Which of the following aqueous solution has the highest feezing point ?

Options

A. `0.1 mHCl`
B. `0.05` m Sucrose
C. `0.1` m glucose
D. `0.1` m Nacl.

Explanation

To determine which aqueous solution has the highest freezing point, we can use the formula for the change in freezing point (ΔTf):

\[ \Delta Tf = i \cdot Kf \cdot m \]

Where:
- \( \Delta Tf \) = change in freezing point
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( Kf \) = cryoscopic constant (a property of the solvent, water in this case)
- \( m \) = molality of the solution

### Step-by-Step Solution:

1. **Identify the van 't Hoff factor (i)** for each solute:
- **HCl**: Dissociates into \( H^+ \) and \( Cl^- \) → \( i = 2 \)
- **Sucrose**: Does not dissociate → \( i = 1 \)
- **Glucose**: Does not dissociate → \( i = 1 \)
- **NaCl**: Dissociates into \( Na^+ \) and \( Cl^- \) → \( i = 2 \)

2. **Determine the molality (m)** for each solution:
- **HCl**: 0.1 molal
- **Sucrose**: 0.05 molal
- **Glucose**: 0.1 molal
- **NaCl**: 0.1 molal

3. **Calculate ΔTf for each solution**:
- **HCl**:
\[ \Delta Tf = i \cdot Kf \cdot m = 2 \cdot Kf \cdot 0.1 = 0.2 Kf \]
- **Sucrose**:
\[ \Delta Tf = 1 \cdot Kf \cdot 0.05 = 0.05 Kf \]
- **Glucose**:
\[ \Delta Tf = 1 \cdot Kf \cdot 0.1 = 0.1 Kf \]
- **NaCl**:
\[ \Delta Tf = 2 \cdot Kf \cdot 0.1 = 0.2 Kf \]

4. **Compare the ΔTf values**:
- HCl: \( 0.2 Kf \)
- Sucrose: \( 0.05 Kf \)
- Glucose: \( 0.1 Kf \)
- NaCl: \( 0.2 Kf \)

5. **Determine which solution has the highest freezing point**:
- The freezing point depression (ΔTf) is highest for both HCl and NaCl (both have \( 0.2 Kf \)).
- Since the freezing point depression is the change from the pure solvent, the solutions with the highest freezing points are those with the lowest ΔTf values.

### Conclusion:
The solutions with the highest freezing points are HCl and NaCl.