Decreasing order of freezing point of \(0.1\, \ M\) aqueous solution of following will be:
(A) \(K_4[Fe(CN)_6]\)
(B) \(K_2SO_4\)
(C) \(NH_2CONH_2\)
(D) \(AlCl_3\)
(E) \(HCl\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (C) > (E) > (B) > (D) > (A)
Depression of freezing point is a colligative property, ΔTf = i Kf m. At the same concentration (0.1 M), the solute with the larger van’t Hoff factor i gives more particles, a larger ΔTf and so a lower freezing point. Placing each solution, highest freezing point first:
(C) \( \ce{NH2CONH2} \) — first. Urea is a non-electrolyte, i = 1.
(E) HCl — second. \( \ce{HCl -> H+ + Cl-} \), i = 2.
(B) \( \ce{K2SO4} \) — third. \( \ce{K2SO4 -> 2K+ + SO4^2-} \), i = 3.
(D) \( \ce{AlCl3} \) — fourth. \( \ce{AlCl3 -> Al^3+ + 3Cl-} \), i = 4.
(A) \( \ce{K4[Fe(CN)6]} \) — last (lowest freezing point). \( \ce{K4[Fe(CN)6] -> 4K+ + [Fe(CN)6]^4-} \), i = 5.
NCERT (2026-27 edition):
Abnormal Molar Masses – “… Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows: … Depression of Freezing point, ΔTf = i Kf m”