For a `0.01 M CH_(2)COOH` solution, `wedge_(m)=7.8 Omega^(-1)cm^(2) mol^(-1)` if `wedge_(m)^(@)=390Omega^(-1)cm^(2)mol^(-1)` . What is the degree of the dissociation `(alpha)` of acetic acid ?

Options

A. `0.20`
B. `0.48`
C. `0.02`
D. `0.05`

Explanation

To find the degree of dissociation (α) of acetic acid (CH₃COOH) in a 0.01 M solution, we can use the formula:

\[ \alpha = \frac{\Lambda}{\Lambda^0} \]

where:
- \(\Lambda\) is the molar conductance of the solution at the given concentration (in this case, 0.01 M).
- \(\Lambda^0\) is the molar conductance at infinite dilution.

Given:
- \(\Lambda = 7.8 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)
- \(\Lambda^0 = 390 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)

### Step 1: Substitute the values into the formula

\[ \alpha = \frac{7.8 \, \Omega^{-1} \, cm^2 \, mol^{-1}}{390 \, \Omega^{-1} \, cm^2 \, mol^{-1}} \]

### Step 2: Calculate α

Now, perform the division:

\[ \alpha = \frac{7.8}{390} \]

Calculating this gives:

\[ \alpha = 0.02 \]

### Conclusion

Thus, the degree of dissociation (α) of acetic acid in the 0.01 M solution is **0.02**.

### Final Answer:
**0.02**

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