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: Understanding the Concept:
The elevation of boiling point ($\Delta T_b$) is a colligative property proportional to the molality ($m$) of the solute.
Step 2: Key Formula or Approach:
$\Delta T_b = K_b \times m$
Where $\Delta T_b = T_b(\text{solution}) - T_b(\text{pure solvent})$.
Step 3: Detailed Explanation:
Given:
Boiling point of solution $= 319.8\text{ K}$
Boiling point of solvent $= 319.5\text{ K}$
Molality ($m$) $= 0.12\text{ m}$
Calculation of $\Delta T_b$:
\[ \Delta T_b = 319.8\text{ K} - 319.5\text{ K} = 0.3\text{ K} \]
Using the formula $\Delta T_b = K_b \times m$:
\[ 0.3 = K_b \times 0.12 \]
\[ K_b = \frac{0.3}{0.12} \]
\[ K_b = \frac{30}{12} = 2.5\text{ K kg mol}^{-1} \]
Step 4: Final Answer:
The molal elevation constant is $2.5\text{ K kg mol}^{-1}$.