Match List-I with List-II
|
List-I Molecule |
List-II van't Hoff Factor |
|
(A) Glucose |
(I) 3 |
|
(B) \( K_2SO_4 \) |
(II) 5 |
|
(C) \( Ca_3(PO_4)_2 \) |
(III) 1 |
|
(D) \( K_3[Fe(CN)_6] \) |
(IV) 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
The van’t Hoff factor i is the number of particles one formula unit gives in solution. For a strong electrolyte that dissociates completely, i is found by writing the dissociation and counting the ions; a complex ion stays together as one ion. A non-electrolyte does not dissociate, so i = 1.
(A) Glucose — This matches (III) 1. Glucose is a non-electrolyte; it does not dissociate: 1 particle, so i = 1.
(B) \(K_2SO_4\) — This matches (I) 3. Two potassium ions and one sulphate ion: 2 + 1 = 3 ions, so i = 3.
\( \displaystyle \ce{K2SO4 -> 2K^+ + SO4^2-} \)
(C) \(Ca_3(PO_4)_2\) — This matches (II) 5. Three calcium ions and two phosphate ions: 3 + 2 = 5 ions, so i = 5.
\( \displaystyle \ce{Ca3(PO4)2 -> 3Ca^2+ + 2PO4^3-} \)
(D) \(K_3[Fe(CN)_6]\) — This matches (IV) 4. Three potassium ions and one hexacyanidoferrate(III) ion, which does not break up further: 3 + 1 = 4 ions, so i = 4.
\( \displaystyle \ce{K3[Fe(CN)6] -> 3K^+ + [Fe(CN)6]^3-} \)
Shortcut: Glucose, a non-electrolyte, has i = 1, and (A)-(III) appears only in Option (4).