To `500cm^(3)` of water, `3.0 xx 10^(-3) kg` acetic acid is added. If `23%` of acetic acid is dissociated, what will be the depression in freezing point? `K_(f)` and density of water are `1.86 K kg mol^(-1)` and `0.997 g cm^(-3)` respectively.

Explanation

To solve the problem step by step, we will calculate the depression in the freezing point of the solution formed by dissolving acetic acid in water.

### Step 1: Convert the volume of water to liters
Given that the volume of water is \(500 \, \text{cm}^3\):
\[ \text{Volume in liters} = \frac{500 \, \text{cm}^3}{1000} = 0.5 \, \text{L} \]

**Hint:** Remember that \(1 \, \text{L} = 1000 \, \text{cm}^3\).

### Step 2: Calculate the mass of acetic acid
The mass of acetic acid added is given as \(3.0 \times 10^{-3} \, \text{kg}\):
\[ \text{Mass of acetic acid} = 3.0 \times 10^{-3} \, \text{kg} = 3.0 \, \text{g} \]

**Hint:** Convert kilograms to grams by multiplying by \(1000\).

### Step 3: Calculate the number of moles of acetic acid
The molecular weight of acetic acid (\(CH_3COOH\)) is calculated as follows:
\[ \text{Molecular weight} = 12 + 4 + 32 = 60 \, \text{g/mol} \]
Now, calculate the number of moles:
\[ \text{Moles of acetic acid} = \frac{\text{mass}}{\text{molecular weight}} = \frac{3.0 \, \text{g}}{60 \, \text{g/mol}} = 0.05 \, \text{mol} \]

**Hint:** Use the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).

### Step 4: Calculate the mass of the solvent (water)
The mass of the solvent (water) can be calculated using the density of water:
\[ \text{Density of water} = 0.997 \, \text{g/cm}^3 \]
\[ \text{Mass of water} = \text{Volume} \times \text{Density} = 500 \, \text{cm}^3 \times 0.997 \, \text{g/cm}^3 = 498.5 \, \text{g} \]
Now, calculate the mass of the solvent:
\[ \text{Mass of solvent} = 498.5 \, \text{g} - 3.0 \, \text{g} = 495.5 \, \text{g} = 0.4955 \, \text{kg} \]

**Hint:** Remember to subtract the mass of the solute from the total mass of the solution.

### Step 5: Calculate the molality of the solution
Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05 \, \text{mol}}{0.4955 \, \text{kg}} \approx 0.101 \, \text{mol/kg} \]

**Hint:** Molality is calculated using the formula \(m = \frac{n}{W}\), where \(n\) is the number of moles and \(W\) is the mass of the solvent in kg.

### Step 6: Calculate the van 't Hoff factor (\(i\))
Given that \(23\%\) of acetic acid is dissociated, we can find the van 't Hoff factor:
\[ \alpha = 0.23 \quad (\text{fraction dissociated}) \]
\[ i = 1 + \alpha = 1 + 0.23 = 1.23 \]

**Hint:** The van 't Hoff factor accounts for the number of particles the solute dissociates into.

### Step 7: Calculate the depression in freezing point (\(\Delta T_f\))
Using the formula for depression in freezing point:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Substituting the values:
\[ \Delta T_f = 1.23 \cdot 1.86 \, \text{K kg/mol} \cdot 0.101 \, \text{mol/kg} \approx 0.228 \, \text{K} \]

**Hint:** Use the formula \(\Delta T_f = i \cdot K_f \cdot m\) to find the freezing point depression.

### Final Answer
The depression in freezing point is approximately \(0.228 \, \text{K}\).