Match List I with List II
|
LIST I Salts |
LIST II van't Hoff Factor |
| A. Mohr's salt | I. 3 |
| B. Blue Vitriol | II. 8 |
| C. Lime slake | III. 2 |
| D. Alum | IV. 5 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-IV, B-III, C-I, D-II
The van’t Hoff factor i accounts for dissociation or association of the solute. For a strong electrolyte that dissociates completely, i equals the number of ions produced by one formula unit. Water of crystallisation does not count.
(A) Mohr’s salt — This matches (IV) 5. Mohr’s salt is \( \ce{FeSO4} \)·\( \ce{(NH4)2SO4} \)·\( \ce{6H2O} \), a double salt that gives all its ions in water:
\( \ce{FeSO4.(NH4)2SO4.6H2O -> Fe^2+ + 2NH4+ + 2SO4^2- + 6H2O} \): 1 + 2 + 2 = 5 ions, i = 5
(B) Blue Vitriol — This matches (III) 2. Blue vitriol is \( \ce{CuSO4} \)·\( \ce{5H2O} \):
\( \ce{CuSO4.5H2O -> Cu^2+ + SO4^2- + 5H2O} \): 1 + 1 = 2 ions, i = 2
(C) Lime slake — This matches (I) 3. Slaked lime is \( \ce{Ca(OH)2} \):
\( \ce{Ca(OH)2 -> Ca^2+ + 2OH-} \): 1 + 2 = 3 ions, i = 3
(D) Alum — This matches (II) 8. Potash alum is \( \ce{K2SO4} \)·\( \ce{Al2(SO4)3} \)·\( \ce{24H2O} \):
\( \ce{K2SO4.Al2(SO4)3.24H2O -> 2K+ + 2Al^3+ + 4SO4^2- + 24H2O} \): 2 + 2 + 4 = 8 ions, i = 8
NCERT (2026-27 edition):
Abnormal Molar Masses – …
