Van't Hoff Factor
Which solute has van't Hoff factor i ≈ 3 in dilute aqueous solution?
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About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- van't hoff factor
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
Na2SO4
- The Concept: i=Theoretical number of particlesActual number of particles after dissociation.
- Evaluation:
- Glucose: Non-electrolyte. i=1.
- KCl: Dissociates into K+ and Cl−. i=2.
- Na2SO4: Dissociates into 2Na+ and 1SO42−. Total particles = 3. Thus, i=3.
- AlCl3: Dissociates into 1Al3+ and 3Cl−. Total particles = 4. Thus, i=4.
- Prerequisite: Complete dissociation is assumed for strong electrolytes in very dilute solutions.
This hard difficulty chemistry question is from the chapter solutions, covering the topic of van't hoff factor. It appeared in the 2025 exam.
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