Question

Calculate the molal elevation constant of solvent if boiling point of $0.12\text{ m}$ solution is $319.8\text{ K}$ (Boling point of solvent $= 319.5\text{ K}$ )}

• A. $2.0\text{ K kg mol}^{-1}$
• B. $3.0\text{ K kg mol}^{-1}$
• C. $2.5\text{ K kg mol}^{-1}$
• D. $3.5\text{ K kg mol}^{-1}$

Hint:

$K_b$ is also called the ebullioscopic constant and is characteristic of the solvent.

Answer 1

The Correct Option is C

Step 1: Concept Elevation of boiling point ($\Delta T_b$) is proportional to molality ($m$), expressed as $\Delta T_b = K_b \cdot m$.

Step 2: Meaning $\Delta T_b$ is the difference between the boiling point of the solution ($T_b$) and the pure solvent ($T_b^\circ$).

Step 3: Analysis $\Delta T_b = 319.8 - 319.5 = 0.3 \text{ K}$.
Given $m = 0.12 \text{ m}$, $K_b = \frac{\Delta T_b}{m} = \frac{0.3}{0.12}$.

Step 4: Conclusion $K_b = \frac{30}{12} = 2.5 \text{ K kg mol}^{-1}$. Final Answer: (C)