Question

Which from following solutions exhibits minimum boiling point elevation under identical conditions? (Assume complete dissociation)}

• A. 0.2 m KCl
• B. 0.1 m NaCl
• C. 1 mAlCl$_3$
• D. 0.05 mMgCl$_2$

Hint:

Boiling point elevation is a colligative property. Higher the concentration of particles ($i \times m$), higher the elevation.

Answer 1

The Correct Option is B

Step 1: Formula
$\Delta T_b = i \times K_b \times m$. Since $K_b$ is constant, $\Delta T_b \propto i \times m$.
Step 2: Calculate effective molality ($i \times m$)
- (A) $KCl: i=2, m=0.2 \to 2 \times 0.2 = 0.4$ - (B) $NaCl: i=2, m=0.1 \to 2 \times 0.1 = 0.2$ - (C) $AlCl_3: i=4, m=1 \to 4 \times 1 = 4.0$ - (D) $MgCl_2: i=3, m=0.05 \to 3 \times 0.05 = 0.15$
Step 3: Conclusion
Wait, let's re-check the values. (B) is $0.2$ and (D) is $0.15$. Among the given options, $0.15$ is the smallest. However, standard MHT-CET keys often cite $0.1$ m NaCl as the conceptual minimum if $0.05$ isn't correctly processed. Let's strictly follow the calculation: $0.15$ (D) is less than $0.2$ (B). *Correction:* Looking at the options provided, if (D) is $0.05 \times 3 = 0.15$, it is the minimum. If the question intended (B) as the answer, there might be a typo in the molality of (D). Based on the provided values, (D) is the minimum.
Final Answer: (D)