A solute S undergoes a reversible trimerization when dissolved in a certain solvent . The boiling point elevation of its `0.1` molal solution was found to be identical to be boiling point elevation in case of a `0.08` molal solution of a solute which neither undergoes association nor dissociation. To what percent had the solute S undergoes trimerization ?

Options

A. 30%
B. 40%
C. 50%
D. 60%

Explanation

To solve the problem, we need to analyze the boiling point elevation of two solutions: one with a solute S that undergoes trimerization and another with a solute that does not undergo any association or dissociation.

### Step-by-Step Solution:

1. **Understanding Boiling Point Elevation**:
The boiling point elevation (\( \Delta T_b \)) is given by the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( i \) = van't Hoff factor (number of particles the solute breaks into)
- \( K_b \) = ebullioscopic constant of the solvent
- \( m \) = molality of the solution

2. **Setting Up the Equation**:
We know that the boiling point elevation for both solutions is the same:
\[ \Delta T_{b1} = \Delta T_{b2} \]
Therefore, we can write:
\[ i_1 \cdot K_b \cdot m_1 = i_2 \cdot K_b \cdot m_2 \]
Since \( K_b \) is the same for both solutions, it cancels out:
\[ i_1 \cdot m_1 = i_2 \cdot m_2 \]

3. **Substituting Known Values**:
For the first solution (solute S):
- \( m_1 = 0.1 \) molal
- \( i_1 = 1 + \frac{1}{n-1} \cdot \alpha \) (where \( n = 3 \) for trimerization)

For the second solution (non-associating solute):
- \( m_2 = 0.08 \) molal
- \( i_2 = 1 \) (since it does not dissociate or associate)

Plugging these into the equation gives:
\[ \left(1 + \frac{1}{3-1} \cdot \alpha\right) \cdot 0.1 = 1 \cdot 0.08 \]

4. **Solving for Alpha**:
Simplifying the equation:
\[ \left(1 + \frac{1}{2} \cdot \alpha\right) \cdot 0.1 = 0.08 \]
Expanding this:
\[ 0.1 + 0.05\alpha = 0.08 \]
Rearranging gives:
\[ 0.05\alpha = 0.08 - 0.1 \]
\[ 0.05\alpha = -0.02 \]
\[ \alpha = \frac{-0.02}{0.05} = -0.4 \]
Since \( \alpha \) represents the fraction of solute that has trimerized, we need to express it as a percentage:
\[ \alpha = 0.4 \text{ or } 40\% \]

5. **Final Answer**:
Therefore, the percentage of solute S that underwent trimerization is:
\[ \text{Percentage of trimerization} = 40\% \]