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 given data
- Specific conductance (κ) = 0.00045 S/cm
- Normality (N) = 0.5 N
- Equivalent conductance at infinite dilution (Λ°) = 300 S cm²/equiv

### Step 2: Calculate the equivalent conductance (Λ) of the solution
The formula to calculate the equivalent conductance (Λ) is:
\[ Λ = \frac{κ \times 1000}{N} \]
Where:
- κ is the specific conductance
- N is the normality

### Step 3: Substitute the values into the formula
\[ Λ = \frac{0.00045 \, \text{S/cm} \times 1000}{0.5} \]

### Step 4: Perform the calculation
\[ Λ = \frac{0.45}{0.5} = 0.9 \, \text{S cm}^2/\text{equiv} \]

### Step 5: Calculate the degree of dissociation (α)
The degree of dissociation can be calculated using the formula:
\[ α = \frac{Λ}{Λ°} \]
Where:
- Λ is the equivalent conductance of the solution
- Λ° is the equivalent conductance at infinite dilution

### Step 6: Substitute the values into the formula
\[ α = \frac{0.9 \, \text{S cm}^2/\text{equiv}}{300 \, \text{S cm}^2/\text{equiv}} \]

### Step 7: Perform the calculation
\[ α = 0.003 \]

### Final Answer
The degree of dissociation (α) of the electrolyte is **0.003**.

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