The equivalent conductivites of acetic acid at `298K` at the concentration of `0.1M` and `0.001M` are `5.20` and `49.2S cm^(2) eq^(-1)` respectively. Calculate the degree of dissociation of acetic acid at the these concentrations. Given that, `lambda^(prop) (H^(+))` and `lambda^(prop) (CH_(3)COO^(-))` are `349.8` and `40.9 ohm^(-1) cm^(2) mol^(-1)` respectively.

Explanation

To solve the problem of calculating the degree of dissociation of acetic acid at two different concentrations, we will follow these steps:

### Step 1: Understand the Given Data
We have the following data:
- Equivalent conductivity of acetic acid at 0.1 M: \( \Lambda_{0.1} = 5.20 \, \text{S cm}^2 \text{eq}^{-1} \)
- Equivalent conductivity of acetic acid at 0.001 M: \( \Lambda_{0.001} = 49.2 \, \text{S cm}^2 \text{eq}^{-1} \)
- Molar conductivity at infinite dilution for \( H^+ \): \( \lambda^{\text{prop}}(H^+) = 349.8 \, \text{S cm}^2 \text{mol}^{-1} \)
- Molar conductivity at infinite dilution for \( CH_3COO^- \): \( \lambda^{\text{prop}}(CH_3COO^-) = 40.9 \, \text{S cm}^2 \text{mol}^{-1} \)
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