The conductivity of 0.001 mol `L^(-1)` solution of `CH_(3)COOH` is `3.905xx10^(-5)S cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`. `("Given":lamda_((H^(+)))^(@)=349.65 S cm^(2)mol^(-1)andlamda^(@)(CH_(3)COO^(-))=40.9 D cm^(2)mol^(-1))`

Explanation

To solve the problem, we need to calculate the molar conductivity and the degree of dissociation (α) of the acetic acid solution. Let's break this down step by step.

### Step 1: Calculate Molar Conductivity (λm)

The formula for molar conductivity (λm) is given by:

\[ \lambda_m = \frac{1000 \cdot \kappa}{C} \]

Where:
- \( \kappa \) is the conductivity of the solution (in S cm\(^{-1}\))
- \( C \) is the concentration of the solution (in mol L\(^{-1}\))

Given:
- \( \kappa = 3.905 \times 10^{-5} \, \text{S cm}^{-1} \)
- \( C = 0.001 \, \text{mol L}^{-1} \)

Substituting the values into the formula:

\[ \lambda_m = \frac{1000 \cdot 3.905 \times 10^{-5}}{0.001} \]

Calculating this gives:

\[ \lambda_m = \frac{3.905 \times 10^{-2}}{0.001} = 39.05 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 2: Calculate Molar Conductivity at Infinite Dilution (λm,∞)

The molar conductivity at infinite dilution for acetic acid (CH₃COOH) is the sum of the molar conductivities of its ions:

\[ \lambda_{m, \infty} = \lambda_{H^+} + \lambda_{CH_3COO^-} \]

Given:
- \( \lambda_{H^+} = 349.65 \, \text{S cm}^2 \text{mol}^{-1} \)
- \( \lambda_{CH_3COO^-} = 40.9 \, \text{S cm}^2 \text{mol}^{-1} \)

Substituting the values:

\[ \lambda_{m, \infty} = 349.65 + 40.9 = 390.55 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 3: Calculate Degree of Dissociation (α)

The degree of dissociation (α) is calculated using the formula:

\[ \alpha = \frac{\lambda_m}{\lambda_{m, \infty}} \]

Substituting the values we calculated:

\[ \alpha = \frac{39.05}{390.55} \]

Calculating this gives:

\[ \alpha \approx 0.1 \]

### Final Answers

- Molar Conductivity (λm) = 39.05 S cm² mol⁻¹
- Degree of Dissociation (α) = 0.1