The limiting molar conductivities of HCl, `CH_(3)COONa` and NaCl are respectively 425, 90 and 250 mho `cm^(2)mol^(-1)` at `25^(@)C`.t he molar conductivity of 0.1 M `CH_(3)COOH` solution is 7.8 mho `cm^(2)mol^(-1)` at the same temperature. The degree of dissociation of 0.1 M acetic acid solution at the same temperature is

Options

A. `0.10`
B. `0.02`
C. `0.15`
D. `0.03`

Explanation

To find the degree of dissociation of 0.1 M acetic acid (CH₃COOH) solution, we can follow these steps:

### Step 1: Calculate the limiting molar conductivity of acetic acid (CH₃COOH)

The limiting molar conductivity (Λ°) of a weak electrolyte can be calculated using the formula:
\[ \Lambda° = \Lambda°_HCl + \Lambda°_{CH₃COO^-} + \Lambda°_{Na^+} \]

Where:
- \(\Lambda°_{HCl} = 425 \, mho \, cm^2 mol^{-1}\)
- \(\Lambda°_{CH₃COO^-} = 90 \, mho \, cm^2 mol^{-1}\)
- \(\Lambda°_{Na^+} = 250 \, mho \, cm^2 mol^{-1}\)

However, since CH₃COOH is a weak acid, we will not directly use the above formula but rather focus on its degree of dissociation.

### Step 2: Use the molar conductivity of the acetic acid solution

The molar conductivity (Λ) of the 0.1 M acetic acid solution is given as:
\[ \Lambda = 7.8 \, mho \, cm^2 mol^{-1} \]

### Step 3: Relate the molar conductivity to degree of dissociation

The molar conductivity of a weak electrolyte can be expressed as:
\[ \Lambda = \alpha \Lambda° \]
Where:
- \(\alpha\) is the degree of dissociation
- \(\Lambda°\) is the limiting molar conductivity of the weak acid.

### Step 4: Calculate the limiting molar conductivity of acetic acid

The limiting molar conductivity of acetic acid can be calculated as:
\[ \Lambda°_{CH₃COOH} = \Lambda°_{H^+} + \Lambda°_{CH₃COO^-} \]
Where:
- \(\Lambda°_{H^+} = 425 \, mho \, cm^2 mol^{-1}\)
- \(\Lambda°_{CH₃COO^-} = 90 \, mho \, cm^2 mol^{-1}\)

Thus,
\[ \Lambda°_{CH₃COOH} = 425 + 90 = 515 \, mho \, cm^2 mol^{-1} \]

### Step 5: Calculate the degree of dissociation (α)

Now we can find the degree of dissociation using the molar conductivity:
\[ \alpha = \frac{\Lambda}{\Lambda°} \]
Substituting the values:
\[ \alpha = \frac{7.8}{515} \]

### Step 6: Calculate the numerical value

Calculating the above gives:
\[ \alpha \approx 0.0151 \]

### Step 7: Convert to percentage

To express the degree of dissociation as a percentage:
\[ \alpha \times 100 \approx 1.51\% \]

### Final Answer

The degree of dissociation of 0.1 M acetic acid solution at 25°C is approximately **1.51%**.

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