The increasing order of the elevation in boiling points of the following 0.1 m aqueous solutions will be:
(assuming that the electrolytic solute is completely dissociated)
A. Potassium sulphate
B. Sodium chloride
C. Glucose
D. Ammonium phosphate
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → C, B, A, D
Elevation of boiling point is a colligative property: it depends on the number of solute particles. For an electrolyte, the van’t Hoff factor i gives the number of particles formed from one formula unit:
\( \displaystyle \Delta T_b=i\,K_b\,m \)
All solutions are 0.1 m in the same solvent (water), so ΔTb is proportional to i.
Matching the given solutes (complete dissociation):
(C) Glucose: a non-electrolyte, i = 1
(B) Sodium chloride: NaCl → \( \ce{Na^{+}} \) + \( \ce{Cl^{-}} \), i = 2
(A) Potassium sulphate: \( \ce{K2SO4} \) → \( \ce{2K^{+}} \) + \( \ce{SO4^{2-}} \), i = 3
(D) Ammonium phosphate: \( \ce{(NH4)3PO4} \) → \( \ce{3NH4^{+}} \) + \( \ce{PO4^{3-}} \), i = 4
Shortcut: Glucose (C) gives the fewest particles and ammonium phosphate (D) the most. Only Option (1) starts with (C) and ends with (D).