0.1 mol of sugar was dissolved in 1 kg of water. The freezing point of the solution was found to be 272.814 K. What conclusion would you draw about the molecular state of sugar? `K_(f)` for water is `"1.86 K kg mol"^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Depression in Freezing Point
The depression in freezing point (ΔTf) can be calculated using the formula:
\[ \Delta T_f = T_f^0 - T_f \]
Where:
- \( T_f^0 \) is the freezing point of pure water (273 K)
- \( T_f \) is the freezing point of the solution (272.814 K)

Substituting the values:
\[ \Delta T_f = 273 \, \text{K} - 272.814 \, \text{K} = 0.186 \, \text{K} \]

### Step 2: Calculate the Molality of the Solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Given that 0.1 moles of sugar are dissolved in 1 kg of water:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.1 \, \text{mol}}{1 \, \text{kg}} = 0.1 \, \text{mol/kg} \]

### Step 3: Use the Freezing Point Depression Formula
The freezing point depression can also be expressed as:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( i \) is the van 't Hoff factor (which indicates the number of particles the solute breaks into)
- \( K_f \) is the freezing point depression constant for water (1.86 K kg/mol)

Rearranging the formula to find \( i \):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

Substituting the known values:
\[ i = \frac{0.186 \, \text{K}}{1.86 \, \text{K kg/mol} \cdot 0.1 \, \text{mol/kg}} \]

Calculating \( i \):
\[ i = \frac{0.186}{0.186} = 1 \]

### Step 4: Conclusion about the Molecular State of Sugar
Since the van 't Hoff factor \( i \) is equal to 1, this indicates that the sugar does not dissociate into ions in solution. Therefore, we conclude that sugar is present in the molecular state (as a covalent solid) and does not ionize in solution.

### Final Answer
The molecular state of sugar is that it is present as a covalent solid and does not dissociate into ions.

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