The freezing point of equimolal solution will be highest for :

Options

A. `C_(6)H_(5)NH_(3)Cl`
B. `Ca(NO_(3))_(2)`
C. `La(NO_(3))_(2)`
D. `C_(6)H_(12)O_(6)`

Answer

To solve the question regarding which equimolar solution has the highest freezing point, we can follow these steps:

### Step 1: Understand the Freezing Point Depression
The freezing point depression (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = K_f \cdot i \cdot m \]
Where:
- \( K_f \) is the cryoscopic constant (depends on the solvent),
- \( i \) is the van 't Hoff factor (number of particles the solute dissociates into),
- \( m \) is the molality of the solution.

### Step 2: Relate Freezing Point to Depression
The freezing point of the solution (\( T_f \)) is related to the freezing point of the pure solvent (\( T_{f, \text{pure}} \)) by:
\[ T_f = T_{f, \text{pure}} - \Delta T_f \]
This means that as \( \Delta T_f \) increases (due to a higher \( i \)), the freezing point of the solution decreases.

### Step 3: Analyze the Given Solutions
We need to determine the van 't Hoff factor (\( i \)) for each of the given solutes:

1. **C6H5NH3Cl (Aniline hydrochloride)**:
- Dissociates into: \( C6H5NH3^+ + Cl^- \)
- \( i = 2 \)

2. **Ca(NO3)2 (Calcium nitrate)**:
- Dissociates into: \( Ca^{2+} + 2NO3^- \)
- \( i = 3 \)

3. **La(NO3)3 (Lanthanum nitrate)**:
- Dissociates into: \( La^{3+} + 3NO3^- \)
- \( i = 4 \)

4. **C6H12O6 (Glucose)**:
- Does not dissociate (non-electrolyte).
- \( i = 1 \)

### Step 4: Compare the Values of \( i \)
From the calculations:
- \( i \) for C6H5NH3Cl = 2
- \( i \) for Ca(NO3)2 = 3
- \( i \) for La(NO3)3 = 4
- \( i \) for C6H12O6 = 1

### Step 5: Determine the Highest Freezing Point
Since the freezing point decreases with an increase in \( i \), the solution with the lowest \( i \) will have the highest freezing point. Here, glucose has the lowest \( i = 1 \).

### Conclusion
Thus, the freezing point of the equimolar solution will be highest for **C6H12O6 (glucose)**.

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