A solution of sucrose (molar mass `=342 g mol^(-1)`) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (`K_(f)` for water = `1.86K kg mol^(-1)`)

Options

A. `-0.372^(@)C`
B. `-0.520^(@)C`
C. `+0.372^(@)C`
D. `-0.570^(@)C`

Explanation

To solve the problem of finding 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 = 68.5 g
- Molar mass of sucrose = 342 g/mol

\[ \text{Number of moles of sucrose} = \frac{68.5 \, \text{g}}{342 \, \text{g/mol}} \approx 0.200 \, \text{mol} \]

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

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

\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{mass of solvent (kg)}} = \frac{0.200 \, \text{mol}}{1 \, \text{kg}} = 0.200 \, \text{mol/kg} \]

### Step 3: Calculate the depression in freezing point (ΔTf)
The depression in freezing point can be calculated using the formula:

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

Where:
- \( i \) = van 't Hoff factor (for sucrose, \( i = 1 \))
- \( K_f \) = freezing point depression constant for water = 1.86 K kg/mol
- \( m \) = molality of the solution = 0.200 mol/kg

\[ \Delta T_f = 1 \cdot 1.86 \, \text{K kg/mol} \cdot 0.200 \, \text{mol/kg} = 0.372 \, \text{K} \]

### Step 4: Calculate the freezing point of the solution
The freezing point of pure water is 0 °C. The freezing point of the solution will be lowered by the depression calculated in the previous step.

\[ \text{Freezing point of solution} = 0 \, \text{°C} - 0.372 \, \text{K} = -0.372 \, \text{°C} \]

### Final Answer
The freezing point of the sucrose solution is approximately -0.372 °C.

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