Question

An aqueous solution freezes at -0.186°C, then elevation in boiling point is (\(K_b = 0.512\), \(K_f = 1.86\))

• A. 0.0512°C
• B. 100.0512°C
• C. -0.0512°C
• D. None of these

Hint:

Freezing point depression and boiling point elevation are colligative properties.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
\(\Delta T_f = K_f \cdot m\), \(\Delta T_b = K_b \cdot m\).
Step 2: Detailed Explanation:
\(\Delta T_f = 0.186°C\)
\(m = 0.186/1.86 = 0.1\ \mathrm{molal}\)
\(\Delta T_b = 0.512 \times 0.1 = 0.0512°C\)
Step 3: Final Answer:
Elevation is 0.0512°C.