Arrange the following equimolal aqueous solutions in increasing order…

The correct answer is Option (4) → D, B, A, C

The elevation of boiling point is a colligative property. It depends on the number of solute particles, so for electrolytes the van’t Hoff factor i appears in the formula:

    \( \displaystyle \Delta T_b=i\,K_b\,m \)

For equimolal solutions in the same solvent, K_b and m are the same, so the boiling point rises with i. Counting the particles:

• A. Potassium sulphate: \( \ce{K2SO4} \) → \( \ce{2K^{+}} \) + \( \ce{SO4^{2-}} \): 2 + 1 = 3 ions, so i = 3.

• B. Sodium chloride: NaCl → \( \ce{Na^{+}} \) + \( \ce{Cl^{-}} \): 1 + 1 = 2 ions, so i = 2.

• C. Potash alum: \( \ce{K2SO4} \)·\( \ce{Al2(SO4)3} \)·\( \ce{24H2O} \) → \( \ce{2K^{+}} \) + \( \ce{2Al^{3+}} \) + \( \ce{4SO4^{2-}} \): 2 + 2 + 4 = 8 ions, so i = 8. (Written as \( \ce{KAl(SO4)2} \)·\( \ce{12H2O} \), it still gives 1 + 1 + 2 = 4 ions, more than any other solute here.)

• D. Urea: a non-electrolyte, does not dissociate: 1 particle, i = 1.

    urea < NaCl < \( \ce{K2SO4} \) < potash alum

Shortcut: Urea (D) gives the fewest particles and potash alum (C) the most. Only Option (4) starts with D and ends with C.