When `20g` of naphtholic acid `(C_(11)H_(8)O_(2))` is dissolved in `50g` of benzene `(K_(f) = 1.72 K kg mol^(-1))` a freezing point depression of `2K` is observed. The van'f 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: Identify the given data
- Mass of naphtholic acid (C₁₁H₈O₂) = 20 g
- Mass of benzene = 50 g
- Freezing point depression (ΔTf) = 2 K
- Freezing point depression constant for benzene (Kf) = 1.72 K kg mol⁻¹

### Step 2: Calculate the molar mass of naphtholic acid
The molar mass of naphtholic acid (C₁₁H₈O₂) can be calculated as follows:
- Carbon (C): 12.01 g/mol × 11 = 132.11 g/mol
- Hydrogen (H): 1.008 g/mol × 8 = 8.064 g/mol
- Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol

Total molar mass = 132.11 + 8.064 + 32.00 = 172.174 g/mol (approximately 172 g/mol)

### Step 3: Calculate the number of moles of naphtholic acid
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Number of moles of naphtholic acid} = \frac{20 \text{ g}}{172 \text{ g/mol}} \approx 0.1163 \text{ mol} \]

### Step 4: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} (m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
\[ \text{Mass of benzene in kg} = \frac{50 \text{ g}}{1000} = 0.050 \text{ kg} \]
\[ m = \frac{0.1163 \text{ mol}}{0.050 \text{ kg}} \approx 2.326 \text{ mol/kg} \]

### Step 5: Use the freezing point depression formula
The freezing point depression is given by:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Rearranging to find the van 't Hoff factor (i):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]
Substituting the known values:
\[ i = \frac{2 \text{ K}}{1.72 \text{ K kg mol}^{-1} \cdot 2.326 \text{ mol/kg}} \]
Calculating the denominator:
\[ 1.72 \cdot 2.326 \approx 4.000 \]
Now substituting back:
\[ i = \frac{2}{4.000} = 0.5 \]

### Final Answer
The van 't Hoff factor (i) is **0.5**.

---