Pure benzene freezes t `5.3^(@)C`. A solution of 0.223 g of phenylacetic acid `(C_(6)H_(5)CH_(2)COOH)` I 4.4 g of benzene ` (K_(f) = 5.12 Kkg mol^(-1))` freezes at `4.47^(@)C`. From the observation one can conclude that :

Options

A. phenylacetic acid exists as such in benzene
B. phenylacetic acid undergoes partial ionization in benzene
C. phenylacetic acid undergoes complete ionization in benzene
D. phenylacetic acid dimerises in benzene

Answer

To solve the problem, we need to determine the van 't Hoff factor (i) for phenylacetic acid in benzene and analyze the results to conclude whether phenylacetic acid dimerizes in the solution.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Freezing point of pure benzene (T₀) = 5.3 °C
- Freezing point of the solution (T_f) = 4.47 °C
- Mass of phenylacetic acid (solute) = 0.223 g
- Mass of benzene (solvent) = 4.4 g
- Freezing point depression constant (K_f) for benzene = 5.12 K kg mol⁻¹

2. **Calculate the Freezing Point Depression (ΔT_f):**
\[ \Delta T_f = T₀ - T_f = 5.3 °C - 4.47 °C = 0.83 °C \]

3. **Calculate the Molar Mass of Phenylacetic Acid (C₆H₅CH₂COOH):**
- Carbon (C): 6 × 12.01 g/mol = 72.06 g/mol
- Hydrogen (H): 8 × 1.008 g/mol = 8.064 g/mol
- Oxygen (O): 2 × 16.00 g/mol = 32.00 g/mol
- Total Molar Mass = 72.06 + 8.064 + 32.00 = 112.124 g/mol (correcting from the video)

4. **Calculate the Number of Moles of Phenylacetic Acid:**
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.223 \text{ g}}{112.124 \text{ g/mol}} \approx 0.001989 \text{ mol} \]

5. **Calculate the Mass of Benzene in kg:**
\[ \text{Mass of benzene in kg} = \frac{4.4 \text{ g}}{1000} = 0.0044 \text{ kg} \]

6. **Calculate the Molality (m) of the Solution:**
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} = \frac{0.001989 \text{ mol}}{0.0044 \text{ kg}} \approx 0.451 \text{ mol/kg} \]

7. **Use the Freezing Point Depression Formula:**
\[ \Delta T_f = i \cdot K_f \cdot m \]
Rearranging for i:
\[ i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.83 °C}{5.12 \text{ K kg mol}^{-1} \cdot 0.451 \text{ mol/kg}} \approx 0.36 \]

8. **Interpret the Van 't Hoff Factor (i):**
- The calculated van 't Hoff factor (i ≈ 0.36) is significantly less than 1, indicating that the solute molecules are associating in the solution rather than dissociating into ions.
- This suggests that phenylacetic acid is undergoing dimerization, where two molecules combine to form a single entity.

9. **Conclusion:**
- Since the van 't Hoff factor is less than expected (which would be 1 for a non-dissociating solute), we conclude that phenylacetic acid dimerizes in benzene.

### Final Answer:
Phenylacetic acid dimerizes in benzene.