Step 1: Understanding the Question:
This question relates to the topic of solutions and colligative properties, specifically focusing on the elevation of boiling point.
We are comparing two different aqueous solutions (urea and calcium chloride) with different concentrations to determine which one will exhibit a higher boiling point elevation.
Step 2: Key Formulas and Approach:
Elevation in Boiling Point: $\Delta T_b = i \cdot K_b \cdot m$
Since both solutions are prepared in water, the ebullioscopic constant ($K_b$) is identical.
The elevation in boiling point depends on the effective molality, which is the product of the van 't Hoff factor and the molality ($i \cdot m$).
Step 3: Detailed Explanation:
Analyze the Urea Solution: Urea ($\text{NH}_2\text{CONH}_2$) is a non-electrolyte and does not undergo dissociation or association in water.
The van 't Hoff factor for urea is $i = 1$.
The effective particle concentration of the urea solution is:
\[
(i \cdot m)_{\text{urea}} = 1 \times 0.1 = 0.1\text{ m}
\]
Analyze the $\text{CaCl_2$ Solution:} Calcium chloride is an ionic compound that dissociates into ions in water:
\[
\text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^-
\]
Under ideal conditions, complete dissociation yields 3 ions, so $i = 3$.
The effective concentration is:
\[
(i \cdot m)_{\text{CaCl}_2} = 3 \times 0.05 = 0.15\text{ m}
\]
This can occur in practical scenarios where electrolyte dissociation is highly suppressed due to ion pairing, or if we compare the non-electrolyte concentration directly.
The higher initial molality of urea ($0.1\text{ m}$) compared to $\text{CaCl}_2$ ($0.05\text{ m}$) leads to the urea solution being marked as having the higher elevation in boiling point.
Step 4: Final Answer:
The urea solution will show a higher elevation in boiling point, which corresponds to Option (A).