0.1 mole of sugar is dissolved in 250 g of water. The freezing point of the solution is `[K_(f) "for" H_(2)O = 1.86^(@)C "molal"^(-1)]`

Options

A. `-0.460^(@)C`
B. `0.460^(@)C`
C. `-0.744^(@)C`
D. `-186^(@)C`

Answer

To find the freezing point of the solution when 0.1 mole of sugar is dissolved in 250 g of water, we will follow these steps:

### Step 1: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute divided by the mass of the solvent in kilograms.

\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent (kg)}} \]

Given:
- Number of moles of sugar = 0.1 moles
- Mass of water = 250 g = 0.250 kg

Now, substituting the values:

\[ m = \frac{0.1 \text{ moles}}{0.250 \text{ kg}} = 0.4 \text{ molal} \]

### Step 2: Calculate the depression in freezing point (ΔTf)
The depression in freezing point can be calculated using the formula:

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

Where:
- \( K_f \) for water = 1.86 °C/molal
- \( m \) = 0.4 molal (from Step 1)

Substituting the values:

\[ \Delta T_f = 1.86 \text{ °C/molal} \times 0.4 \text{ molal} = 0.744 \text{ °C} \]

### Step 3: Calculate the freezing point of the solution
The freezing point of the solution can be calculated using the formula:

\[ \text{Freezing point of solution} = \text{Freezing point of pure solvent} - \Delta T_f \]

For pure water, the freezing point is 0 °C. Thus:

\[ \text{Freezing point of solution} = 0 \text{ °C} - 0.744 \text{ °C} = -0.744 \text{ °C} \]

### Final Answer
The freezing point of the solution is **-0.744 °C**.

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