The equivalent conductance of `0.001` M acetic acid is `50 " ohm"^(-1)cm^(2)eq^(-1)`. The maximum value of equivalent conductance is 250 `"ohm"^(-1) cm^(2) eq^(-1)`.What is its degree of ionization?

Options

A. `0.2%`
B. `2%`
C. `20%`
D. `40%`

Answer

To find the degree of ionization (α) of acetic acid, we can use the formula:

\[ \alpha = \frac{\Lambda}{\Lambda_0} \]

where:
- \(\Lambda\) is the equivalent conductance of the solution at a given concentration (in this case, 0.001 M).
- \(\Lambda_0\) is the maximum value of equivalent conductance (at infinite dilution).

Given data:
- \(\Lambda = 50 \, \Omega^{-1} \, cm^2 \, eq^{-1}\)
- \(\Lambda_0 = 250 \, \Omega^{-1} \, cm^2 \, eq^{-1}\)

### Step-by-step Solution:

1. **Write the formula for degree of ionization**:
\[ \alpha = \frac{\Lambda}{\Lambda_0} \]

2. **Substitute the values into the formula**:
\[ \alpha = \frac{50 \, \Omega^{-1} \, cm^2 \, eq^{-1}}{250 \, \Omega^{-1} \, cm^2 \, eq^{-1}} \]

3. **Calculate the value of α**:
\[ \alpha = \frac{50}{250} = \frac{1}{5} = 0.2 \]

4. **Convert α into percentage**:
\[ \text{Degree of ionization (as a percentage)} = \alpha \times 100 = 0.2 \times 100 = 20\% \]

5. **Conclusion**:
The degree of ionization of acetic acid is 20%.

### Final Answer:
The degree of ionization of acetic acid is 20%.

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