`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, we will follow the steps outlined in the video transcript. We will calculate the Van't Hoff factor (i) and the dissociation constant (Ka) of fluoroacetic acid (CH2FCOOH) step by step.

### Step 1: Determine the Molecular Weight of CH2FCOOH
The molecular weight of CH2FCOOH can be calculated as follows:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Fluorine (F): 19 g/mol × 1 = 19 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol

Adding these together:
\[ \text{Molecular Weight} = 24 + 2 + 19 + 32 = 77 \text{ g/mol} \]

### Step 2: Calculate the Molality of the Solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. First, we need to calculate the number of moles of CH2FCOOH:
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molecular weight (g/mol)}} = \frac{19.5 \text{ g}}{77 \text{ g/mol}} \approx 0.253 \text{ moles} \]

Now, we convert the mass of water from grams to kilograms:
\[ \text{Mass of water} = 500 \text{ g} = 0.5 \text{ kg} \]

Now we can calculate the molality:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent (kg)}} = \frac{0.253 \text{ moles}}{0.5 \text{ kg}} = 0.506 \text{ mol/kg} \]

### Step 3: Use the Freezing Point Depression Formula
The freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \( \Delta T_f = 1.0^\circ C \)
- \( K_f = 1.86 \text{ °C kg/mol} \)
- \( m = 0.506 \text{ mol/kg} \)

Substituting the known values into the formula:
\[ 1.0 = i \cdot 1.86 \cdot 0.506 \]

Now we can solve for \( i \):
\[ i = \frac{1.0}{1.86 \cdot 0.506} \approx \frac{1.0}{0.94236} \approx 1.061 \]

### Step 4: Calculate the Degree of Dissociation (α)
The Van't Hoff factor (i) is related to the degree of dissociation (α) as follows:
\[ i = 1 + \alpha \]

From our calculation:
\[ 1.061 = 1 + \alpha \]
\[ \alpha = 1.061 - 1 = 0.061 \]

### Step 5: Calculate the Dissociation Constant (Ka)
The dissociation of fluoroacetic acid can be represented as:
\[ \text{CH}_2\text{FCOOH} \rightleftharpoons \text{CH}_2\text{FCOO}^- + \text{H}^+ \]

The expression for the dissociation constant (Ka) is:
\[ K_a = C \cdot \alpha^2 \]

Where C is the molality of the solution, which we calculated as 0.506 mol/kg.
Now substituting the values:
\[ K_a = 0.506 \cdot (0.061)^2 \]
\[ K_a = 0.506 \cdot 0.003721 \approx 0.00188 \]

### Final Results
- Van't Hoff factor (i) ≈ 1.061
- Dissociation constant (Ka) ≈ 0.00188