Which of the following `0.1M` aqueous solution will have the lowest f.p. ?

Options

A. `K_(2)SO_(4)`
B. `NaCl`
C. Urea
D. Glucose

Answer

To determine which of the following 0.1 M aqueous solutions will have the lowest freezing point, we need to consider the colligative property of freezing point depression. The freezing point depression (\( \Delta T_f \)) is given 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 (the number of particles the solute dissociates into)
- \( K_f \) = freezing point depression constant (specific to the solvent)
- \( m \) = molality of the solution

Since all the solutions are 0.1 M, we can focus on the van 't Hoff factor (\( i \)) for each solute to determine which solution will have the lowest freezing point.

1. **K₂SO₄ (Potassium sulfate)**:
- Dissociates into 2 K⁺ ions and 1 SO₄²⁻ ion.
- Thus, \( i = 2 + 1 = 3 \).

2. **NaCl (Sodium chloride)**:
- Dissociates into 1 Na⁺ ion and 1 Cl⁻ ion.
- Thus, \( i = 1 + 1 = 2 \).

3. **Urea**:
- Urea does not dissociate in solution.
- Thus, \( i = 1 \).

4. **Glucose**:
- Glucose also does not dissociate in solution.
- Thus, \( i = 1 \).

Now, we can summarize the van 't Hoff factors for the solutions:
- K₂SO₄: \( i = 3 \)
- NaCl: \( i = 2 \)
- Urea: \( i = 1 \)
- Glucose: \( i = 1 \)

Since the freezing point depression is directly proportional to the van 't Hoff factor, the solution with the highest \( i \) will have the lowest freezing point.

**Conclusion**:
- K₂SO₄ has the highest \( i \) value of 3, meaning it will cause the greatest depression of the freezing point.
- Therefore, the 0.1 M aqueous solution of K₂SO₄ will have the lowest freezing point.