Question

Which of the following aqueous solutions having same molality exhibits maximum boiling point elevation? (Assume complete dissociation)

• A. $KCl$
• B. $NaCl$
• C. $AlCl_{3}$
• D. $BaCl_{2}$

Hint:

More ions produced = Greater change in colligative properties.

Answer 1

The Correct Option is C

Step 1: Concept
Elevation in boiling point ($\Delta T_{b}$) is a colligative property: $\Delta T_{b} = i \times K_{b} \times m$.

Step 2: Meaning
For the same molality, $\Delta T_{b}$ depends on the van’t Hoff factor ($i$), which is the number of particles after dissociation.

Step 3: Analysis
- $KCl$: $K^{+} + Cl^{-}$ ($i=2$) - $NaCl$: $Na^{+} + Cl^{-}$ ($i=2$) - $BaCl_{2}$: $Ba^{2+} + 2Cl^{-}$ ($i=3$) - $AlCl_{3}$: $Al^{3+} + 3Cl^{-}$ ($i=4$)

Step 4: Conclusion
$AlCl_{3}$ has the highest $i$ value, thus maximum boiling point elevation. Final Answer: (C)