An aqueous solution freezes at `-0.2^(@)C`. What is the molality of the solution ?
Determine also (i) elevation in the boiling point
(ii) lowering in vapour pressure at `25^(@)C`, given that `K_(f)=1.86^(@)C" kg mol"^(-1), K_(b)=0.512^(@)C" kg mol"^(-1)`, and vapour pressure of water at `25^(@)C` 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 calculate the elevation in boiling point and the lowering in vapor pressure.

### Step 1: Calculate the Molality of the Solution
The formula for freezing point depression is given by:

\[ \Delta T_f = K_f \times m \]

Where:
- \(\Delta T_f\) = Freezing point depression
- \(K_f\) = Molar freezing point depression constant
- \(m\) = Molality of the solution

Given:
- \(\Delta T_f = 0.2^\circ C\) (since the freezing point is lowered from \(0^\circ C\) to \(-0.2^\circ C\))
- \(K_f = 1.86^\circ C \, \text{kg/mol}\)

Rearranging the formula to find molality \(m\):

\[ m = \frac{\Delta T_f}{K_f} \]

Substituting the values:

\[ m = \frac{0.2}{1.86} \approx 0.1075 \, \text{mol/kg} \]

### Step 2: Calculate the Elevation in Boiling Point
The formula for elevation in boiling point is given by:

\[ \Delta T_b = K_b \times m \]

Where:
- \(\Delta T_b\) = Elevation in boiling point
- \(K_b\) = Molar elevation of boiling point constant

Given:
- \(K_b = 0.512^\circ C \, \text{kg/mol}\)

Substituting the values:

\[ \Delta T_b = 0.512 \times 0.1075 \approx 0.055 \, ^\circ C \]

### Step 3: Calculate the Lowering in Vapor Pressure
The formula for lowering in vapor pressure is given by:

\[ \Delta P = \frac{\Delta T_f \times M \times P_0}{K_f \times 1000} \]

Where:
- \(\Delta P\) = Lowering in vapor pressure
- \(M\) = Molar mass of water (approximately \(18 \, \text{g/mol}\))
- \(P_0\) = Vapor pressure of pure water at \(25^\circ C\)

Given:
- \(\Delta T_f = 0.2\)
- \(K_f = 1.86\)
- \(P_0 = 23.756 \, \text{mm}\)

Substituting the values:

\[ \Delta P = \frac{0.2 \times 18 \times 23.756}{1.86 \times 1000} \]

Calculating:

\[ \Delta P \approx \frac{0.2 \times 18 \times 23.756}{1860} \approx 0.046 \, \text{mm} \]

### Final Answers
1. **Molality of the solution**: \(0.1075 \, \text{mol/kg}\)
2. **Elevation in boiling point**: \(0.055 \, ^\circ C\)
3. **Lowering in vapor pressure**: \(0.046 \, \text{mm}\)