For an aqueous solution, the freezing point is – 0.186°C. Elevation of the boiling point of the same solution is (\(K_f = \)1.86° mol–1 kg and \(K_b = \)0.512° mol–1 kg)
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 0.0512°.
To determine the elevation of the boiling point (\(\Delta T_b\)) of the solution, we can use the relationship between the depression of the freezing point (\(\Delta T_f\)) and the elevation of the boiling point. Given the constants \(K_f\) and \(K_b\) for water, we can calculate the molality of the solution from the freezing point depression and then use this to find the boiling point elevation.
Given:
Depression of freezing point, \(\Delta T_f = 0.186^\circ \text{C}\)
Freezing point depression constant, \(K_f = 1.86^\circ \text{C mol}^{-1} \text{kg}\)
Using the formula for freezing point depression:
\(\Delta T_f = K_f \times m\)
Solving for molality (\(m\)):
\(m = \frac{\Delta T_f}{K_f} = \frac{0.186}{1.86} = 0.1 \text{ mol/kg}\)
Given:
Boiling point elevation constant, \(K_b = 0.512^\circ \text{C mol}^{-1} \text{kg}\)
Molality (\(m\)) from the previous step is \(0.1 \text{ mol/kg}\)
Using the formula for boiling point elevation:
\(\Delta T_b = K_b \times m = 0.512 \times 0.1 = 0.0512^\circ \text{C}\)
Thus, the elevation of the boiling point for the given aqueous solution is \(0.0512^\circ \text{C}\).
Therefore, the correct answer is: 0.0512°