Match List I with List II List I List II A. K_2SO_4 (aq) with 60% dis…

The correct answer is Option (4) → A-III, B-I, C-IV, D-II

Take 1 mol of an electrolyte that gives n ions per formula unit, with degree of dissociation α. At equilibrium (1 − α) mol stays undissociated and nα mol of ions form:

    \( \displaystyle i = \frac{(1-\alpha) + n\alpha}{1} = 1 + (n-1)\alpha \)

• A. \( \ce{K2SO4} \) (aq) with 60% dissociation — This matches III. i = 2.2. \( \ce{K2SO4 -> 2K+ + SO4^2-} \), n = 3: i = 1 + 2(0.60) = 2.2.

• B. \( \ce{K3[Fe(CN)6]} \) (aq) with 90% dissociation — This matches I. i = 3.7. \( \ce{K3[Fe(CN)6] -> 3K+ + [Fe(CN)6]^3-} \), n = 4: i = 1 + 3(0.90) = 3.7.

• C. \( \ce{AlCl3} \) (aq) with 80% dissociation — This matches IV. i = 3.4. \( \ce{AlCl3 -> Al^3+ + 3Cl-} \), n = 4: i = 1 + 3(0.80) = 3.4.

• D. \( \ce{K2HgI4} \) (aq) with 40% dissociation — This matches II. i = 1.8. \( \ce{K2HgI4 -> 2K+ + [HgI4]^2-} \), n = 3: i = 1 + 2(0.40) = 1.8.

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
Here abnormal molar mass is the experimentally determined molar mass and calculated colligative properties are obtained by assuming that the non-volatile solute is neither associated nor dissociated. In case of association, value of i is less than unity while for dissociation it is greater than unity.”