객관식
At the freezing point, the liquid form of a substance is in equilibrium with the solid form of the substance. Which statement best describes the relationship between the liquid and solid forms at this temperature?
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14. Solutions
Freezing Point Depression
Multiple Choice
Which aqueous solution will have the lowest freezing point, assuming all solutes behave ideally and each solution has the same molal concentration?
A
NaCl(aq)
B
glucose(aq)
C
urea(aq)
D
CaCl(aq)
검증된 단계별 안내1
Understand that the freezing point depression of a solution depends on the number of solute particles in solution, not just the concentration of the solute itself. This is described by the formula for freezing point depression: \(\Delta T_f = i \cdot K_f \cdot m\), where \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_f\) is the freezing point depression constant of the solvent, and \(m\) is the molality of the solution.
Identify the van't Hoff factor (\(i\)) for each solute: glucose and urea are molecular compounds that do not dissociate, so \(i = 1\). NaCl dissociates into Na\(^+\) and Cl\(^-\) ions, so \(i = 2\). CaCl\(_2\) dissociates into one Ca\(^{2+}\) ion and two Cl\(^-\) ions, so \(i = 3\).
Since all solutions have the same molality (\(0.10\,m\)) and the same solvent (water), and assuming ideal behavior, the freezing point depression depends directly on the van't Hoff factor \(i\). The greater the \(i\), the more particles in solution, and the greater the freezing point depression.
Compare the values of \(i\) for each solute: glucose and urea (\(i=1\)), NaCl (\(i=2\)), and CaCl\(_2\) (\(i=3\)). The solution with CaCl\(_2\) will have the greatest number of dissolved particles per formula unit, leading to the largest freezing point depression.
Conclude that the \(0.10\,m\) CaCl\(_2\) solution will have the lowest freezing point because it produces the highest freezing point depression due to its higher van't Hoff factor.
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