Arrange the following solutions according to their increasing boiling point.
A. 0.1 M Glucose
B. 0.1 M \(NaCl\)
C. 0.1 M \(K_4[Fe(CN)_6]\)
D. 0.1 M \(MgCl_2\)
E. 0.1 M \(AlCl_3\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A < B < D < E < C
Elevation of boiling point is a colligative property, ΔTb = i Kb m. All solutions are 0.1 M in water, so ΔTb depends only on the van’t Hoff factor i, the number of particles each formula unit gives. The larger i, the higher the boiling point.
A. Glucose (first, lowest) — a non-electrolyte: 1 particle, i = 1.
B. NaCl (second) — \( \ce{NaCl -> Na+ + Cl-} \): 2 ions, i = 2.
D. MgCl2 (third) — \( \ce{MgCl2 -> Mg^2+ + 2Cl-} \): 1 + 2 = 3 ions, i = 3.
E. AlCl3 (fourth) — \( \ce{AlCl3 -> Al^3+ + 3Cl-} \): 1 + 3 = 4 ions, i = 4.
C. K4[Fe(CN)6] (last, highest) — \( \ce{K4[Fe(CN)6] -> 4K+ + [Fe(CN)6]^4-} \): 4 + 1 = 5 ions, i = 5.
Shortcut: glucose (A) has the lowest boiling point and K4[Fe(CN)6] (C) the highest. Only Option (2) starts with A and ends with C.
NCERT (2026-27 edition):
Abnormal Molar Masses – “… Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows: Relative lowering of vapour pressure of solvent, (p1° – p1)/p1° = i.n2/n1; Elevation of Boiling point, ΔTb = i Kb m; Depression of Freezing point, ΔTf = i Kf m; Osmotic pressure of solution, Π = i n2 R T / V.”