Which of the following solutions has the maximum freezing point?

Options

A. 1 molal of NaCl solution
B. 1 molal of KCl solution
C. `1 molal of CaCl_(2)` solution
D. 1 molal of urea solution

Explanation

To determine which of the given solutions has the maximum freezing point, we need to analyze the relationship between the freezing point depression (\( \Delta T_f \)) and the van 't Hoff factor (\( i \)).

### Step-by-Step Solution:

1. **Understand Freezing Point Depression**:
The freezing point depression is given by the formula:
\[ \Delta T_f = i \cdot k_f \cdot m \]
where:
- \( \Delta T_f \) = freezing point depression
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( k_f \) = molal freezing point constant (a property of the solvent)
- \( m \) = molality of the solution

2. **Relationship Between Freezing Point and Depression**:
The freezing point of the solution (\( T_f \)) can be expressed as:
\[ T_f = T^0_f - \Delta T_f \]
where \( T^0_f \) is the freezing point of the pure solvent. Thus, a greater \( \Delta T_f \) results in a lower freezing point of the solution.

3. **Identify the Van 't Hoff Factors**:
We need to evaluate the van 't Hoff factor (\( i \)) for each solute:
- For NaCl: \( i = 2 \) (dissociates into Na\(^+\) and Cl\(^-\))
- For KCl: \( i = 2 \) (dissociates into K\(^+\) and Cl\(^-\))
- For CaCl\(_2\): \( i = 3 \) (dissociates into Ca\(^{2+}\) and 2 Cl\(^-\))
- For Urea: \( i = 1 \) (does not dissociate)

4. **Compare the Values of \( i \)**:
From the values calculated:
- NaCl: \( i = 2 \)
- KCl: \( i = 2 \)
- CaCl\(_2\): \( i = 3 \)
- Urea: \( i = 1 \)

5. **Determine the Maximum Freezing Point**:
Since the freezing point depression is directly proportional to \( i \), the solution with the lowest \( i \) will have the maximum freezing point.
- Urea has the lowest \( i \) value of 1.

6. **Conclusion**:
Therefore, the solution with the maximum freezing point is the one containing urea.

### Final Answer:
The solution with the maximum freezing point is **urea**.

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