What is the value of van’t Hoff factor “i” for the complete dissociation of K2SO4?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.0
The van’t Hoff factor (i) compares the number of particles actually present in solution with the number of formula units dissolved:
\( \displaystyle i=\frac{\text{total number of moles of particles after dissociation}}{\text{number of moles of particles before dissociation}} \)
For complete dissociation, i equals the number of ions produced by one formula unit. Potassium sulphate gives three ions:
\( \displaystyle \ce{K2SO4 -> 2K^+ + SO4^2-} \)
\( \displaystyle i=2+1=3 \)
So the van’t Hoff factor for complete dissociation of \( \ce{K2SO4} \) is 3.0.
NCERT (2026-27 edition):
Abnormal Molar Masses – …
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 25-05 S2 | 30-05 S2 | – | – | – |