Question

Calculate the molal elevation constant of water if boiling point of 0.7 m solution of non volatile solute is 373.5 K .

[Boiling point of water \(=373.15 \mathrm{~K}\) ]

• A. \(0.4\text{ }\mathrm{K}\text{ }\mathrm{kg}\text{ }\mathrm{mol}^{-1}\)
• B. \(0.25\text{ }\mathrm{K}\text{ }\mathrm{kg}\text{ }\mathrm{mol}^{-1}\)
• C. \(0.5\text{ }\mathrm{K}\text{ }\mathrm{kg}\text{ }\mathrm{mol}^{-1}\)
• D. \(0.60\text{ }\mathrm{K}\text{ }\mathrm{kg}\text{ }\mathrm{mol}^{-1}\)

Hint:

Use the formula elevation in boiling point equals Kb times molality.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a non-volatile solute, \(\Delta T_b = K_b \, m\).

Step 2: Elevation in boiling point
\(\Delta T_b = 373.5 - 373.15 = 0.35\ \text{K}\).

Step 3: Calculation
\(K_b = \dfrac{\Delta T_b}{m} = \dfrac{0.35}{0.7} = 0.5\ \text{K kg mol}^{-1}\).

Final Answer:
The molal elevation constant is 0.5 K kg per mole. \[ \boxed{\text{(C) }0.5\text{ K kg mol}^{-1}} \]