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. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the change in freezing point (ΔTf)
The freezing point depression (ΔTf) can be calculated using the formula:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]
Where:
- \(T_f^{\text{pure}} = 5.3^\circ C\) (freezing point of pure benzene)
- \(T_f^{\text{solution}} = 4.47^\circ C\) (freezing point of the solution)

Calculating ΔTf:
\[ \Delta T_f = 5.3 - 4.47 = 0.83^\circ C \]

### Step 2: Use the freezing point depression formula
The freezing point depression can also be expressed as:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \(K_f = 5.12 \, \text{K kg mol}^{-1}\) (freezing point depression constant for benzene)
- \(m\) is the molality of the solution
- \(i\) is the van 't Hoff factor

### Step 3: Calculate the molality (m)
Molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
First, we need to calculate the moles of phenylacetic acid:
- Molar mass of phenylacetic acid \(C_6H_5CH_2COOH\) = 136 g/mol
- Mass of phenylacetic acid = 0.223 g

Calculating moles of phenylacetic acid:
\[ \text{moles} = \frac{0.223 \, \text{g}}{136 \, \text{g/mol}} \approx 0.00164 \, \text{mol} \]

Now, convert the mass of benzene to kg:
- Mass of benzene = 4.4 g = 0.0044 kg

Now calculate the molality:
\[ m = \frac{0.00164 \, \text{mol}}{0.0044 \, \text{kg}} \approx 0.3727 \, \text{mol/kg} \]

### Step 4: Substitute values into the freezing point depression formula
Now we can substitute the values into the freezing point depression formula:
\[ 0.83 = i \cdot 5.12 \cdot 0.3727 \]

### Step 5: Solve for the van 't Hoff factor (i)
Rearranging the equation to solve for \(i\):
\[ i = \frac{0.83}{5.12 \cdot 0.3727} \approx 0.43 \]

### Step 6: Analyze the van 't Hoff factor (i)
The van 't Hoff factor \(i\) indicates the number of particles the solute breaks into in solution. Since \(i < 1\), this suggests that the phenylacetic acid is dimerizing in benzene.

### Conclusion
From the calculations, we conclude that phenylacetic acid undergoes dimerization in benzene, which is consistent with the value of \(i\) being less than 1.

### Final Answer
The correct conclusion is that phenylacetic acid undergoes dimerization in benzene.

---