Equimolal aqueous solutions of potassium nitrate, fructose and potassium ferricyanide were prepared. The expected values for van't Hoff factor 'i' are respectively
[Assume complete dissociation wherever applicable]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2,1,4
The van’t Hoff factor i is the ratio of the number of particles actually present in solution to the number of formula units dissolved. For complete dissociation, i equals the number of ions given by one formula unit; for a non-electrolyte, i = 1.
\( \displaystyle \ce{KNO3 -> K^+ + NO3^-}\quad\Rightarrow\quad i=2 \)
\( \displaystyle \text{Fructose, }\ce{C6H12O6}\text{ (non-electrolyte)}\quad\Rightarrow\quad i=1 \)
\( \displaystyle \ce{K3[Fe(CN)6] -> 3K^+ + [Fe(CN)6]^3-}\quad\Rightarrow\quad i=4 \)
Potassium ferricyanide is a coordination compound: the complex ion \( \ce{[Fe(CN)6]^{3-}} \) stays together as one particle and does not break into \( \ce{Fe^{3+}} \) and \( \ce{CN^{-}} \) ions.
So the van’t Hoff factors are 2, 1 and 4 respectively.
Shortcut: Fructose does not ionise, so its value is 1, and it is the second compound. Only Option (4) has 1 in the middle.