The molar conductivity of acetic acid at infinite dilution is `387omega^(-1)cm^(2)mol^(-1)`. At the same temperature, but at a concentration of 1 mole in 1000 litres, it is 55 `Omega^(-1)cm^(2)mol^(-1)`. What is the % age dissociation of 0.001 M acetic acid?

Explanation

To find the percentage dissociation of 0.001 M acetic acid, we can follow these steps:

### Step 1: Understand the given data
- Molar conductivity at infinite dilution (Λ°) of acetic acid = 387 Ω^(-1) cm^2 mol^(-1)
- Molar conductivity (Λ) at 0.001 M concentration = 55 Ω^(-1) cm^2 mol^(-1)

### Step 2: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:

\[ \alpha = \frac{\Lambda}{\Lambda^\circ} \]

Where:
- Λ = Molar conductivity at the given concentration
- Λ° = Molar conductivity at infinite dilution

Substituting the values:

\[ \alpha = \frac{55 \, \Omega^{-1} \, cm^2 \, mol^{-1}}{387 \, \Omega^{-1} \, cm^2 \, mol^{-1}} \]

### Step 3: Perform the calculation
Calculating α:

\[ \alpha = \frac{55}{387} \approx 0.142 \]

### Step 4: Convert α to percentage
To find the percentage dissociation, we multiply α by 100:

\[ \text{Percentage dissociation} = \alpha \times 100 = 0.142 \times 100 \approx 14.2\% \]

### Final Answer
The percentage dissociation of 0.001 M acetic acid is approximately **14.2%**.

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