The decreasing order of value of van't Hoff factor for the following in aqueous medium is
(A) \( K_3[Fe(CN)_6] \)
(B) \( NaCl \)
(C) \( K_2SO_4 \)
(D) \( C_6H_{12}O_6 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C), (B), (D)
The van’t Hoff factor i is the number of particles one formula unit gives in solution. For a strong electrolyte in a dilute aqueous solution it is the number of ions formed on complete dissociation; a non-electrolyte gives one particle, so i = 1.
(A) \( \ce{K3[Fe(CN)6]} \): the complex ion stays intact and only the three \( \ce{K+} \) ions separate. 3 + 1 = 4 ions, so i = 4.
\( \displaystyle \ce{K3[Fe(CN)6] -> 3K+ + [Fe(CN)6]^3-} \)
(B) \( \ce{NaCl} \): 1 + 1 = 2 ions, so i = 2.
\( \displaystyle \ce{NaCl -> Na+ + Cl-} \)
(C) \( \ce{K2SO4} \): 2 + 1 = 3 ions, so i = 3.
\( \displaystyle \ce{K2SO4 -> 2K+ + SO4^2-} \)
(D) \( \ce{C6H12O6} \): glucose is a non-electrolyte and does not dissociate: 1 particle, so i = 1.
In decreasing order of the van’t Hoff factor:
\( \displaystyle \ce{K3[Fe(CN)6]}\ (4) \gt \ce{K2SO4}\ (3) \gt \ce{NaCl}\ (2) \gt \ce{C6H12O6}\ (1) \)
Shortcut: Only Options (1) and (2) start with (A) and end with (D). \( \ce{K2SO4} \) gives more ions than NaCl, so (C) comes before (B): Option (1).