Which one of the following salts would have the same value of van't H…

The correct answer is Option (2) → \(K_3[Fe(CN)_6]\)

For a strong electrolyte that dissociates completely, the van’t Hoff factor i equals the number of ions formed from one formula unit. A complex ion stays as one unit in solution.

    \( \ce{Al(NO3)3 -> Al^3+ + 3NO3^-} \): 1 + 3 = 4 ions, so i = 4

The same count for each option:

    \( \ce{Al2(SO4)3 -> 2Al^3+ + 3SO4^2-} \): 2 + 3 = 5 ions, i = 5
    \( \ce{K3[Fe(CN)6] -> 3K+ + [Fe(CN)6]^3-} \): 3 + 1 = 4 ions, i = 4
    \( \ce{K2SO4 -> 2K+ + SO4^2-} \): 2 + 1 = 3 ions, i = 3
    \( \ce{KCl -> K+ + Cl-} \): 1 + 1 = 2 ions, i = 2

Only \( \ce{K3[Fe(CN)6]} \) gives 4 ions; the \( \ce{CN-} \) ligands are not released, because \( \ce{[Fe(CN)6]^3-} \) is one ion.

NCERT (2026-27 edition):
Abnormal Molar Masses – “… In 1880 van’t Hoff introduced a factor i, known as the van’t Hoff factor, to account for the extent of dissociation or association. This factor i is defined as:
i = Normal molar mass / Abnormal molar mass
= Observed colligative property / Calculated colligative property
i = Total number of moles of particles after association/dissociation / Number of moles of particles before association/dissociation”