From a measurement of the freezing point depression of benzene, the molecular weight of acetic acid in a benzene solution was determined to be `100`. The percentage association of acetic acid is

Options

A. `79%`
B. `93%`
C. `80%`
D. `100%`

Explanation

To solve the problem of determining the percentage association of acetic acid in a benzene solution based on the freezing point depression, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Normal Molar Mass of Acetic Acid**:
- The molecular formula of acetic acid is \( \text{CH}_3\text{COOH} \).
- Calculate the molar mass:
- Carbon (C): \( 12 \, \text{g/mol} \times 2 = 24 \, \text{g/mol} \)
- Hydrogen (H): \( 1 \, \text{g/mol} \times 4 = 4 \, \text{g/mol} \)
- Oxygen (O): \( 16 \, \text{g/mol} \times 2 = 32 \, \text{g/mol} \)
- Total molar mass = \( 24 + 4 + 32 = 60 \, \text{g/mol} \).

2. **Determine the Van 't Hoff Factor (i)**:
- The observed molar mass of acetic acid in the solution is given as \( 100 \, \text{g/mol} \).
- The Van 't Hoff factor \( i \) can be calculated using the formula:
\[ i = \frac{\text{Normal Molar Mass}}{\text{Observed Molar Mass}} = \frac{60}{100} = 0.6 \]

3. **Relate the Van 't Hoff Factor to Association**:
- The Van 't Hoff factor \( i \) indicates the degree of association. When \( i < 1 \), it suggests that the solute is associating.
- The relationship between \( i \) and the degree of association \( \alpha \) is given by:
\[ i = 1 - \frac{\alpha}{2} \]

4. **Solve for the Degree of Association \( \alpha \)**:
- Substitute \( i = 0.6 \) into the equation:
\[ 0.6 = 1 - \frac{\alpha}{2} \]
- Rearranging gives:
\[ \frac{\alpha}{2} = 1 - 0.6 = 0.4 \]
- Therefore, multiplying both sides by 2:
\[ \alpha = 0.8 \]

5. **Calculate the Percentage Association**:
- To find the percentage association, multiply \( \alpha \) by 100:
\[ \text{Percentage Association} = \alpha \times 100 = 0.8 \times 100 = 80\% \]

### Final Answer:
The percentage association of acetic acid in the benzene solution is **80%**.