To determine which solution has the lowest freezing point, we can use the concept of freezing point depression, which is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \(\Delta T_f\) is the change in freezing point (how much lower the freezing point is),
- \(i\) is the van 't Hoff factor (the number of particles the solute breaks into),
- \(K_f\) is the freezing point depression constant (which is specific to the solvent, and we assume it's the same for all since they are in water),
- \(m\) is the molality of the solution.
We will analyze each option based on \(i\):
-
Pure water:
- Freezing point: 0 °C
- \(i = 1\) (1 particle)
-
0.5 M ionic NaCl:
- NaCl dissociates into Na\(^+\) and Cl\(^-\), so:
- \(i = 2\)
- Effective concentration = \(0.5 \times 2 = 1.0\)
-
0.5 M ionic CaCl2:
- CaCl2 dissociates into Ca\(^{2+}\) and 2 Cl\(^-\), so:
- \(i = 3\)
- Effective concentration = \(0.5 \times 3 = 1.5\)
-
0.5 M ionic AlCl3:
- AlCl3 dissociates into Al\(^{3+}\) and 3 Cl\(^-\), so:
- \(i = 4\)
- Effective concentration = \(0.5 \times 4 = 2.0\)
-
0.5 M molecular C12H22O11 (sucrose):
- Sucrose does not dissociate, so:
- \(i = 1\)
- Effective concentration = \(0.5 \times 1 = 0.5\)
Now we summarize the effective concentrations:
- Pure water: 0
- NaCl: 1.0
- CaCl2: 1.5
- AlCl3: 2.0
- C12H22O11: 0.5
The solution with the highest effective concentration (and thus the lowest freezing point) is 0.5 M ionic AlCl3.
Thus, D. 0.5 M ionic AlCl3 has the lowest freezing point.