Van't Hoff Factor
A 0.1 m aqueous solution of BaCl₂ shows a depression of freezing point of 0.480 K. If Kf for water is 1.86 K kg mol⁻¹, the van't Hoff factor (i) for BaCl₂ in this solution is:
Select the correct option:
Solution
2.58
The observed freezing point depression is ΔTf = i × Kf × m. Substituting: 0.480 = i × 1.86 × 0.1 = i × 0.186. Therefore, i = 0.480 / 0.186 = 2.58. For complete dissociation of BaCl₂ → Ba²⁺ + 2Cl⁻, the theoretical van't Hoff factor would be 3 (one formula unit yields 3 ions). The observed value of 2.58 is less than 3, indicating incomplete dissociation or ion-pair formation in the solution. This is typical behaviour for strong electrolytes at moderate concentrations where interionic attractions reduce the effective number of particles.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More van't hoff factor Practice Questions
About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- van't hoff factor
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
2.58
The observed freezing point depression is ΔTf = i × Kf × m. Substituting: 0.480 = i × 1.86 × 0.1 = i × 0.186. Therefore, i = 0.480 / 0.186 = 2.58. For complete dissociation of BaCl₂ → Ba²⁺ + 2Cl⁻, the theoretical van't Hoff factor would be 3 (one formula unit yields 3 ions). The observed value of 2.58 is less than 3, indicating incomplete dissociation or ion-pair formation in the solution. This is typical behaviour for strong electrolytes at moderate concentrations where interionic attractions reduce the effective number of particles.
This medium difficulty chemistry question is from the chapter solutions, covering the topic of van't hoff factor. It appeared in the 2025 exam.
Looking for more practice? Explore all chemistry questions or browse solutions questions on RankGuru.