Which of the following aqueous solution will exhibit highest boiling point?
Correct Answer :
0.01 M Na2SO4
0.01 M Na2SO4
Solution :
The correct answer is 0.01 M Na2SO4.
Underlying Concept:
Boiling point elevation is a colligative property, which means it depends on the total number of solute particles (ions or molecules) present in the solution rather than their chemical identity.
The elevation in boiling point () is given by the formula:
where:
• is the van 't Hoff factor (the number of particles into which the solute dissociates),
• is the molal boiling point elevation constant of the solvent (water), and
• is the molarity (concentration) of the solution.
Since the solvent (water) is the same for all options, remains constant. Therefore, the boiling point elevation is directly proportional to the product of the van 't Hoff factor and the molar concentration of the solution ():
The solution with the highest value of will undergo the greatest elevation in boiling point and will exhibit the highest boiling point.
Step-by-Step Calculation:
1. 0.01 M Urea:
Urea is a non-electrolyte and does not dissociate in water.
• Van 't Hoff factor () = 1
•
2. 0.01 M KNO3:
Potassium nitrate dissociates completely into two ions:
KNO3 → K+ + NO3-
• Van 't Hoff factor () = 2
•
3. 0.01 M Na2SO4:
Sodium sulfate dissociates completely into three ions:
Na2SO4 → 2Na+ + SO42-
• Van 't Hoff factor () = 3
•
4. 0.015 M C6H12O6 (Glucose):
Glucose is a non-electrolyte and does not dissociate in water.
• Van 't Hoff factor () = 1
•
Conclusion:
Comparing the effective particle concentrations ():
• 0.01 M Urea = 0.01 M
• 0.01 M KNO3 = 0.02 M
• 0.01 M Na2SO4 = 0.03 M
• 0.015 M C6H12O6 = 0.015 M
Since 0.01 M Na2SO4 has the highest value of (0.03 M), it produces the maximum number of solute particles in the solution, resulting in the highest elevation in boiling point. Therefore, it has the highest boiling point among the given options.
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