Pure benzene freezes t `5.3^(@)C`. A solution of 0.223 g of phenylacetic acid `(C_(6)H_(5)CH_(2)COOH)` in 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 conclude whether it undergoes dimerization based on the calculated value. Let's break down the solution step by step.

### Step 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
- Cryoscopic constant (K_f) for benzene = 5.12 K kg mol⁻¹

### Step 2: Calculate the depression in freezing point (ΔT_f)
\[ ΔT_f = T₀ - T_f = 5.3 °C - 4.47 °C = 0.83 °C \]

### Step 3: Calculate the number of moles of phenylacetic acid
First, we need to find the molar mass of phenylacetic acid (C₆H₅CH₂COOH):
- C: 6 × 12.01 g/mol = 72.06 g/mol
- H: 8 × 1.008 g/mol = 8.064 g/mol
- O: 2 × 16.00 g/mol = 32.00 g/mol

Total molar mass = 72.06 + 8.064 + 32.00 = 112.124 g/mol (corrected from the transcript)

Now, calculate the number of moles of phenylacetic acid:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.223 \text{ g}}{112.124 \text{ g/mol}} \approx 0.001989 \text{ mol} \]

### Step 4: Calculate the molality of the solution
Convert the mass of benzene from grams to kilograms:
\[ \text{Mass of benzene in kg} = \frac{4.4 \text{ g}}{1000} = 0.0044 \text{ kg} \]

Now, calculate the molality (m):
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.001989 \text{ mol}}{0.0044 \text{ kg}} \approx 0.451 \text{ mol/kg} \]

### Step 5: Use the freezing point depression formula
The freezing point depression can be expressed as:
\[ ΔT_f = i \cdot K_f \cdot m \]
Substituting the known values:
\[ 0.83 °C = i \cdot 5.12 \text{ K kg mol}^{-1} \cdot 0.451 \text{ mol/kg} \]

### Step 6: Solve for the van't Hoff factor (i)
Rearranging the equation to solve for i:
\[ i = \frac{0.83 °C}{5.12 \cdot 0.451} \approx 0.5 \]

### Step 7: Conclusion about the behavior of phenylacetic acid
The van't Hoff factor (i) is 0.5, which indicates that the number of particles in solution is less than expected. This suggests that phenylacetic acid is undergoing dimerization, where two molecules combine to form one particle in solution.

### Final Answer
Phenylacetic acid dimerizes in benzene.

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