 0.03
C. 0.003
D. 0.3

Answer

To find the degree of dissociation (α) of the electrolyte, we can follow these steps:

### Step 1: Understand the relationship between specific conductance, normality, and equivalent conductance.
The specific conductance (κ) is related to the equivalent conductance (λ) and normality (N) by the formula:
\[ \lambda_m = \frac{\kappa}{N} \]
where:
- \( \lambda_m \) is the equivalent conductance at the given concentration,
- \( \kappa \) is the specific conductance,
- \( N \) is the normality of the solution.

### Step 2: Substitute the given values into the formula.
Given:
- Specific conductance, \( \kappa = 0.00045 \, S \, cm^{-1} \)
- Normality, \( N = 0.5 \, N \)

Substituting the values:
\[ \lambda_m = \frac{0.00045 \, S \, cm^{-1}}{0.5 \, N} = 0.0009 \, S \, cm^2 \, eq^{-1} \]

### Step 3: Calculate the degree of dissociation (α).
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\lambda_m}{\lambda_{m}^{\infty}} \]
where:
- \( \lambda_{m}^{\infty} \) is the equivalent conductance at infinite dilution.

Given:
- \( \lambda_{m}^{\infty} = 300 \, S \, cm^2 \, eq^{-1} \)

Substituting the values:
\[ \alpha = \frac{0.0009 \, S \, cm^2 \, eq^{-1}}{300 \, S \, cm^2 \, eq^{-1}} = \frac{0.0009}{300} = 0.000003 \]

### Step 4: Finalize the result.
Thus, the degree of dissociation (α) is:
\[ \alpha = 0.003 \]

### Conclusion:
The degree of dissociation of the electrolyte is **0.003**.

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