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

Options

A. `0.5`
B. `2.0`
C. `1.0`
D. `3.0`

Answer

To solve the problem, we need to find the van't Hoff factor (i) for naphthenoic acid when it is dissolved in benzene. We will use the formula for freezing point depression.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of naphthenoic acid (C₁₁H₈O₂), \( W_a = 20 \, \text{g} \)
- Mass of benzene (solvent), \( W_b = 50 \, \text{g} \)
- Freezing point depression, \( \Delta T_f = 2 \, \text{K} \)
- Freezing point depression constant for benzene, \( K_f = 1.72 \, \text{K kg/mol} \)

2. **Calculate the Molecular Weight of Naphthenoic Acid:**
- The molecular formula is C₁₁H₈O₂.
- Calculate the molecular weight (M):
\[ M = (11 \times 12) + (8 \times 1) + (2 \times 16) = 132 + 8 + 32 = 172 \, \text{g/mol} \]

3. **Calculate the Number of Moles of Naphthenoic Acid:**
- Use the formula:
\[ \text{Number of moles} = \frac{W_a}{M} = \frac{20 \, \text{g}}{172 \, \text{g/mol}} \approx 0.1163 \, \text{mol} \]

4. **Convert the Mass of Benzene to Kilograms:**
- Convert grams to kilograms:
\[ W_b = 50 \, \text{g} = 0.050 \, \text{kg} \]

5. **Calculate the Molality (m):**
- Molality is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{\text{Number of moles}}{W_b} = \frac{0.1163 \, \text{mol}}{0.050 \, \text{kg}} \approx 2.326 \, \text{mol/kg} \]

6. **Use the Freezing Point Depression Formula:**
- The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
- Rearranging to find the van't Hoff factor (i):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

7. **Substituting the Values:**
- Substitute the known values into the equation:
\[ i = \frac{2 \, \text{K}}{1.72 \, \text{K kg/mol} \cdot 2.326 \, \text{mol/kg}} \approx \frac{2}{4.003} \approx 0.499 \]

8. **Final Result:**
- Rounding off, we find:
\[ i \approx 0.5 \]

### Conclusion:
The van't Hoff factor \( i \) for naphthenoic acid in benzene is approximately **0.5**.