The degree of dissociation of weak electrolyde is inversely proportional to the square root fo concentration. It is called Ostwald's dilution law.
`alpha = sqrt((K_(a))/(c))` As the tempertaure increases, degree of dissociation will increase.
`(alpha_(1))/(alpha_(2)) = sqrt((K_(a_(1)))/(K_(a_(2))))` if concentration is same.
`(alpha_(1))/(alpha_(2)) = sqrt((c_(2))/(c_(1)))` if acid is same.
`0.01M CH_(3)COOH` has `4.24%` degree of dissociation, the degree of dissociation of `0.1M CH_(3)COOH` will be

Options

A. `1.33%`
B. `4.24%`
C. `5.24%`
D. `0.33%`

Explanation

To solve the problem, we will use Ostwald's dilution law, which relates the degree of dissociation of a weak electrolyte to its concentration. Here are the steps to find the degree of dissociation of 0.1 M CH₃COOH given that 0.01 M CH₃COOH has a degree of dissociation of 4.24%.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Degree of dissociation (α₁) for 0.01 M CH₃COOH = 4.24%
- Concentration (C₁) = 0.01 M
- Concentration (C₂) = 0.1 M
- We need to find the degree of dissociation (α₂) for 0.1 M CH₃COOH.

2. **Use the Ostwald's Dilution Law Formula:**
Since the acid is the same, we can use the formula:
\[ \frac{\alpha_1}{\alpha_2} = \sqrt{\frac{C_2}{C_1}} \]

3. **Substitute the Known Values:**
\[ \frac{4.24}{\alpha_2} = \sqrt{\frac{0.1}{0.01}} \]
Simplifying the right side:
\[ \sqrt{\frac{0.1}{0.01}} = \sqrt{10} \approx 3.162 \]

4. **Set Up the Equation:**
\[ \frac{4.24}{\alpha_2} = 3.162 \]

5. **Solve for α₂:**
Rearranging the equation gives:
\[ \alpha_2 = \frac{4.24}{3.162} \]
Performing the calculation:
\[ \alpha_2 \approx 1.34\% \]

6. **Conclusion:**
The degree of dissociation of 0.1 M CH₃COOH is approximately 1.34%.

### Final Answer:
The degree of dissociation of 0.1 M CH₃COOH is **1.34%**.