The equivalent conductivity of N/10 solution of acitic acid at `25^(@)C` is `14.3 ohm^(-1) cm^(2) eq^(-1)`. Calculate the degree of dissociation of `CH_(3)COOH` if `Lambda_(oo CH_(3)COOH)` is 390.71.

Options

A. `3.66%`
B. `3.9%`
C. `2.12%`
D. `0.008%`

Explanation

To solve the problem, we need to calculate the degree of dissociation (α) of acetic acid (CH₃COOH) using the given equivalent conductivity values. Here’s a step-by-step solution:

### Step 1: Write down the given data
- Equivalent conductivity of N/10 solution of acetic acid (λ) = 14.3 ohm⁻¹ cm² eq⁻¹
- Molar conductivity at infinite dilution (λ₀) for acetic acid = 390.71 ohm⁻¹ cm² eq⁻¹

### Step 2: Use the formula for degree of dissociation
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\lambda}{\lambda_0} \]
where:
- λ is the equivalent conductivity of the solution
- λ₀ is the molar conductivity at infinite dilution

### Step 3: Substitute the values into the formula
Now, substituting the values we have:
\[ \alpha = \frac{14.3}{390.71} \]

### Step 4: Calculate the degree of dissociation
Perform the calculation:
\[ \alpha = \frac{14.3}{390.71} \approx 0.0366 \]

### Step 5: Convert to percentage
To express the degree of dissociation as a percentage, multiply by 100:
\[ \alpha \times 100 = 0.0366 \times 100 \approx 3.66\% \]

### Conclusion
The degree of dissociation of acetic acid (CH₃COOH) in the N/10 solution at 25°C is approximately **3.66%**.

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