Question

each solute is dissolved in 1 l of solvent. which solution will exhibit the greatest drop in freezing point compared to plain water?
1.0 mole cabr₂ (dissociates into 3 particles)
1.0 mole ch₃oh (dissociates into 1 particle)
1.0 mole ki (dissociates into 2 particles)

Explanation:

Step1: Recall Freezing Point Depression

Freezing point depression depends on the number of solute particles. The formula is $\Delta T_f = iK_fm$, where $i$ is the van't Hoff factor (number of particles per formula unit), $K_f$ is the cryoscopic constant, and $m$ is molality. Since volume is 1 L and solvent is water (density ~1 kg/L), molality $m$ is approximately moles of solute (1.0 mol here). So, $\Delta T_f$ is proportional to $i$.

Step2: Calculate $i$ for Each Solute

• For $\ce{CaBr2}$: $i = 3$ (dissociates into 3 particles).
• For $\ce{CH3OH}$: $i = 1$ (dissociates into 1 particle).
• For $\ce{KI}$: $i = 2$ (dissociates into 2 particles).

Step3: Compare $i$ Values

Higher $i$ means greater $\Delta T_f$. Among 3, 1, 2, 3 is the largest. So $\ce{CaBr2}$ has the greatest drop.

Answer:

1.0 mole $\ce{CaBr2}$ (dissociates into 3 particles)