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 find the Van't Hoff factor (i) for naphthoic acid when it is dissolved in benzene, we can use the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot m \cdot i \]

Where:
- \(\Delta T_f\) = freezing point depression
- \(K_f\) = cryoscopic constant of the solvent (benzene in this case)
- \(m\) = molality of the solution
- \(i\) = Van't Hoff factor

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

1. **Find the number of moles of naphthoic acid (C₁₁H₈O₂)**:
- Molar mass of naphthoic acid (C₁₁H₈O₂) = 11(12.01) + 8(1.008) + 2(16.00) = 172.18 g/mol
- Mass of naphthoic acid = 20 g
- Moles of naphthoic acid = \(\frac{20 \text{ g}}{172.18 \text{ g/mol}} = 0.116 \text{ mol}\)

2. **Convert the mass of benzene to kilograms**:
- Mass of benzene = 50 g = 0.050 kg

3. **Calculate molality (m)**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.116 \text{ mol}}{0.050 \text{ kg}} = 2.32 \text{ mol/kg} \]

### Step 2: Use the freezing point depression formula
Now we can plug the values into the freezing point depression formula:

\[ \Delta T_f = K_f \cdot m \cdot i \]

Given:
- \(\Delta T_f = 2 \text{ K}\)
- \(K_f = 1.72 \text{ K kg mol}^{-1}\)
- \(m = 2.32 \text{ mol/kg}\)

Substituting the values:

\[ 2 = 1.72 \cdot 2.32 \cdot i \]

### Step 3: Solve for the Van't Hoff factor (i)
Rearranging the equation to solve for \(i\):

\[ i = \frac{2}{1.72 \cdot 2.32} \]

Calculating the denominator:

\[ 1.72 \cdot 2.32 = 3.9984 \]

Now substituting back:

\[ i = \frac{2}{3.9984} \approx 0.5004 \]

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