Of the following 0.2 m aqueous solutions, which one will exhibit largest freezing point depression?

Options

A. Giucose
B. Sucrose
C. Urea
D. NaCI

Answer

To determine which 0.2 m aqueous solution will exhibit the largest freezing point depression, we can use the formula for freezing point depression:

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

Where:
- \(\Delta T_f\) = freezing point depression
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(K_f\) = freezing point depression constant (depends on the solvent, usually water)
- \(m\) = molality of the solution

Given that all solutions are 0.2 m, we need to consider the van 't Hoff factor (\(i\)) for each solute:

1. **Glucose (C₆H₁₂O₆)**:
- Glucose does not dissociate in solution, so \(i = 1\).
- Therefore, \(i \cdot m = 1 \cdot 0.2 = 0.2\).

2. **Sucrose (C₁₂H₂₂O₁₁)**:
- Sucrose also does not dissociate in solution, so \(i = 1\).
- Therefore, \(i \cdot m = 1 \cdot 0.2 = 0.2\).

3. **Urea (NH₂CONH₂)**:
- Urea does not dissociate in solution, so \(i = 1\).
- Therefore, \(i \cdot m = 1 \cdot 0.2 = 0.2\).

4. **Sodium Chloride (NaCl)**:
- Sodium chloride dissociates into two ions: Na⁺ and Cl⁻, so \(i = 2\).
- Therefore, \(i \cdot m = 2 \cdot 0.2 = 0.4\).

Now, we can compare the values of \(i \cdot m\) for each solute:

- Glucose: \(0.2\)
- Sucrose: \(0.2\)
- Urea: \(0.2\)
- Sodium Chloride: \(0.4\)

Since sodium chloride has the highest value of \(i \cdot m\), it will exhibit the largest freezing point depression.

**Final Answer**: Sodium Chloride (NaCl) will exhibit the largest freezing point depression.