Which has maximum freezing point?

Options

A. `6g` urea solution in `100g H_(2)O`
B. `6g` acetic acid solution in `100gH_(2)O`
C. `6g` sodium chloride in `100g H_(2)O`
D. All have equal freezing point

Explanation

To determine which solution has the maximum freezing point among the given options (water with urea, acetic acid, and sodium chloride), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Freezing Point Depression**:
- The freezing point of a solution is lowered when a solute is added. This phenomenon is described by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
- Where:
- \(\Delta T_f\) = change in freezing point
- \(i\) = Van 't Hoff factor (number of particles the solute dissociates into)
- \(K_f\) = freezing point depression constant (specific to the solvent)
- \(m\) = molality of the solution

2. **Identifying the Solutes**:
- The solutes in the question are urea, acetic acid, and sodium chloride. Water is the solvent in all cases.

3. **Calculating the Van 't Hoff Factor (i)**:
- **Urea**: Urea does not dissociate in solution, so \(i = 1\).
- **Acetic Acid**: Acetic acid partially dissociates in solution, so \(i > 1\) (but less than 2).
- **Sodium Chloride (NaCl)**: Sodium chloride dissociates completely into two ions (Na\(^+\) and Cl\(^-\)), so \(i = 2\).

4. **Calculating Molality (m)**:
- The molality is calculated as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
- Given that the mass of the solvent (water) is the same in all cases (let's assume it's 1 kg for simplicity), we only need to consider the moles of each solute.

5. **Determining Moles of Each Solute**:
- **Urea**:
- Molecular mass of urea = 60 g/mol
- For 6 g of urea:
\[ \text{Moles of urea} = \frac{6 \text{ g}}{60 \text{ g/mol}} = 0.1 \text{ mol} \]
- **Acetic Acid**:
- Molecular mass of acetic acid = 60 g/mol
- For 6 g of acetic acid:
\[ \text{Moles of acetic acid} = \frac{6 \text{ g}}{60 \text{ g/mol}} = 0.1 \text{ mol} \]
- **Sodium Chloride**:
- Molecular mass of NaCl = 58.5 g/mol
- For 6 g of NaCl:
\[ \text{Moles of NaCl} = \frac{6 \text{ g}}{58.5 \text{ g/mol}} \approx 0.102 \text{ mol} \]

6. **Calculating Molality for Each Solute**:
- Since the mass of the solvent is 1 kg, the molality for each solute is approximately equal to the number of moles:
- Urea: \(m = 0.1\)
- Acetic Acid: \(m = 0.1\)
- Sodium Chloride: \(m \approx 0.102\)

7. **Comparing \(\Delta T_f\)**:
- For urea:
\[ \Delta T_f = 1 \cdot K_f \cdot 0.1 \]
- For acetic acid:
\[ \Delta T_f > 1 \cdot K_f \cdot 0.1 \quad (\text{since } i > 1) \]
- For sodium chloride:
\[ \Delta T_f = 2 \cdot K_f \cdot 0.102 \]

8. **Conclusion**:
- The solution with urea has the smallest \(\Delta T_f\), which means it has the maximum freezing point among the three solutions.

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