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)` ):
(a)0.6
(b)0.7
(c)0.75
(d)0.8

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) for the dimerization of a carboxylic acid in benzene, given the elevation of boiling point and the ebullioscopic constant of benzene.

### Step-by-Step Solution:

1. **Understanding the Dimerization Reaction:**
The dimerization of a carboxylic acid can be represented as:
\[ 2 \text{RCOOH} \rightleftharpoons \text{(RCOOH)}_2 \]
Here, 2 moles of the acid associate to form 1 mole of the dimer.

2. **Initial Concentration:**
Let's assume we start with 1 mole of the carboxylic acid in 1 kg of benzene (1 molal solution). Initially, we have:
- Moles of acid = 1
- Moles of dimer = 0

3. **Change in Concentration due to Association:**
If α is the degree of association, then:
- Moles of acid that dimerize = α
- Moles of acid remaining = \(1 - α\)
- Moles of dimer formed = \(\frac{α}{2}\)

Therefore, the total number of moles after dimerization is:
\[ n_{\text{total}} = (1 - α) + \frac{α}{2} = 1 - \frac{α}{2} \]

4. **Van't Hoff Factor (i):**
The van't Hoff factor (i) is defined as the ratio of the actual number of particles in solution after dissociation or association to the number of formula units initially dissolved. Thus:
\[ i = \frac{n_{\text{total}}}{n_{\text{initial}}} = \frac{1 - \frac{α}{2}}{1} \]
Therefore:
\[ i = 1 - \frac{α}{2} \]

5. **Using the Elevation of Boiling Point:**
The elevation of boiling point (ΔT_b) is given by the formula:
\[ ΔT_b = i \cdot K_b \cdot m \]
Where:
- \(ΔT_b = 1.518 \, \text{K}\)
- \(K_b = 2.53 \, \text{K kg mol}^{-1}\)
- \(m = 1 \, \text{molal}\)

Plugging in the values:
\[ 1.518 = i \cdot 2.53 \cdot 1 \]
Rearranging gives:
\[ i = \frac{1.518}{2.53} \approx 0.6 \]

6. **Finding Degree of Association (α):**
Now we can substitute \(i\) back into the equation:
\[ 0.6 = 1 - \frac{α}{2} \]
Rearranging gives:
\[ \frac{α}{2} = 1 - 0.6 = 0.4 \]
Therefore:
\[ α = 0.8 \]

7. **Conclusion:**
The degree of association (α) for the dimerization of the carboxylic acid in benzene is:
\[ \boxed{0.8} \]