Which of the following aqueous solution will have highest elevation of boiling point?

Question

Prepp Team · Accepted Answer

Understanding Boiling Point Elevation in Aqueous Solutions

Boiling point elevation is a colligative property, which means it depends on the number of solute particles present in a solution, not on their chemical identity. When a non-volatile solute is added to a solvent, the boiling point of the solution is higher than that of the pure solvent.

The elevation in boiling point ($\Delta T_b$) is given by the formula:

$\Delta T_b = i \cdot K_b \cdot m$

Where:

$\Delta T_b$ is the boiling point elevation.
$i$ is the van't Hoff factor, representing the number of particles the solute dissociates into in the solution.
$K_b$ is the ebullioscopic constant of the solvent (for water, $K_b$ is approximately $0.512 \text{ }^\circ\text{C/m}$).
$m$ is the molality of the solution (moles of solute per kilogram of solvent).

In this question, all solutions are 1 M aqueous solutions. For dilute aqueous solutions, the molality ($m$) is approximately equal to the molarity ($M$). The solvent is water for all cases, so $K_b$ is the same. Therefore, the boiling point elevation ($\Delta T_b$) will be directly proportional to the van't Hoff factor ($i$) and the molarity ($M$). Since molarity is the same (1 M) for all options, the solution with the highest van't Hoff factor ($i$) will have the highest elevation of boiling point.

Analyzing the van't Hoff Factor (i) for Each Solution

Let's determine the ideal van't Hoff factor ($i$) for each electrolyte based on its dissociation in water:

1 M NaOH: Sodium hydroxide (NaOH) is a strong electrolyte. It dissociates into one sodium ion ($\text{Na}^+$) and one hydroxide ion ($\text{OH}^-$).
$\text{NaOH(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)}$
Number of particles = 1 + 1 = 2. So, $i \approx 2$.
1 M Na$_2$SO$_4$: Sodium sulfate ($\text{Na}_2\text{SO}_4$) is a strong electrolyte. It dissociates into two sodium ions ($\text{Na}^+$) and one sulfate ion ($\text{SO}_4^{2-}$).
$\text{Na}_2\text{SO}_4\text{(aq)} \rightarrow 2\text{Na}^+\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}$
Number of particles = 2 + 1 = 3. So, $i \approx 3$.
1 M NH$_4$NO$_3$: Ammonium nitrate ($\text{NH}_4\text{NO}_3$) is a strong electrolyte. It dissociates into one ammonium ion ($\text{NH}_4^+$) and one nitrate ion ($\text{NO}_3^-$).
$\text{NH}_4\text{NO}_3\text{(aq)} \rightarrow \text{NH}_4^+\text{(aq)} + \text{NO}_3^-\text{(aq)}$
Number of particles = 1 + 1 = 2. So, $i \approx 2$.
1 M KNO$_3$: Potassium nitrate ($\text{KNO}_3$) is a strong electrolyte. It dissociates into one potassium ion ($\text{K}^+$) and one nitrate ion ($\text{NO}_3^-$).
$\text{KNO}_3\text{(aq)} \rightarrow \text{K}^+\text{(aq)} + \text{NO}_3^-\text{(aq)}$
Number of particles = 1 + 1 = 2. So, $i \approx 2$.

Comparing van't Hoff Factors

Let's summarize the approximate van't Hoff factors for each solution:

Solution	Solute Formula	Dissociation	Ideal $i$ Value
1 M NaOH	NaOH	$\text{Na}^+ + \text{OH}^-$	2
1 M Na$_2$SO$_4$	Na$_2$SO$_4$	$2\text{Na}^+ + \text{SO}_4^{2-}$	3
1 M NH$_4$NO$_3$	NH$_4$NO$_3$	$\text{NH}_4^+ + \text{NO}_3^-$	2
1 M KNO$_3$	KNO$_3$	$\text{K}^+ + \text{NO}_3^-$	2

Determining the Highest Boiling Point Elevation

Since the boiling point elevation is directly proportional to the van't Hoff factor $i$ (assuming similar molality), the solution with the highest $i$ will have the highest boiling point elevation. Comparing the $i$ values:

NaOH: $i \approx 2$
Na$_2$SO$_4$: $i \approx 3$
NH$_4$NO$_3$: $i \approx 2$
KNO$_3$: $i \approx 2$

The 1 M Na$_2$SO$_4$ solution has the highest van't Hoff factor ($i \approx 3$). Therefore, it will exhibit the highest elevation of boiling point among the given solutions.

Revision Table: Colligative Properties & Boiling Point Elevation

Concept	Description	Relevance to Boiling Point Elevation
Colligative Properties	Properties of solutions that depend on the ratio of the number of solute particles to the number of solvent particles, not on the nature of the chemical species present.	Boiling point elevation is a colligative property.
Boiling Point Elevation ($\Delta T_b$)	The increase in the boiling point of a solvent when a solute is dissolved in it.	Given by $\Delta T_b = i \cdot K_b \cdot m$.
van't Hoff Factor ($i$)	Ratio of the actual concentration of particles produced when the substance is dissolved to the theoretical concentration predicted by its mass. For ideal electrolytes, it equals the number of ions formed upon dissociation.	Directly proportional to boiling point elevation. Higher $i$ means higher $\Delta T_b$.
Molality ($m$)	Moles of solute per kilogram of solvent.	Concentration term used in the boiling point elevation formula.

Additional Information: Real vs. Ideal van't Hoff Factor

The ideal van't Hoff factors calculated above (2, 3) assume complete dissociation of the electrolytes. In reality, due to interionic attractions, the actual van't Hoff factor ($i_{actual}$) is often slightly less than the ideal van't Hoff factor ($i_{ideal}$), especially at higher concentrations. However, for comparing the relative boiling point elevations of solutions of similar concentration, assuming ideal behavior based on the number of ions formed is a reasonable approach and generally predicts the correct trend.

Prepp Team · Answer

1 M NaOH

Prepp Team · Answer

1 M NH 4 NO 3

Prepp Team · Answer

1 M KNO 3

Solution	Solute Formula	Dissociation	Ideal \(i\) Value
1 M NaOH	NaOH	\(\text{Na}^+ + \text{OH}^-\)	2
1 M Na\(_2\)SO\(_4\)	Na\(_2\)SO\(_4\)	\(2\text{Na}^+ + \text{SO}_4^{2-}\)	3
1 M NH\(_4\)NO\(_3\)	NH\(_4\)NO\(_3\)	\(\text{NH}_4^+ + \text{NO}_3^-\)	2
1 M KNO\(_3\)	KNO\(_3\)	\(\text{K}^+ + \text{NO}_3^-\)	2