Equivalent conductance of 1M `CH_(3)COOH` is `10ohm^(-1) cm^(2) "equiv"^(-1)` and that at inifinite dilution is 200 `ohm^(-1) cm^(2) "equiv"^(-1)`. Hence, % ionisation of `CH_(3)COOJ` is:

Options

A. `5%`
B. `2%`
C. `4%`
D. `1%`

Explanation

To find the percentage ionization of acetic acid (CH₃COOH), we can use the formula for percentage ionization based on equivalent conductance:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Equivalent conductance (Λ) of 1M CH₃COOH = 10 ohm⁻¹ cm² equiv⁻¹
- Equivalent conductance at infinite dilution (Λ₀) = 200 ohm⁻¹ cm² equiv⁻¹

2. **Use the Formula for Percentage Ionization**:
The percentage ionization (α) can be calculated using the formula:
\[ \text{Percentage Ionization} = \left( \frac{\Lambda}{\Lambda_0} \right) \times 100 \]
where:
- Λ = equivalent conductance at a specific concentration (1M in this case)
- Λ₀ = equivalent conductance at infinite dilution

3. **Substitute the Values**:
Substitute the given values into the formula:
\[ \text{Percentage Ionization} = \left( \frac{10}{200} \right) \times 100 \]

4. **Calculate the Value**:
\[ \text{Percentage Ionization} = \left( 0.05 \right) \times 100 = 5\% \]

5. **Conclusion**:
The percentage ionization of CH₃COOH is 5%.

### Final Answer:
The percentage ionization of CH₃COOH is **5%**.

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