Question

Which has the lowest boiling point at 1 atm pressure?

• A. \(0.1\,M\;KCl\)
• B. \(0.1\,M\;\text{urea}\)
• C. \(0.1\,M\;CaCl_2\)
• D. \(0.1\,M\;AlCl_3\)

Hint:

For colligative properties, always compare the total number of particles formed in solution. Non-electrolytes like glucose and urea have \(i=1\), so they show minimum effect on boiling point and freezing point compared to ionic compounds of same concentration.

Answer 1

The Correct Option is B

Concept: Boiling point elevation is a colligative property. It depends upon the number of solute particles present in the solution and is given by: \[ \Delta T_b = iK_bm \] where:
• \( \Delta T_b \) = elevation in boiling point
• \( i \) = van't Hoff factor
• \( K_b \) = molal elevation constant
• \( m \) = molality Greater the number of particles formed in solution, greater is the boiling point elevation.

Step 1: Understanding the relation between boiling point and number of particles.
Pure solvents have the lowest boiling point. When a solute is added, the boiling point increases. The increase depends upon the value of van't Hoff factor \(i\). More dissociation \( \Rightarrow \) more particles \( \Rightarrow \) higher boiling point. Thus, the solution producing the least number of particles will have the lowest boiling point.

Step 2: Calculating relative number of particles for each solution.

• \(KCl \rightarrow K^+ + Cl^- \) Number of ions formed \(=2\) Therefore, \(i \approx 2\)
• \(CaCl_2 \rightarrow Ca^{2+} + 2Cl^- \) Number of ions formed \(=3\) Therefore, \(i \approx 3\)
• \(AlCl_3 \rightarrow Al^{3+} + 3Cl^- \) Number of ions formed \(=4\) Therefore, \(i \approx 4\)
• Urea is a non-electrolyte and does not dissociate in water. Therefore, \[ i = 1 \]

Step 3: Comparing boiling point elevations.
Since: \[ \Delta T_b \propto i \] Relative boiling point elevation: \[ \text{urea} < KCl < CaCl_2 < AlCl_3 \] Thus, urea causes the minimum elevation in boiling point. Hence, it has the lowest boiling point among the given solutions.