Which one of the following aqueous solutions will exhibit highest boiling point?
(a) 0.01 M Na2SO4
(

Q: Why do we multiply concentration by the van't Hoff factor?

The concentration given is for the formula unit. However, colligative properties depend on the *total* number of particles. The van't Hoff factor ($i$) tells us how many particles one formula unit breaks into, giving us the true 'effective' concentration.

Q: Why is $i=1$ for urea and glucose?

Urea and glucose are covalent molecular compounds that dissolve as intact molecules. They do not break apart into ions in water, so one molecule yields exactly one dissolved particle.

Q: Does the identity of the ions matter for the boiling point?

No! Colligative properties are 'democratic.' A sodium ion has the exact same effect on the boiling point as a massive sulfate ion or a tiny chloride ion. Only the *number* of particles matters.

Q: What if the concentrations were all identical?

If all solutions had the same concentration, you would simply look for the compound with the highest van't Hoff factor ($i$). In this case, $Na_2SO_4$ ($i=3$) would still win.

Q: Would the answer change if we were asked for the lowest freezing point?

No. Freezing point depression ($\Delta T_f$) is also proportional to $i \times C$. The solution with the highest effective concentration will have the most *depressed* (lowest) freezing point.

Question

Which one of the following aqueous solutions will exhibit highest boiling point?

Options:

(A) 0.01 M $Na_2SO_4$

(B) 0.01 M $KNO_3$

(C) 0.015 M urea

(D) 0.015 M glucose

Sigma Percentile · Accepted Answer

### The Battle of Boiling Points: Unmasking Colligative Properties

Imagine you are a chef trying to boil water for pasta. You throw in a handful of salt, and suddenly, the water stops boiling for a moment. Why? Because you've just witnessed a **colligative property** in action! Colligative properties are fascinating because they don't care *what* you dissolve in the solvent; they only care about *how much* of it is dissolved.

In this problem, we are asked to find which of the four given solutions will have the highest boiling point. To solve this, we need to understand the mathematics of boiling point elevation.

### The Master Equation

The elevation in boiling point, denoted as $\Delta T_b$, is given by the formula:

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

Here, $K_b$ is the molal elevation constant (which is the same for all aqueous solutions since the solvent is water), $m$ is the molality (which is approximately equal to molarity $C$ for dilute solutions), and $i$ is the **van't Hoff factor**.

Because $K_b$ is a constant, the elevation in boiling point is directly proportional to the product of the van't Hoff factor and the concentration:

$$\Delta T_b \propto i 	imes C$$

This product, $i 	imes C$, is what we call the **effective concentration** of particles in the solution. The solution with the highest effective concentration will have the highest boiling point.

### Counting the Pieces: The van't Hoff Factor

The van't Hoff factor ($i$) is simply the number of particles a single formula unit breaks into when dissolved. Let's analyze our contenders:

**1. Sodium Sulfate ($Na_2SO_4$)**
This is a strong electrolyte. When it hits the water, it shatters completely:

$$Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-}$$

One molecule yields 3 ions. Therefore, $i = 3$. 
Its effective concentration is $3 	imes 0.01 = \mathbf{0.03 	ext{ M}}$.

**2. Potassium Nitrate ($KNO_3$)**
Another strong electrolyte, but it breaks into fewer pieces:

$$KNO_3 ightarrow K^+ + NO_3^-$$

One molecule yields 2 ions. Therefore, $i = 2$.
Its effective concentration is $2 	imes 0.01 = \mathbf{0.02 	ext{ M}}$.

**3. Urea and Glucose**
These are covalent, organic molecules. They are non-electrolytes, meaning they dissolve as intact, whole molecules. They do not dissociate into ions. Therefore, for both of them, $i = 1$.
Their effective concentration is simply $1 	imes 0.015 = \mathbf{0.015 	ext{ M}}$.

### The Grand Finale

Now, we just compare the effective concentrations we calculated:

$$0.03 > 0.02 > 0.015$$

Sodium sulfate ($Na_2SO_4$) generates the highest concentration of actual particles floating around in the water. Because it has the most particles, it disrupts the escape of water molecules into the vapor phase the most, requiring the highest temperature to finally boil.

Thus, the $0.01 	ext{ M } Na_2SO_4$ solution claims the crown for the highest boiling point!

Animated Solution for Chemistry - Solutions: Which one of the following aqueous solutions will exhibit highest boiling point?

Select Answer:

Visualized Solution

$Δ T_{b} = i \cdot K_{b} \cdot m$

$van’t Hoff Factor (i)$

$Option (a): Na_{2} SO_{4}$

$Option (b): KNO_{3}$

$Options (c) & (d): Urea and Glucose$

$Conclusion$

$The Way Forward$

The Sigma Insight: Colligative Properties

The Battle of Boiling Points

The Master Equation

Counting the Pieces

The Grand Finale

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Animated Solution for Chemistry - Solutions: Which one of the following aqueous solutions will exhibit highest boiling point?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔTb​=i⋅Kb​⋅m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } van’t Hoff Factor (i)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Option (a): Na2​SO4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Option (b): KNO3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Options (c) & (d): Urea and Glucose

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward