An aqueous solution freezes at `-0.2^(@)C`. What is the molality of the soluiton ? Determines also (i) elevation in the boiling point (ii) lowering of vapour pressure at 298 K, given that `K_(f)=1.86^(@)"kg mol"^(-1),K_(b)=0.512^(@)" kg mol"^(-1)` and vapour pressure of water at 298 K is 23.756 mm.

Explanation

To solve the problem step by step, we will first calculate the molality of the solution using the freezing point depression formula, then we will determine the elevation in boiling point and the lowering of vapor pressure.

### Step 1: Calculate the Molality of the Solution

1. **Identify the freezing point depression (ΔTf)**:
- The freezing point of pure water is 0°C. The solution freezes at -0.2°C.
- Therefore, ΔTf = 0 - (-0.2) = 0.2°C.

2. **Use the freezing point depression formula**:
\[ \Delta T_f = K_f \cdot m \]
Where:
- ΔTf = 0.2°C
- Kf = 1.86 kg/mol

3. **Rearranging the formula to find molality (m)**:
\[ m = \frac{\Delta T_f}{K_f} = \frac{0.2}{1.86} \]

4. **Calculate molality**:
\[ m \approx 0.1075 \text{ mol/kg} \]

### Step 2: Calculate the Elevation in Boiling Point (ΔTb)

1. **Use the relationship between freezing point depression and boiling point elevation**:
\[ \frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b} \]
Rearranging gives:
\[ \Delta T_b = \Delta T_f \cdot \frac{K_b}{K_f} \]

2. **Substituting the known values**:
- Kb = 0.512 kg/mol
- ΔTf = 0.2°C
\[ \Delta T_b = 0.2 \cdot \frac{0.512}{1.86} \]

3. **Calculate ΔTb**:
\[ \Delta T_b \approx 0.055 \text{°C} \]

### Step 3: Calculate the Lowering of Vapor Pressure (ΔP)

1. **Use Raoult's law**:
\[ \Delta P = P^0 \cdot X_{solute} \]
Where:
- \( P^0 \) = vapor pressure of pure solvent (water) = 23.756 mmHg
- \( X_{solute} \) = mole fraction of solute

2. **Calculate the mole fraction of solute**:
- First, calculate the number of moles of solute (n2) using molality:
\[ n_2 = m \cdot W_{solvent} \text{ (assuming 1 kg of solvent)} \]
\[ n_2 = 0.1075 \text{ moles} \]

3. **Calculate the number of moles of solvent (n1)**:
- For 1 kg of water:
\[ n_1 = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \]

4. **Calculate the mole fraction of solute (X_{solute})**:
\[ X_{solute} = \frac{n_2}{n_1 + n_2} = \frac{0.1075}{55.56 + 0.1075} \approx 0.00193 \]

5. **Calculate ΔP**:
\[ \Delta P = 23.756 \cdot 0.00193 \approx 0.0459 \text{ mmHg} \]

### Final Answers:
- Molality of the solution: **0.1075 mol/kg**
- Elevation in boiling point: **0.055°C**
- Lowering of vapor pressure: **0.0459 mmHg**