`1.22 g` of benzoic acid is dissolved in (i) 100g acetone `(K_(b)` for acetone `=1.7)` and (ii) 100 g benzene `(K_(b)` for benzene `= 2.6)`. The elevation in boiling points `T_(b)` is `0.17^(@)C` and `0.13^(@)C` respectively.
(a) What are the molecular weights of benzoic acid in both the solutions?
(b) What do you deduce out of it in terms of structure of benzoic acid?

Explanation

To solve the problem, we need to calculate the molecular weight of benzoic acid in both solutions (acetone and benzene) using the formula for boiling point elevation. The formula is given by:

\[ \Delta T_b = K_b \times m \times i \]

Where:
- \(\Delta T_b\) is the elevation in boiling point.
- \(K_b\) is the ebullioscopic constant of the solvent.
- \(m\) is the molality of the solution.
- \(i\) is the van 't Hoff factor (which indicates the number of particles the solute breaks into).

### Step-by-step Solution:

**(a) Molecular Weight of Benzoic Acid in Acetone:**

1. **Given Data:**
- Mass of benzoic acid (solute) = 1.22 g
- Mass of acetone (solvent) = 100 g = 0.1 kg
- \(K_b\) for acetone = 1.7 °C kg/mol
- \(\Delta T_b\) for acetone = 0.17 °C

2. **Using the boiling point elevation formula:**
\[ \Delta T_b = K_b \times m \times i \]
Rearranging gives:
\[ m = \frac{\Delta T_b}{K_b \times i} \]

3. **Assuming benzoic acid does not dissociate in acetone, \(i = 1\):**
\[ m = \frac{0.17}{1.7 \times 1} = 0.1 \text{ mol/kg} \]

4. **Calculating moles of benzoic acid:**
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \implies \text{moles of solute} = m \times \text{mass of solvent in kg} \]
\[ \text{moles of solute} = 0.1 \times 0.1 = 0.01 \text{ moles} \]

5. **Calculating molecular weight of benzoic acid:**
\[ \text{Molecular Weight} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1.22 \text{ g}}{0.01 \text{ moles}} = 122 \text{ g/mol} \]

**(b) Molecular Weight of Benzoic Acid in Benzene:**

1. **Given Data:**
- Mass of benzene (solvent) = 100 g = 0.1 kg
- \(K_b\) for benzene = 2.6 °C kg/mol
- \(\Delta T_b\) for benzene = 0.13 °C

2. **Using the boiling point elevation formula:**
\[ m = \frac{\Delta T_b}{K_b \times i} \]

3. **Assuming benzoic acid dimerizes in benzene, \(i = 2\):**
\[ m = \frac{0.13}{2.6 \times 2} = \frac{0.13}{5.2} = 0.025 \text{ mol/kg} \]

4. **Calculating moles of benzoic acid:**
\[ \text{moles of solute} = m \times \text{mass of solvent in kg} = 0.025 \times 0.1 = 0.0025 \text{ moles} \]

5. **Calculating molecular weight of benzoic acid:**
\[ \text{Molecular Weight} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1.22 \text{ g}}{0.0025 \text{ moles}} = 488 \text{ g/mol} \]

### Summary of Results:
- Molecular weight of benzoic acid in acetone = **122 g/mol**
- Molecular weight of benzoic acid in benzene = **488 g/mol**

### Conclusion:
The significant difference in molecular weight indicates that benzoic acid dimerizes in benzene (as shown by the higher molecular weight), while it remains as a monomer in acetone.