Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depressi

Q: Why don't glucose and glycine dissociate?

Glucose and glycine are primarily covalent molecular compounds. When they dissolve in water, they separate into individual molecules, but those molecules do not break apart into charged ions.

Q: Does $\text{HSO}_4^-$ dissociate further?

While $\text{HSO}_4^-$ is a weak acid and can dissociate slightly into $\text{H}^+$ and $\text{SO}_4^{2-}$, the primary strong dissociation into $\text{K}^+$ and $\text{HSO}_4^-$ already gives $i \approx 2$. This is already higher than the $i=1$ of the other options, making it the clear winner.

Q: What exactly is the van't Hoff factor ($i$)?

It is the ratio of the actual number of particles in solution after dissociation (or association) to the number of formula units initially dissolved. For a non-electrolyte, $i=1$. For a salt like $\text{NaCl}$, $i=2$.

Q: Why does a higher number of particles cause a larger freezing point depression?

Solute particles disrupt the formation of the solid solvent lattice. More particles mean more disruption, requiring a lower temperature to force the solvent molecules into a solid structure.

Q: Would the answer change if the concentrations were different?

Absolutely! The total freezing point depression depends on the product of $i$ and $m$. If a non-electrolyte had a much higher concentration than the salt, it could potentially cause a larger depression.

Question

Which one of the following $0.10	ext{ M}$ aqueous solutions will exhibit the largest freezing point depression?

Options:

(A) Hydrazine

(B) Glucose

(C) Glycine

(D) $	ext{KHSO}_4$

Sigma Percentile · Accepted Answer

### The Core Concept: Colligative Properties

Imagine you are trying to freeze a pure solvent, like water. The molecules neatly arrange themselves into a solid crystal lattice at exactly $0^\circ	ext{C}$. Now, what happens if you throw some solute particles into the mix? These foreign particles get in the way, disrupting the formation of the lattice. To force the solvent molecules to freeze despite this disruption, you have to lower the temperature even further. This phenomenon is known as **freezing point depression**.

The beauty of freezing point depression is that it is a **colligative property**. This means it doesn't care *what* the solute particles are—it only cares *how many* there are. The mathematical relationship is elegantly simple:

$$\Delta T_f = i \cdot K_f \cdot m$$

Here, $\Delta T_f$ is the depression in freezing point, $K_f$ is the cryoscopic constant of the solvent, $m$ is the molality of the solution, and $i$ is the **van't Hoff factor**.

### Analyzing the Contenders

In our problem, we are given four different $0.10	ext{ M}$ aqueous solutions. Since the concentration ($m$) is identical for all of them, and they are all in the same solvent (water, so $K_f$ is constant), the freezing point depression will be directly proportional to the van't Hoff factor:

$$\Delta T_f \propto i$$

The van't Hoff factor ($i$) represents the number of particles a solute breaks into when dissolved. Let's evaluate our options:

1. **Hydrazine, Glucose, and Glycine:** These are primarily covalent, molecular compounds. When you dissolve them in water, they separate into individual molecules, but those molecules remain intact. They do not dissociate into ions. Therefore, one formula unit yields exactly one particle in solution. For all of these, **$i = 1$**.

2. **Potassium Hydrogen Sulfate ($	ext{KHSO}_4$):** This is an ionic salt. The moment it hits the water, the ionic bonds shatter, and it completely dissociates into its constituent ions:

$$	ext{KHSO}_4 \longrightarrow 	ext{K}^+ + 	ext{HSO}_4^-$$

For every single formula unit of $	ext{KHSO}_4$ you dissolve, you get *two* distinct particles floating around in the solution. Therefore, for this salt, **$i = 2$**.

### The Winning Solution

Since $	ext{KHSO}_4$ produces twice as many particles in solution compared to the other options, it will cause twice the amount of disruption to the freezing process. Consequently, it will exhibit the **largest freezing point depression**.

This is a classic JEE concept: always check if your solute is an electrolyte or a non-electrolyte before comparing colligative properties!

Animated Solution for Chemistry - Solutions: Which one of the following $0.10 M$ aqueous solutions will exhibit the largest freezing point depression?

Select Answer:

Visualized Solution

\Delta T_f = i \cdot K_f \cdot m

\Delta T_f \propto i

i = 1

i = 2

\text{Maximum } \Delta T_f

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Core Concept

Analyzing the Contenders

The Winning Solution

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Animated Solution for Chemistry - Solutions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?