When 9.45 g of `CICH_2COOH` is added to 500 mL of water, its freezing point drops by `0.5^@C`. The dissociation constant of `CICH_2COOH` is `"x"xx10^(-3)`. The value of x is `"_____"`
(Rounded off to the nearest integer)
`[K_(f(H_2O))=1.86" K kg mol"^(-1)]`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the molality of the solution
First, we need to find the number of moles of `CICH_2COOH` (Chloroacetic acid).

1. **Calculate the molar mass of `CICH_2COOH`:**
- Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
- Hydrogen (H): 1.008 g/mol × 4 = 4.032 g/mol
- Chlorine (Cl): 35.45 g/mol × 1 = 35.45 g/mol
- Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol
- Total molar mass = 24.02 + 4.032 + 35.45 + 32.00 = 95.50 g/mol

2. **Calculate the number of moles of `CICH_2COOH`:**
- Moles = mass (g) / molar mass (g/mol) = 9.45 g / 95.50 g/mol = 0.0990 mol

3. **Calculate the molality (m):**
- Molality (m) = moles of solute / mass of solvent (kg)
- Mass of water = 500 mL × 1 g/mL = 500 g = 0.500 kg
- Molality (m) = 0.0990 mol / 0.500 kg = 0.198 mol/kg

### Step 2: Use the freezing point depression formula
The freezing point depression can be calculated using the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
where:
- \(\Delta T_f\) = freezing point depression = 0.5 °C
- \(K_f\) = freezing point depression constant for water = 1.86 K kg mol⁻¹
- \(m\) = molality = 0.198 mol/kg
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)

### Step 3: Rearranging the formula to find \(i\)
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]
Substituting the known values:
\[ i = \frac{0.5 °C}{1.86 K kg mol^{-1} \cdot 0.198 mol/kg} = \frac{0.5}{0.36828} \approx 1.36 \]

### Step 4: Relate \(i\) to the dissociation constant
For a weak acid like `CICH_2COOH`, the dissociation can be represented as:
\[ CICH_2COOH \rightleftharpoons CICH_2O^- + H^+ \]
This means that `CICH_2COOH` dissociates into 2 particles (1 `CICH_2O^-` and 1 `H^+`), so:
\[ i = 1 + \alpha \]
where \(\alpha\) is the degree of dissociation.

Thus, we can write:
\[ 1 + \alpha = 1.36 \implies \alpha = 0.36 \]

### Step 5: Calculate the dissociation constant \(K_a\)
The dissociation constant \(K_a\) can be calculated using the formula:
\[ K_a = \frac{c \cdot \alpha^2}{1 - \alpha} \]
where \(c\) is the initial concentration of the acid in mol/L.

1. **Calculate the concentration \(c\):**
- Concentration \(c\) = moles of solute / volume of solution (L) = 0.0990 mol / 0.5 L = 0.198 mol/L

2. **Substituting values into the \(K_a\) formula:**
\[ K_a = \frac{0.198 \cdot (0.36)^2}{1 - 0.36} = \frac{0.198 \cdot 0.1296}{0.64} \approx \frac{0.0257}{0.64} \approx 0.0402 \]

### Step 6: Convert to the required form
Since the problem asks for \(K_a\) in the form of \(x \times 10^{-3}\):
\[ K_a \approx 0.0402 \approx 40.2 \times 10^{-3} \]
Thus, \(x \approx 40\).

### Final Answer
The value of \(x\) is **40**.

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