One molal solution of a carboxylic acid in benzene shows the elevation of boiling point of 1.518 K. The degree of association for simerization of the acid in benzene is (`K_(b)` for beznene = `2.53 K kg mol^(-1)` ):

Options

A. 0.6
B. 0.7
C. 0.75
D. 0.8

Answer

To solve the problem, we need to determine the degree of association (α) of the carboxylic acid in benzene based on the given boiling point elevation.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Elevation of boiling point (ΔTb) = 1.518 K
- ebullioscopic constant (Kb) for benzene = 2.53 K kg mol⁻¹
- Molality (m) of the solution = 1 mol/kg

2. **Use the Formula for Boiling Point Elevation:**
The formula for boiling point elevation is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- ΔTb = boiling point elevation
- i = van 't Hoff factor (degree of dissociation)
- Kb = ebullioscopic constant
- m = molality of the solution

3. **Substituting Known Values:**
Substitute the known values into the equation:
\[ 1.518 = i \cdot 2.53 \cdot 1 \]

4. **Calculate the van 't Hoff Factor (i):**
Rearranging the equation to solve for i:
\[ i = \frac{1.518}{2.53} \approx 0.6 \]

5. **Determine the Degree of Association (α):**
For a carboxylic acid that dimerizes, the association can be represented as:
\[ 2 \text{RCOOH} \rightleftharpoons \text{(RCOOH)}_2 \]
Before association:
- Number of moles = 1 (initially)
- After association:
- Number of moles = \(1 - \alpha + \frac{\alpha}{2}\)

The van 't Hoff factor (i) can be expressed as:
\[ i = 1 - \alpha + \frac{\alpha}{2} \]
Substituting the value of i:
\[ 0.6 = 1 - \alpha + \frac{\alpha}{2} \]

6. **Solve for α:**
Rearranging the equation:
\[ 0.6 = 1 - \alpha + 0.5\alpha \]
\[ 0.6 = 1 - 0.5\alpha \]
\[ 0.5\alpha = 1 - 0.6 \]
\[ 0.5\alpha = 0.4 \]
\[ \alpha = 0.8 \]

7. **Conclusion:**
The degree of association (α) for the carboxylic acid in benzene is 0.8.

### Final Answer:
The degree of association for the carboxylic acid in benzene is **0.8**.