The resistance of 0.01 M `CH_(2)COOH` solution was found to be 2220 ohm in a conductivity cell having cell constant 0.366 `cm^(-1)`. Calculate:
(i) molar conductivity `(wedge_(m))" of "0.01 M CH_(3)COOH`
(ii) `wedge_(m)^(oo)`
(iii) degree of dissociation, `alpha` and
(iv) dissociation constant of the acid.
`[lambda^(0) (H^(+))=349.1"ohm"^(-1)cm^(2)"mol"^(-1), lambda^(0)(CH_(3)COO^(-)) =40.9 "ohm"^(-1)cm^(2) "mol"^(-1)]`

Explanation

To solve the problem step by step, let's break it down into the required parts:

### Given Data:
- Resistance (R) = 2220 ohms
- Cell constant (k) = 0.366 cm⁻¹
- Concentration (C) = 0.01 M
- Conductivity of ions:
- λ⁰(H⁺) = 349.1 ohm⁻¹ cm² mol⁻¹
- λ⁰(CH₃COO⁻) = 40.9 ohm⁻¹ cm² mol⁻¹

### (i) Calculate Molar Conductivity (λₘ) of 0.01 M CH₃COOH

1. **Calculate the conductivity (K)**:
\[ K = \frac{1}{R} = \frac{1}{2220} \text{ S} = 0.00045045 \text{ S} \]

2. **Calculate molar conductivity (λₘ)**:
\[ λₘ = K \times \text{cell constant} \times 1000 \]
\[ λₘ = 0.00045045 \times 0.366 \times 1000 = 16.48 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \]

### (ii) Calculate λₘ⁰ (Molar Conductivity at Infinite Dilution)

1. **Using Kohlrausch's Law**:
\[ λₘ⁰ = λ⁰(H⁺) + λ⁰(CH₃COO⁻) \]
\[ λₘ⁰ = 349.1 + 40.9 = 390 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \]

### (iii) Calculate Degree of Dissociation (α)

1. **Using the formula for degree of dissociation**:
\[ α = \frac{λₘ}{λₘ⁰} \]
\[ α = \frac{16.48}{390} = 0.0422 \]

### (iv) Calculate Dissociation Constant (Kₐ)

1. **Using the formula for dissociation constant**:
\[ K_a = \frac{C \cdot α^2}{1 - α} \]
\[ K_a = \frac{0.01 \cdot (0.0422)^2}{1 - 0.0422} \]
\[ K_a = \frac{0.01 \cdot 0.00178484}{0.9578} \approx 1.86 \times 10^{-5} \text{ mol}^{-1} \]

### Summary of Results:
1. Molar Conductivity (λₘ) = 16.48 ohm⁻¹ cm² mol⁻¹
2. Molar Conductivity at Infinite Dilution (λₘ⁰) = 390 ohm⁻¹ cm² mol⁻¹
3. Degree of Dissociation (α) = 0.0422
4. Dissociation Constant (Kₐ) = 1.86 × 10⁻⁵ mol⁻¹