The freezing point depression, which indicates how much the freezing point of a solution is lowered, is proportional to the number of solute particles in the solution. The equation that describes this relationship is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \(\Delta T_f\) is the change 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 a property of the solvent, typically water), and
- \(m\) is the molality of the solution.
To evaluate the freezing point of each solution, we need to consider the van 't Hoff factor (\(i\)) for each solute:
- Sucrose (C₁₂H₂₂O₁₁): It is a non-ionic, molecular solute, hence \(i = 1\).
- Sodium chloride (NaCl): It dissociates into 2 ions (Na⁺ and Cl⁻), so \(i = 2\).
- Calcium bromide (CaBr₂): This dissociates into 3 ions (Ca²⁺ and 2Br⁻), so \(i = 3\).
- Aluminum bromide (AlBr₃): This dissociates into 4 ions (Al³⁺ and 3Br⁻), so \(i = 4\).
- Potassium chloride (KCl): It dissociates into 2 ions (K⁺ and Cl⁻), so \(i = 2\).
Now we can summarize the effective concentration of particles in each solution:
- Sucrose: \(1.0 , \text{M} \times 1 = 1.0 , \text{particles}\)
- NaCl: \(1.0 , \text{M} \times 2 = 2.0 , \text{particles}\)
- CaBr₂: \(1.0 , \text{M} \times 3 = 3.0 , \text{particles}\)
- AlBr₃: \(1.0 , \text{M} \times 4 = 4.0 , \text{particles}\)
- KCl: \(1.0 , \text{M} \times 2 = 2.0 , \text{particles}\)
Since a higher number of particles results in a greater depression of the freezing point, the solution with the least number of total particles has the highest freezing point.
Among the options:
- Sucrose has the least impact on freezing point since it generates only 1 particle.
- NaCl, KCl both increase the freezing point lower than sucrose due to having 2 particles.
- CaBr₂ and AlBr₃ will have even lower freezing points due to having 3 and 4 particles, respectively.
Thus, the solution with the highest freezing point is:
A. 1.0 M molecular sucrose (C₁₂H₂₂O₁₁).