The freezing point of equimolal aqueous solution will be highest for

Options

A. `C_6H_5N^(+) H_3Cl^(-)` (aniline hydrochloride)
B. `Mg(NO_3)_2`
C. `La(NO_3)_3`
D. `NH_2CONH_2` (Urea)

Answer

To determine which equimolal aqueous solution has the highest freezing point, we need to analyze the van 't Hoff factor (i) for each solute. The freezing point depression is given by the formula:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(\Delta T_f\) is the depression in freezing point,
- \(i\) is the van 't Hoff factor (the number of particles the solute dissociates into),
- \(K_f\) is the cryoscopic constant (which is the same for all solutions in this case),
- \(m\) is the molality of the solution (which is also the same since we are dealing with equimolal solutions).

### Step-by-Step Solution:

1. **Identify the Solutes and Their Dissociation**:
- **Urea (NH2CO NH2)**: This is a non-electrolyte and does not dissociate in solution. Thus, \(i = 1\).
- **C6H5NH3Cl (Anilinium chloride)**: This dissociates into two ions: \(C6H5NH3^+\) and \(Cl^-\). Therefore, \(i = 2\).
- **Mg(NO3)2**: This dissociates into three ions: \(Mg^{2+}\) and \(2NO3^-\). Thus, \(i = 3\).
- **La(NO3)3**: This dissociates into four ions: \(La^{3+}\) and \(3NO3^-\). Therefore, \(i = 4\).

2. **Calculate the van 't Hoff Factor (i)**:
- Urea: \(i = 1\)
- Anilinium chloride: \(i = 2\)
- Magnesium nitrate: \(i = 3\)
- Lanthanum nitrate: \(i = 4\)

3. **Determine the Effect on Freezing Point**:
- The greater the value of \(i\), the lower the freezing point of the solution (since \(\Delta T_f\) increases).
- Conversely, the smaller the value of \(i\), the higher the freezing point of the solution.

4. **Conclusion**:
- Among the given solutes, urea has the lowest \(i\) value of 1, meaning it will cause the least depression in freezing point.
- Therefore, the freezing point of the equimolal aqueous solution will be highest for urea.

### Final Answer:
The freezing point of the equimolal aqueous solution will be highest for **Urea (NH2CO NH2)**.

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