Question

Statement-I: \( K_b \) is more than \( K_f \) for water.
Statement-II: When we add a non-volatile solute in water, elevation in boiling point is more than depression in freezing point.

• A. Both Statement I and Statement II are correct.
• B. Statement I is correct but Statement II is incorrect.
• C. Statement I is incorrect but Statement II is correct.
• D. Both Statement I and Statement II are incorrect.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question compares the ebullioscopic constant ($K_b$) and cryoscopic constant ($K_f$) for water and their respective colligative effects for a solution.
Step 2: Detailed Explanation:
For water, the values of the constants are:
$K_b \approx 0.512$ K kg mol$^{-1}$
$K_f \approx 1.86$ K kg mol$^{-1}$
Clearly, $K_b<K_f$ for water. Therefore, Statement-I is incorrect.

For the same molal solution of a non-volatile solute:
Elevation in boiling point ($\Delta T_b$) = $K_b \times m$
Depression in freezing point ($\Delta T_f$) = $K_f \times m$
Since $K_b<K_f$, for the same molality $m$, $\Delta T_b<\Delta T_f$.
Thus, the elevation in boiling point is less than the depression in freezing point. Statement-II is incorrect.
Step 3: Final Answer:
Both Statement I and Statement II are incorrect.