When `20 g` of naphthoic acid `(C_(11)H_(8)O_(2))` is dissolved in `50 g` of benzene `(K_(f)=1.72 K kg mol^(-1))`, a freezing point depression of `2 K` is observed. The Van't Hoff factor (i) is

Options

A. 0.5
B. 1
C. 2
D. 3

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Molecular Weight of Naphthoic Acid
Naphthoic acid has the formula \( C_{11}H_{8}O_{2} \). To find its molecular weight, we sum the atomic weights of all the atoms in the formula.

- Carbon (C): 12 g/mol × 11 = 132 g/mol
- Hydrogen (H): 1 g/mol × 8 = 8 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol

**Total Molecular Weight:**
\[ M = 132 + 8 + 32 = 172 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of Naphthoic Acid
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{20 \text{ g}}{172 \text{ g/mol}} \approx 0.1163 \text{ mol} \]

### Step 3: Calculate the Molality of the Solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The mass of benzene is given as 50 g, which is 0.050 kg.

\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Substituting the values:
\[ \text{Molality} = \frac{0.1163 \text{ mol}}{0.050 \text{ kg}} \approx 2.326 \text{ mol/kg} \]

### Step 4: Use the Freezing Point Depression Formula
The freezing point depression (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f = 2 \text{ K} \)
- \( K_f = 1.72 \text{ K kg mol}^{-1} \)
- \( m \) is the molality calculated in the previous step.

Rearranging the formula to find \( i \):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]
Substituting the values:
\[ i = \frac{2 \text{ K}}{1.72 \text{ K kg mol}^{-1} \cdot 2.326 \text{ mol/kg}} \approx \frac{2}{4.003} \approx 0.499 \]

### Final Answer
The Van't Hoff factor \( i \) is approximately \( 0.5 \).

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