Question:
\(0.1\text{ m}\) of urea and \(0.05\text{ m}\) of \(\text{CaCl}_2\) are dissolved separately in equal volumes of water. Which solution will have higher elevation in boiling point?
\(0.1\text{ m}\) of urea and \(0.05\text{ m}\) of \(\text{CaCl}_2\) are dissolved separately in equal volumes of water. Which solution will have higher elevation in boiling point?
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To prevent an easy miscalculation, always compare the final product of \((i \times m)\) directly instead of looking at the raw molalities. Even though the urea solution has double the starting concentration (\(0.1\text{ m}\) vs \(0.05\text{ m}\)), the triple ion generation of \(\text{CaCl}_2\) completely overtakes it.
Updated On: May 25, 2026
- Urea solution
- $\text{CaCl}_2$ solution
- Both will show equal elevation
- None will show elevation
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The Correct Option is A
Solution and Explanation
Concept:
Elevation in boiling point (\(\Delta T_b\)) is a colligative property, meaning it depends directly on the total number of solute particles present in the solution rather than the identity of the solute.
The mathematical relationship is given by the formula:
\[
\Delta T_b = i \cdot K_b \cdot m
\]
Where:
• \(i\) = van 't Hoff factor (total number of particles formed after dissociation or association)
• \(K_b\) = Molal boiling point elevation constant (or ebullioscopic constant of the solvent)
• \(m\) = Molality of the solution (\(\text{moles of solute / kg of solvent}\)) Since both solutes are dissolved in the same solvent (water), \(K_b\) remains constant. Therefore, the elevation in boiling point is directly proportional to the effective particle concentration: \[ \Delta T_b \propto (i \cdot m) \]
Step 1: Evaluating the effective concentration ($i \cdot m$) for Urea solution.
Urea (\(\text{NH}_2\text{CONH}_2\)) is an organic non-electrolyte compound. It does not dissociate or associate when dissolved in water: \[ i_{\text{urea}} = 1 \] Given molality \(m_{\text{urea}} = 0.1\text{ m}\). Calculating the effective particle factor: \[ (i \cdot m)_{\text{urea}} = 1 \times 0.1 = 0.1\text{ m} \]
Step 2: Evaluating the effective concentration ($i \cdot m$) for $\text{CaCl}_2$ solution.
Calcium chloride (\(\text{CaCl}_2\)) is an ionic salt that undergoes complete dissociation in aqueous solutions: \[ \text{CaCl}_2(aq) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq) \] Counting the total moles of ions produced per formula unit: 1 calcium ion + 2 chloride ions = 3 ions. \[ i_{\text{CaCl}_2} = 3 \] Given molality \(m_{\text{CaCl}_2} = 0.05\text{ m}\). Calculating the effective particle factor: \[ (i \cdot m)_{\text{CaCl}_2} = 3 \times 0.05 = 0.15\text{ m} \]
Step 3: Comparing the values to find the higher elevation.
Comparing the effective particle concentrations calculated: \[ (i \cdot m)_{\text{CaCl}_2} = 0.15\text{ m} \quad > \quad (i \cdot m)_{\text{urea}} = 0.1\text{ m} \] Since \(0.15 > 0.1\), the \(\text{CaCl}_2\) solution contains a higher absolute concentration of solute particles in the solvent, thereby leading to a greater elevation in its boiling point: \[ \Delta T_{b_{\text{CaCl}_2}} > \Delta T_{b_{\text{urea}}} \]
• \(i\) = van 't Hoff factor (total number of particles formed after dissociation or association)
• \(K_b\) = Molal boiling point elevation constant (or ebullioscopic constant of the solvent)
• \(m\) = Molality of the solution (\(\text{moles of solute / kg of solvent}\)) Since both solutes are dissolved in the same solvent (water), \(K_b\) remains constant. Therefore, the elevation in boiling point is directly proportional to the effective particle concentration: \[ \Delta T_b \propto (i \cdot m) \]
Step 1: Evaluating the effective concentration ($i \cdot m$) for Urea solution.
Urea (\(\text{NH}_2\text{CONH}_2\)) is an organic non-electrolyte compound. It does not dissociate or associate when dissolved in water: \[ i_{\text{urea}} = 1 \] Given molality \(m_{\text{urea}} = 0.1\text{ m}\). Calculating the effective particle factor: \[ (i \cdot m)_{\text{urea}} = 1 \times 0.1 = 0.1\text{ m} \]
Step 2: Evaluating the effective concentration ($i \cdot m$) for $\text{CaCl}_2$ solution.
Calcium chloride (\(\text{CaCl}_2\)) is an ionic salt that undergoes complete dissociation in aqueous solutions: \[ \text{CaCl}_2(aq) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq) \] Counting the total moles of ions produced per formula unit: 1 calcium ion + 2 chloride ions = 3 ions. \[ i_{\text{CaCl}_2} = 3 \] Given molality \(m_{\text{CaCl}_2} = 0.05\text{ m}\). Calculating the effective particle factor: \[ (i \cdot m)_{\text{CaCl}_2} = 3 \times 0.05 = 0.15\text{ m} \]
Step 3: Comparing the values to find the higher elevation.
Comparing the effective particle concentrations calculated: \[ (i \cdot m)_{\text{CaCl}_2} = 0.15\text{ m} \quad > \quad (i \cdot m)_{\text{urea}} = 0.1\text{ m} \] Since \(0.15 > 0.1\), the \(\text{CaCl}_2\) solution contains a higher absolute concentration of solute particles in the solvent, thereby leading to a greater elevation in its boiling point: \[ \Delta T_{b_{\text{CaCl}_2}} > \Delta T_{b_{\text{urea}}} \]
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