Which of the following solutions will exhibit the same elevation in their boiling points?
A. 0.03 M NaCl solution
B. 0.06 M Glucose solution
C. 0.01 M Aluminum Sulfate Solution
D. 0.02 M Calcium Chloride Solution
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A, B and D only
Elevation of 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 formed from one formula unit:
\( \displaystyle \Delta T_b=i\,K_b\,m \)
For dilute aqueous solutions, the molality is almost equal to the molarity. So all four solutions (same solvent, same Kb) show the same elevation when the product i × concentration is the same. Assuming complete dissociation:
(A) 0.03 M NaCl solution — \( \ce{NaCl -> Na+ + Cl-} \): 1 + 1 = 2 ions, i = 2, so i × C = 2 × 0.03 = 0.06.
(B) 0.06 M Glucose solution — glucose does not dissociate: 1 particle, i = 1, so i × C = 0.06.
(C) 0.01 M Aluminum Sulfate Solution — \( \ce{Al2(SO4)3 -> 2Al^3+ + 3SO4^2-} \): 2 + 3 = 5 ions, i = 5, so i × C = 5 × 0.01 = 0.05.
(D) 0.02 M Calcium Chloride Solution — \( \ce{CaCl2 -> Ca^2+ + 2Cl-} \): 1 + 2 = 3 ions, i = 3, so i × C = 3 × 0.02 = 0.06.