Before You Begin
Question #44714
The Question
A saturated aqueous solution of cerium(III) sulfate, \ce{Ce2(SO4)3}, is prepared at in of water. The excess un-dissolved solid is rapidly filtered off at . The filtrate is then cooled to determine its freezing point. Assuming ideal behavior and complete dissociation of the salt, what is the freezing point depression () of this filtrate?
Given: Solubility of \ce{Ce2(SO4)3} at (assume independent of temperature)
Options
- A.
0.093 K
- B.
0.372 K
- C.
0.186 K
- D.
0.0186 K
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Exception_Exposure
Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.
Concepts Behind This Question
Solubility of Solids in Liquids|Gibbs Energy and Spontaneity|Freezing Point Depression (Pure Solvent)
Why This Question Is Tricky
Common Mistakes
This problem tests the intersection of Le Chatelier's Principle, the Van't Hoff equation, and colligative properties.
The 'Chemical Exception' is retrograde solubility: because the dissolution of \ce{Ce2(SO4)3} is exothermic, its solubility decreases at higher temperatures.
Students commonly assume solubility always increases with temperature, leading them to incorrectly double the solubility instead of halving it (Trap B).
Furthermore, the 'Hidden Variable' is the phase behavior during cooling: if the dissolution were endothermic, cooling the solution would cause precipitation, altering the molality before freezing.
Because it is exothermic, cooling increases the solubility limit, ensuring the solution becomes unsaturated and the molality remains perfectly constant at the value established at 320 K.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Solutions - Class 12 |
| Concept | Solubility of Solids in Liquids |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |