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

Options

A. `0.2%`
B. `2%`
C. `20%`
D. `22%`

Answer

To find the degree of ionization of acetic acid, we can use the relationship between molar conductance at a given concentration and the molar conductance at infinite dilution. The degree of ionization (α) can be calculated using the formula:

\[ \alpha = \frac{\Lambda_m}{\Lambda_{m}^{\infty}} \]

Where:
- \(\Lambda_m\) is the molar conductance at the given concentration (0.001 M in this case).
- \(\Lambda_{m}^{\infty}\) is the molar conductance at infinite dilution (maximum value).

### Step-by-Step Solution:

1. **Identify the given values**:
- Molar conductance at 0.001 M acetic acid, \(\Lambda_m = 50 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)
- Maximum molar conductance, \(\Lambda_{m}^{\infty} = 250 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)

2. **Substitute the values into the formula**:
\[ \alpha = \frac{50}{250} \]

3. **Calculate the degree of ionization**:
\[ \alpha = \frac{50}{250} = 0.2 \]

4. **Convert to percentage**:
\[ \alpha \, (\text{in percentage}) = 0.2 \times 100 = 20\% \]

Thus, the degree of ionization of acetic acid is **20%**.

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