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Question #44520
The Question
Two volatile liquids, \ce{A} and \ce{B}, form a non-ideal solution exhibiting a positive deviation from Raoult's law. At , the vapor pressures of pure \ce{A} and pure \ce{B} are and , respectively. A student incorrectly assumes that the mixture behaves ideally and that the vapor pressures of both pure liquids follow a temperature coefficient (i.e., vapor pressure doubles for every rise in temperature). Experimentally, the mixture forms a minimum-boiling azeotrope at with a liquid mole fraction of \ce{A}, . Calculate the magnitude of the difference between the true mole fraction of \ce{A} in the vapor phase () at the azeotropic point and the vapor mole fraction of \ce{A} () calculated by the student at for the same liquid composition.
Options
- A.
0.25
- B.
0.50
- C.
0.75
- D.
0.00
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Numerical_Precision
Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.
Concepts Behind This Question
Raoult’s Law Ideal vs Non-Ideal|Temperature Coefficient (Q10)|Graphical Slopes Analysis
Why This Question Is Tricky
Common Mistakes
This problem tests the conceptual boundary between kinetic approximations () and thermodynamic phase equilibria.
The trap lures students into calculating partial pressures using an extrapolated ideal Raoult's Law model ().
However, the hidden variable is the definition of an azeotrope itself: at the azeotropic point, the activity coefficients () deviate non-linearly to ensure the vapor composition exactly matches the liquid composition ().
Failing to recognize that the true vapor fraction is fixed by the azeotropic condition renders the ideal extrapolation invalid.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Solutions - Class 12 |
| Concept | Raoult’s Law Ideal vs Non-Ideal |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |