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

To find the elevation in boiling point, we need to follow these steps and use the required formulae for colligative properties. The given problem involves both freezing point depression and boiling point elevation.

Step 1: Understand the Problem

Given data:

• Freezing point of solution: \(-0.186^\circ\text{C}\)
• Freezing point depression constant, \(K_f = 1.86 \, \text{K kg/mol}\)
• Boiling point elevation constant, \(K_b = 0.512 \, \text{K kg/mol}\)

Step 2: Calculate Molality

Use the formula for freezing point depression:

\(\Delta T_f = i \cdot K_f \cdot m\)

Here, \(\Delta T_f = 0.186^\circ \text{C}\), and assuming complete dissociation or non-electrolytic solution, \(i = 1\) (since the van 't Hoff factor is not provided, assume simplest case).

Rearranging to find molality, \(m\):

\(m = \frac{\Delta T_f}{K_f}\) \(m = \frac{0.186}{1.86}\) \(m = 0.1 \, \text{mol/kg}\)

Step 3: Calculate Boiling Point Elevation

Use the formula for boiling point elevation:

\(\Delta T_b = i \cdot K_b \cdot m\)

Substituting the known values (again assuming \(i = 1\)), we get:

\(\Delta T_b = 1 \cdot 0.512 \cdot 0.1\) \(\Delta T_b = 0.0512^\circ \text{C}\)

Conclusion

The elevation in boiling point is \(0.0512^\circ \text{C}\).

Thus, the correct answer is \(0.0512^\circ\text{C}\).