Of the following four aqueous solutions, total number of those solutions whose freezing point is low

Q: Why does ethanol have $i=1$?

Ethanol is a covalent, non-electrolyte molecule. When it dissolves in water, it stays intact as whole molecules rather than breaking apart into ions. Thus, 1 mole of ethanol yields exactly 1 mole of dissolved particles.

Q: How do we find the van't Hoff factor for a salt?

For strong electrolytes in dilute solutions, we assume 100% dissociation. You simply count the total number of cations and anions produced from one formula unit. For example, $\text{KCl}$ gives one $\text{K}^+$ and one $\text{Cl}^-$, so $i=2$.

Q: Does a higher $\Delta T_f$ mean a higher or lower freezing point?

A higher $\Delta T_f$ means a *lower* freezing point. $\Delta T_f$ is the *depression* or the amount the freezing point drops from the pure solvent's freezing point. So, a larger drop means a colder final temperature.

Q: What if the concentrations were different for each solution?

If the molarities were different, you would need to calculate the effective molarity ($i \times M$) for each solution individually and compare those products. The solution with the highest $i \times M$ will have the lowest freezing point.

Q: Are we assuming 100% dissociation here?

Yes! In standard JEE problems involving strong electrolytes like these salts, unless a degree of dissociation ($\alpha$) is explicitly given, we always assume complete 100% dissociation.

Question

Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of $0.10	ext{ M } 	ext{C}_2	ext{H}_5	ext{OH}$ is ………… . (Integer answer)
(i) $0.10	ext{ M } 	ext{Ba}_3(	ext{PO}_4)_2$
(ii) $0.10	ext{ M } 	ext{Na}_2	ext{SO}_4$
(iii) $0.10	ext{ M } 	ext{KCl}$
(iv) $0.10	ext{ M } 	ext{Li}_3	ext{PO}_4$

Sigma Percentile · Accepted Answer

### The Mystery of Freezing Point Depression

Have you ever wondered why we sprinkle salt on icy roads in winter? It's all about **freezing point depression**, a fascinating colligative property.

When we add a solute to a pure solvent, it disrupts the solvent's ability to form a solid lattice, effectively lowering the temperature at which it freezes. The mathematical relationship is beautifully simple:

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

Here, $m$ is the molality, $K_f$ is the cryoscopic constant, and $i$ is the **van't Hoff factor**.

### The Power of the van't Hoff Factor

The van't Hoff factor, $i$, represents the number of particles a solute breaks into when dissolved. Colligative properties depend *only* on the number of particles, not their identity.

A higher value of $i$ means more particles in the solution. More particles lead to a greater depression in the freezing point ($\Delta T_f$), which means the final freezing point ($T_f$) drops even lower.

### Analyzing the Contenders

Let's look at our reference solution: $0.10	ext{ M } 	ext{C}_2	ext{H}_5	ext{OH}$ (Ethanol). Ethanol is a covalent compound. It dissolves but does not dissociate into ions. Therefore, its van't Hoff factor is exactly $1$. Its effective concentration is simply $0.10	ext{ M}$.

Now, let's evaluate the four challengers. They are all strong electrolytes, meaning they dissociate completely in water.

**1. Barium Phosphate:**

$$ 	ext{Ba}_3(	ext{PO}_4)_2 ightarrow 3	ext{Ba}^{2+} + 2	ext{PO}_4^{3-} $$

This yields $5$ ions, so $i = 5$. The effective concentration is $5 	imes 0.10 = 0.50	ext{ M}$.

**2. Sodium Sulfate:**

$$ 	ext{Na}_2	ext{SO}_4 ightarrow 2	ext{Na}^+ + 	ext{SO}_4^{2-} $$

This yields $3$ ions, so $i = 3$. The effective concentration is $3 	imes 0.10 = 0.30	ext{ M}$.

**3. Potassium Chloride:**

$$ 	ext{KCl} ightarrow 	ext{K}^+ + 	ext{Cl}^- $$

This yields $2$ ions, so $i = 2$. The effective concentration is $2 	imes 0.10 = 0.20	ext{ M}$.

**4. Lithium Phosphate:**

$$ 	ext{Li}_3	ext{PO}_4 ightarrow 3	ext{Li}^+ + 	ext{PO}_4^{3-} $$

This yields $4$ ions, so $i = 4$. The effective concentration is $4 	imes 0.10 = 0.40	ext{ M}$.

### The Final Verdict

Every single one of the salt solutions has an effective concentration greater than $0.10	ext{ M}$. Because they produce more particles than ethanol, they will all cause a greater depression in the freezing point.

Consequently, all **four** solutions will have a freezing point lower than that of the ethanol solution. The chemistry of ions perfectly explains this macroscopic phenomenon!

Enter Numerical Value:

Visualized Solution

\text{Understanding Freezing Point Depression}

\text{Role of van't Hoff Factor } (i)

\text{Ethanol: A Non-Electrolyte}

\text{Analyzing } \text{Ba}_3(\text{PO}_4)_2

\text{Analyzing } \text{Na}_2\text{SO}_4

\text{Analyzing KCl}

\text{Analyzing } \text{Li}_3\text{PO}_4

\text{Conclusion}

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Mystery of Freezing Point Depression

The Power of the van't Hoff Factor

Analyzing the Contenders

The Final Verdict

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