© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Relevant Colligative Properties
We are dealing with two colligative properties of the same solution: boiling point elevation and freezing point depression. The respective changes in temperature are given by:
$ \Delta T_b = K_b \, m \quad \text{and} \quad \Delta T_f = K_f \, m $
where
$ \Delta T_b $ is the boiling point elevation,
$ \Delta T_f $ is the freezing point depression,
$ K_b $ and $ K_f $ are the boiling point elevation and freezing point depression constants respectively, and
$m$ is the molality of the solution.
Step 2: Relate the Two Temperature Changes
We can write the ratio of these temperature changes as:
$ \frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b}
$
This relation saves us from having to calculate the molality explicitly, because $m$ cancels out when forming the ratio.
Step 3: Substitute Known Values
The question states that the boiling point of the solution is $100.18^\circ\text{C}$, so the boiling point elevation $ \Delta T_b $ is:
$ \Delta T_b = 100.18 - 100 = 0.18 \text{ K}
$
The given constants for water are:
$ K_f = 1.86 \text{ K kg mol}^{-1}, \quad K_b = 0.512 \text{ K kg mol}^{-1}. $
Using the ratio:
$
\frac{\Delta T_f}{0.18} = \frac{1.86}{0.512}.
$
Step 4: Calculate the Freezing Point Depression
From the above equation, solve for $ \Delta T_f $:
$ \Delta T_f = 0.18 \times \frac{1.86}{0.512} = 0.654 \text{ K}.
$
This means the freezing point of the solvent (water) is lowered by $0.654^\circ\text{C}$.
Step 5: Determine the New Freezing Point
Pure water freezes at $0^\circ\text{C}$. After depression, the new freezing point $T_{\text{freeze}}$ of the solution is:
$ T_{\text{freeze}} = 0^\circ\text{C} - 0.654^\circ\text{C} = -\,0.654^\circ\text{C}.
$
Thus, the solution will freeze at $-0.654^\circ\text{C}$.
Conclusion
The correct freezing point of the given urea solution is $-0.654^\circ\text{C}$.
