A solution of sucrose (molar mass =342) is prepared by dissolving 68.4 g in 1000 g of water. Calculate
The freezing point of solution.

Options

A. `273`
B. `373`
C. `272.628`
D. `271.628`

Explanation

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

### Step 1: Calculate the number of moles of sucrose.
To find the number of moles (n) of sucrose, we use the formula:

\[ n = \frac{\text{mass of sucrose (g)}}{\text{molar mass of sucrose (g/mol)}} \]

Given:
- Mass of sucrose = 68.4 g
- Molar mass of sucrose = 342 g/mol

\[ n = \frac{68.4 \, \text{g}}{342 \, \text{g/mol}} = 0.2 \, \text{moles} \]

### Step 2: Calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent:

\[ m = \frac{n}{\text{mass of solvent (kg)}} \]

Given:
- Mass of water (solvent) = 1000 g = 1 kg

\[ m = \frac{0.2 \, \text{moles}}{1 \, \text{kg}} = 0.2 \, \text{mol/kg} \]

### Step 3: Use the freezing point depression formula.
The freezing point depression (\(\Delta T_f\)) can be calculated using the formula:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(i\) = van 't Hoff factor (for sucrose, a non-electrolyte, \(i = 1\))
- \(K_f\) = freezing point depression constant for water = 1.86 °C kg/mol
- \(m\) = molality of the solution

Substituting the values:

\[ \Delta T_f = 1 \cdot 1.86 \, \text{°C kg/mol} \cdot 0.2 \, \text{mol/kg} = 0.372 \, \text{°C} \]

### Step 4: Calculate the freezing point of the solution.
The freezing point of pure water is 0 °C. Therefore, the freezing point of the solution (\(T_f\)) can be calculated as:

\[ T_f = 0 \, \text{°C} - \Delta T_f \]

Substituting the value of \(\Delta T_f\):

\[ T_f = 0 \, \text{°C} - 0.372 \, \text{°C} = -0.372 \, \text{°C} \]

### Final Answer:
The freezing point of the sucrose solution is approximately \(-0.372 \, \text{°C}\).

---