Arrange the 0.1 molar aqueous solutions of the following salts in increasing order of their elevation in boiling points:
A. Ammonium phosphate.
B. Ammonium sulphate.
C. Aluminium sulphate.
D. Barium sulphate.
Choose the correct answer from the option given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → D, B, A, C
Elevation in boiling point is a colligative property: it depends on the number of solute particles, not on their nature. For an electrolyte, the van’t Hoff factor i counts the particles each formula unit gives:
\( \displaystyle \Delta T_b=i\,K_b\,m \)
All four solutions have the same concentration (0.1 M) in the same solvent, so ΔTb depends only on i. Assuming complete dissociation:
(A) Ammonium phosphate: \( \ce{(NH4)3PO4 -> 3NH4^+ + PO4^3-} \) — 3 + 1 = 4 ions, i = 4
(B) Ammonium sulphate: \( \ce{(NH4)2SO4 -> 2NH4^+ + SO4^2-} \) — 2 + 1 = 3 ions, i = 3
(C) Aluminium sulphate: \( \ce{Al2(SO4)3 -> 2Al^3+ + 3SO4^2-} \) — 2 + 3 = 5 ions, i = 5
(D) Barium sulphate: \( \ce{BaSO4 -> Ba^2+ + SO4^2-} \) — 1 + 1 = 2 ions, i = 2 (BaSO4 is in fact only sparingly soluble; the question treats it as fully dissolved)
Increasing order of i, and so of elevation in boiling point:
\( \displaystyle 2<3<4<5 \)
\( \displaystyle \mathrm{D}<\mathrm{B}<\mathrm{A}<\mathrm{C} \)
Shortcut: Options (3) and (4) both start with D (i = 2) and end with C (i = 5). They differ only in A and B; B (i = 3) comes before A (i = 4), which gives Option (4).