3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated for dissociated).
`("Given : Molar mass of benzoic acid = 122 g mol"^(-1),K_(f)" for benzene = 4.9 K kg mol"^(-1))`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. **Calculate the number of moles of benzoic acid:**
\[ \text{Moles of benzoic acid} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{3.9 \, \text{g}}{122 \, \text{g/mol}} = 0.03196 \, \text{mol} \]

2. **Convert the mass of benzene to kilograms:**
\[ \text{Mass of benzene} = 49 \, \text{g} = 0.049 \, \text{kg} \]

3. **Calculate the molality:**
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.03196 \, \text{mol}}{0.049 \, \text{kg}} = 0.651 \, \text{mol/kg} \]

### Step 2: Use the freezing point depression formula
The freezing point depression (\(\Delta T_f\)) is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \(\Delta T_f = 1.62 \, \text{K}\)
- \(K_f = 4.9 \, \text{K kg/mol}\)
- \(m = 0.651 \, \text{mol/kg}\)

### Step 3: Solve for the van't Hoff factor (i)
Rearranging the formula to find \(i\):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]
Substituting the values:
\[ i = \frac{1.62 \, \text{K}}{4.9 \, \text{K kg/mol} \cdot 0.651 \, \text{mol/kg}} = \frac{1.62}{3.1949} \approx 0.507 \]

### Step 4: Determine the nature of the solute
- If \(i > 1\), the solute dissociates.
- If \(i < 1\), the solute associates.
- If \(i = 1\), there is no association or dissociation.

Since \(i = 0.507 < 1\), we conclude that the solute (benzoic acid) is associated.

### Final Answer
- The van't Hoff factor \(i\) is approximately **0.507**.
- The nature of the solute is **associated**.

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