The conducitivity of a 0.01 M solution of acetic acid at 298 K is `1.65 xx 10^(-4) S cm^(-3)` . Calculate
(i) Molar conductivity of the solution
(ii) degree of dissociation of `CH_3 COOH`
(iii) dissociation constant for acetic acid
Given that
`lambda^(@) (H^(+)) = 349.1` and `lambda^(@) (CH_(3) COO^(-)) = 40.9 S cm^(2) mol^(-1)`.

Explanation

To solve the problem step by step, we will calculate the molar conductivity, degree of dissociation, and dissociation constant for acetic acid based on the given data.

### Given Data:
- Conductivity (κ) of 0.01 M acetic acid = \(1.65 \times 10^{-4} \, \text{S cm}^{-1}\)
- Concentration (C) = 0.01 M
- \(\lambda^0 (H^+) = 349.1 \, \text{S cm}^2 \text{mol}^{-1}\)
- \(\lambda^0 (CH_3COO^-) = 40.9 \, \text{S cm}^2 \text{mol}^{-1}\)

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