`19.5g` of `CHM_(2)FCOOH` is dissolved in `500g ` of water . The depression in the freezing point of water observed is `1.0^(@)C`. Calculate the Van't Hoff factor and dissociation constant of fluoroacetic acid.

Explanation

To solve the problem step by step, we will calculate the Van't Hoff factor (i) and the dissociation constant (Ka) for fluoroacetic acid (CH₂FCOOH) based on the given data.

### Step 1: Gather the Given Data
- Mass of solute (fluoroacetic acid, CH₂FCOOH), W₂ = 19.5 g
- Mass of solvent (water), W₁ = 500 g
- Depression in freezing point (ΔTf) = 1.0 °C
- Freezing point depression constant for water (Kf) = 1.86 °C kg/mol
- Molecular mass of fluoroacetic acid (CH₂FCOOH) = 14 (C) + 2 (H) + 19 (F) + 12 (C) + 16 (O) = 78 g/mol

### Step 2: Calculate Molality (m)
Molality (m) is calculated using the formula:
\[ m = \frac{\text{mass of solute (g)} \times 1000}{\text{molecular mass of solute (g/mol)} \times \text{mass of solvent (kg)}} \]
Convert the mass of solvent from grams to kilograms:
\[ \text{mass of solvent (kg)} = \frac{500 \text{ g}}{1000} = 0.5 \text{ kg} \]
Now substitute the values:
\[ m = \frac{19.5 \times 1000}{78 \times 0.5} = \frac{19500}{39} = 500 \text{ mol/kg} \]

### Step 3: Use the Freezing Point Depression Formula
The freezing point depression is given by:
\[ \Delta Tf = i \cdot Kf \cdot m \]
Substituting the known values:
\[ 1.0 = i \cdot 1.86 \cdot 500 \]
Now solve for i:
\[ i = \frac{1.0}{1.86 \cdot 500} = \frac{1.0}{930} \approx 1.0753 \]

### Step 4: Calculate the Degree of Dissociation (α)
The Van't Hoff factor (i) is related to the degree of dissociation (α) by the formula:
\[ i = 1 + \alpha \]
Substituting the value of i:
\[ 1.0753 = 1 + \alpha \]
Thus,
\[ \alpha = 1.0753 - 1 = 0.0753 \]

### Step 5: Calculate the Dissociation Constant (Ka)
The dissociation of fluoroacetic acid can be represented as:
\[ \text{CH}_2\text{FCOOH} \rightleftharpoons \text{CH}_2\text{FCOO}^- + \text{H}^+ \]
Let C be the initial concentration of the acid, which is equal to the molality calculated earlier (0.5 mol/kg). The dissociation constant (Ka) is given by:
\[ K_a = C \cdot \alpha^2 \]
Substituting the values:
\[ K_a = 0.5 \cdot (0.0753)^2 = 0.5 \cdot 0.00567 \approx 0.002835 \]
Expressing it in scientific notation:
\[ K_a \approx 2.835 \times 10^{-3} \]

### Final Answers
- Van't Hoff factor (i) ≈ 1.0753
- Dissociation constant (Ka) ≈ 2.835 × 10⁻³