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 depression in 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 depression in 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 of sucrose, we use the formula:

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

Given:
- Mass of sucrose = 68.5 g
- Molar mass of sucrose = 342 g/mol

Calculating the number of moles:

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

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

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

Calculating molality:

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

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

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

Where:
- \(K_f\) for water = 1.86 K kg/mol
- \(m\) = molality of the solution

Substituting the values:

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

### Step 4: Convert the depression in freezing point to degrees Celsius.
Since the depression in freezing point is the same in Kelvin and degrees Celsius, we have:

\[ \Delta T_f = 0.372 \, \text{°C} \]

### Conclusion
The depression in freezing point of the sucrose solution is \(-0.372 \, \text{°C}\).

### Final Answer
The correct option is \(-0.372 \, \text{°C}\).

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