A solution of sucross (molar mass = 342 g `"mol"^(-1)`) has been prepared by dissolving 68.4 g of sucrose in one kg of water. `K_(f)` for water is `"1.86 K kg mol"^(-1)` and vapour pressure of water at 298 K is 0.024 atm.
The freezing point of the solution will be

Options

A. `-0.684^(@)C`
B. `-0.342^(@)C`
C. `-0.372^(@)C`
D. `-0.186^(@)C`

Explanation

To find the freezing point of the sucrose solution, we will follow these steps:

### Step 1: Calculate the number of moles of sucrose.
To find the number of moles of sucrose, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]
Given:
- Mass of sucrose = 68.4 g
- Molar mass of sucrose = 342 g/mol

Calculating the number of moles:
\[ \text{Number of moles} = \frac{68.4 \, \text{g}}{342 \, \text{g/mol}} = 0.2 \, \text{mol} \]

### Step 2: Calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality (m)} = \frac{\text{number of moles of solute}}{\text{mass of solvent (kg)}} \]
Given:
- Mass of water (solvent) = 1 kg

Calculating molality:
\[ \text{Molality} = \frac{0.2 \, \text{mol}}{1 \, \text{kg}} = 0.2 \, \text{mol/kg} \]

### Step 3: Calculate the freezing point depression (ΔTf).
The freezing point depression can be calculated using the formula:
\[ \Delta T_f = K_f \times m \]
Where:
- \( K_f \) for water = 1.86 K kg/mol
- \( m \) = 0.2 mol/kg (from Step 2)

Calculating ΔTf:
\[ \Delta T_f = 1.86 \, \text{K kg/mol} \times 0.2 \, \text{mol/kg} = 0.372 \, \text{K} \]

### Step 4: Calculate the freezing point of the solution.
The freezing point of pure water is 0 °C. The freezing point of the solution can be calculated as:
\[ \text{Freezing point of solution} = 0 \, \text{°C} - \Delta T_f \]
Calculating the freezing point:
\[ \text{Freezing point of solution} = 0 \, \text{°C} - 0.372 \, \text{°C} = -0.372 \, \text{°C} \]

### Final Answer:
The freezing point of the solution is **-0.372 °C**.

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