A conductivity cell is used to measure the conductance of electrolyte . It makes use of conductivity of water which does not contain any ions. The cell constant of conductivity cell is determined.
Calculate `alpha` of `CH_(3)COOH` if `Delta^(oo)._(m)` for `HCl, NaCl, CH_(3)COOHNa` are `426,126,91 S cm^(2) mol^(-1)`, respectively, and `wedge_(m)=14.4 S cm^(2) mol^(-1) ` at `0.015 M` concentration.

Options

A. `0.037`
B. `0.018`
C. `0.37`
D. `0.18`

Explanation

To calculate the degree of dissociation (α) of acetic acid (CH₃COOH), we will use the provided molar conductivities at infinite dilution for HCl, NaCl, and CH₃COONa, along with the molar conductivity of CH₃COONa at a specific concentration.

### Step-by-Step Solution:

1. **Identify the Molar Conductivities**:
- Molar conductivity at infinite dilution for:
- HCl (Δm°) = 426 S cm² mol⁻¹
- NaCl (Δm°) = 126 S cm² mol⁻¹
- CH₃COONa (Δm°) = 91 S cm² mol⁻¹

2. **Calculate the Molar Conductivity of Acetic Acid (Δm° for CH₃COOH)**:
- The formula for calculating the molar conductivity of acetic acid at infinite dilution is:
\[ Δm°(CH₃COOH) = Δm°(HCl) - Δm°(NaCl) + Δm°(CH₃COO^-) \]
- Here, we need to recognize that the molar conductivity of acetate ions (CH₃COO⁻) is the same as that of CH₃COONa, which is 91 S cm² mol⁻¹.
- Thus, we can rewrite the equation as:
\[ Δm°(CH₃COOH) = 426 - 126 + 91 \]

3. **Perform the Calculation**:
- Now, calculate:
\[ Δm°(CH₃COOH) = 426 - 126 + 91 = 391 \, \text{S cm}^2 \text{mol}^{-1} \]

4. **Calculate the Degree of Dissociation (α)**:
- The degree of dissociation (α) is given by the formula:
\[ α = \frac{Δm(CH₃COOH)}{Δm°(CH₃COOH)} \]
- Where:
- \(Δm(CH₃COOH)\) is the molar conductivity of acetic acid at the given concentration, which is provided as 14.4 S cm² mol⁻¹.
- Substitute the values:
\[ α = \frac{14.4}{391} \]

5. **Perform the Final Calculation**:
- Now, calculate α:
\[ α ≈ 0.037 \]

### Conclusion:
The degree of dissociation (α) of acetic acid (CH₃COOH) is approximately **0.037**.