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: Calculate the Molecular Weight of Naphtholic Acid
The molecular formula of naphtholic acid is \( C_{11}H_{8}O_{2} \). We can calculate its molecular weight as follows:

\[ \text{Molecular Weight} = (11 \times 12) + (8 \times 1) + (2 \times 16) \]

Calculating this gives:

\[ \text{Molecular Weight} = 132 + 8 + 32 = 172 \, \text{g/mol} \]

### Step 2: Calculate the Number of Moles of Naphtholic Acid
Using the formula for moles:

\[ \text{Number of Moles} = \frac{\text{Weight of solute}}{\text{Molecular Weight}} \]

Substituting the values:

\[ \text{Number of Moles} = \frac{20 \, \text{g}}{172 \, \text{g/mol}} \approx 0.1163 \, \text{mol} \]

### Step 3: Use the Freezing Point Depression Formula
The freezing point depression (\( \Delta T_f \)) is given by the formula:

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

Where:
- \( \Delta T_f = 2 \, \text{K} \)
- \( K_f = 1.72 \, \text{K kg/mol} \)
- \( m = \text{molality} = \frac{\text{Number of Moles}}{\text{Weight of solvent in kg}} \)

First, we need to convert the weight of the solvent (benzene) from grams to kilograms:

\[ \text{Weight of solvent} = 50 \, \text{g} = 0.050 \, \text{kg} \]

Now, we can calculate the molality:

\[ m = \frac{0.1163 \, \text{mol}}{0.050 \, \text{kg}} = 2.326 \, \text{mol/kg} \]

### Step 4: Substitute Values into the Freezing Point Depression Formula
Now, we can substitute the values into the freezing point depression formula to solve for \( i \):

\[ 2 = 1.72 \cdot (2.326) \cdot i \]

Calculating \( 1.72 \cdot 2.326 \):

\[ 1.72 \cdot 2.326 \approx 4.003 \]

Now we can solve for \( i \):

\[ 2 = 4.003 \cdot i \implies i = \frac{2}{4.003} \approx 0.499 \]

### Final Answer
Thus, the van 't Hoff factor \( i \) is approximately \( 0.5 \).

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