Correct increasing order of freezing point of 1 M solution : A. Na_3(…

The correct answer is Option (2) → B only

Note: The given answer is as per the official NTA Answer Key. However, statement (D) is also correct, as explained below.

Depression of freezing point is a colligative property: it depends on the number of solute particles. Including the van’t Hoff factor:

    \( \displaystyle \Delta T_f = i\,K_f\,m \)

For solutions of the same concentration (1 M, taken as about 1 m in water), the larger i, the larger ΔT_f and the lower the freezing point. Counting ions for complete dissociation:

• \( \ce{K4[Fe(CN)6] -> 4K+ + [Fe(CN)6]^4-} \): 4 + 1 = 5 ions, i = 5

• \( \ce{Na3PO4 -> 3Na+ + PO4^3-} \): 3 + 1 = 4 ions, i = 4

• \( \ce{K2SO4 -> 2K+ + SO4^2-} \): 2 + 1 = 3 ions, i = 3

• \( \ce{NaCl -> Na+ + Cl-} \): 1 + 1 = 2 ions, i = 2

Increasing freezing point (decreasing i):

    \( \ce{K4[Fe(CN)6]} \) < \( \ce{Na3PO4} \) < \( \ce{K2SO4} \) < NaCl

• (A) \(Na_3(PO_4) > K_2SO_4 > NaCl > K_4[Fe(CN)_6]\) — Incorrect. NaCl has the highest freezing point, not \( \ce{Na3PO4} \); the order and the sign are both wrong.

• (B) \(K_4[Fe(CN)_6] < Na_3(PO_4) < K_2SO_4 < NaCl\) — Correct. This matches i = 5 < 4 < 3 < 2 read as freezing points.

• (C) \(Na_3(PO_4) < K_2SO_4 < NaCl < K_4[Fe(CN)_6]\) — Incorrect. \( \ce{K4[Fe(CN)6]} \) gives the most particles, so it has the lowest freezing point, not the highest.

• (D) \(NaCl > K_2SO_4 > Na_3(PO_4) > K_4[Fe(CN)_6]\) — Is also correct. It states the same relation as (B), read from the other end: NaCl has the highest freezing point and \( \ce{K4[Fe(CN)6]} \) the lowest. No option names (B) and (D) together, so Option (2) is the answer.

• (E) \(K_2SO_4 > NaCl > K_4[Fe(CN)_6] > Na_3(PO_4)\) — Incorrect. NaCl (i = 2) freezes higher than \( \ce{K2SO4} \) (i = 3), and \( \ce{K4[Fe(CN)6]} \) freezes lower than \( \ce{Na3PO4} \).

NCERT (2026-27 edition):
Abnormal Molar Masses – …