Q.Which of the following statements is false?

The key idea is that colligative properties, while dependent on the number of solute particles, also depend on the nature of the solvent.

1. The depression in freezing point ($\Delta T_f$) is given by the formula $\Delta T_f = i \cdot K_f \cdot m$.
2. For sucrose, a non-electrolyte, the van't Hoff factor $i$ is 1. The molality $m$ is stated to be the same for both solutions.
3. However, $K_f$ (the cryoscopic constant) is a characteristic property of the solvent. …

Concept and Intuition

Colligative properties are physical properties of solutions that depend only on the number of solute particles present, not on their identity. These properties include depression in freezing point, elevation in boiling point, relative lowering of vapor pressure, and osmotic pressure. For electrolyte solutions, the solute dissociates into multiple ions, increasing the effective number of particles, which is accounted for by the van't Hoff factor ($i$). Crucially, the constants associated with these properties (like the cryoscopic constant $K_f$ or ebullioscopic constant $K_b$) are characteristic properties of the solvent itself.

We will analyze each statement based on these principles and the relevant formulas.

Step-by-Step Analysis

1. Analyze Statement (i): Two different solutions of sucrose of same molality prepared in different solvents will have the same depression in freezing point.

   • The depression in freezing point ($\Delta T_f$) is given by the formula:

     $\Delta T_f = i K_f m$

     where $i$ is the van't Hoff factor, $K_f$ is the molal depression constant (cryoscopic constant), and $m$ is the molality of the solution.

   • For sucrose, which is a non-electrolyte, it does not dissociate in solution, so its van't Hoff factor $i=1$.
   • The statement specifies that the molality ($m$) is the same for both solutions.
   • However, the solutions are prepared in different solvents. The cryoscopic constant ($K_f$) is a characteristic property of the solvent. Different solvents have different $K_f$ values. For example, $K_f$ for water is $1.86 \text{ K kg/mol}$, while for benzene it is $5.12 \text{ K kg/mol}$.
   • Since $K_f$ will be different for different solvents, even with the same $i$ and $m$, the depression in freezing point ($\Delta T_f$) will be different.
   • Therefore, statement (i) is false.

2. Analyze Statement (ii): The osmotic pressure of a solution is given by the equation $\Pi = CRT$ (where C is the molarity of the solution).

   • The van't Hoff equation for osmotic pressure is generally written as:

     $\Pi = i CRT$

     where $\Pi$ is the osmotic pressure, $i$ is the van't Hoff factor, $C$ is the molarity, $R$ is the gas constant, and $T$ is the absolute temperature.

   • $\Pi = CRT$ is the standard van't Hoff relation for osmotic pressure as introduced in the NCERT syllabus (drawn by analogy with the ideal gas equation). The correction factor $i$ is a separate refinement applied specifically to solutions of electrolytes that dissociate or non-electrolytes that associate — it does not make the base equation itself false.
   • Therefore, statement (ii) is true.

3. Analyze Statement (iii): Decreasing order of osmotic pressure for 0.01 M aqueous solutions of barium chloride, potassium chloride, acetic acid and sucrose is $BaCl_2 > KCl > CH_3COOH > sucrose$.

   • For solutions of the same molarity ($C$), temperature ($T$), and solvent, the osmotic pressure ($\Pi = i CRT$) is directly proportional to the van't Hoff factor ($i$). We need to determine $i$ for each substance:
     • Barium chloride ($BaCl_2$): A strong electrolyte, it dissociates completely into $Ba^{2+}$ and $2Cl^-$. So, $i = 1+2 = 3$.
     • Potassium chloride ($KCl$): A strong electrolyte, it dissociates completely into $K^+$ and $Cl^-$. So, $i = 1+1 = 2$.
     • Acetic acid ($CH_3COOH$): A weak electrolyte, it partially dissociates into $CH_3COO^-$ and $H^+$. Therefore, its van't Hoff factor $i$ will be greater than 1 but less than 2 ($1 < i < 2$).
     • Sucrose: A non-electrolyte, it does not dissociate. So, $i = 1$. …

Method: Conceptual Elimination with Henry’s Law & Colligative Properties

This is a statement-verification problem. The goal is to identify the false statement by checking each option against known principles.

Step 1 — Understand the concept behind each option

• Option (i) — Depression in freezing point depends on molality and the cryoscopic constant ($K_f$) of the solvent.

  $K_f$ is solvent-specific.

  → Same molality in different solvents → different $\Delta T_f$.

• Option (ii) — Osmotic pressure formula:

  $\Pi = iCRT$ (for electrolytes), but for non-electrolytes $i = 1$, so $\Pi = CRT$ is correct for ideal dilute solutions.

• Option (iii) — Osmotic pressure depends on number of particles (van’t Hoff factor $i$).

  For 0.01 M solutions:

  • $BaCl_2 \rightarrow 3$ ions ($i \approx 3$)
  • $KCl \rightarrow 2$ ions ($i \approx 2$)
  • $CH_3COOH$ — weak acid, $i$ slightly > 1
  • Sucrose — non-electrolyte, $i = 1$ → Order: $BaCl_2 > KCl > CH_3COOH > sucrose$ is correct.

• Option (iv) — Raoult’s law: …

Here is a breakdown of the common mistakes students make with this specific Henry’s Law / Colligative Properties question, and how to avoid them.

Mistake 1: Confusing Henry’s Law with Raoult’s Law (Statement D)

The Mistake:

Students often think statement (iv) is false because they mix up the formulas. They remember Henry’s Law ($P = K_H \cdot x$) and think Raoult’s Law is different. They might incorrectly assume Raoult’s Law applies only to non-volatile solutes.

Why it happens:

Both laws have the form $P \propto x$ (pressure is proportional to mole fraction). The difference is subtle:

• Raoult’s Law: $P = P^0 \cdot x$ (for a volatile component in an ideal solution; $P^0$ is the vapour pressure of the pure component).
• Henry’s Law: $P = K_H \cdot x$ (for a gas dissolved in a liquid; $K_H$ is Henry’s constant).

How to Avoid:

• Memorize the keyword: Raoult’s law specifically mentions a volatile component of a solution. Statement (iv) says exactly that. Therefore, it is true.
• Create a comparison table: | Law | Applies to | Constant | | :--- | :--- | :--- | | Raoult’s Law | Volatile solvent in a solution | $P^0$ (vapour pressure of pure solvent) | | Henry’s Law | Gas dissolved in a liquid | $K_H$ (Henry’s constant) |

Mistake 2: Forgetting the van’t Hoff Factor ($i$) for Osmotic Pressure (Statement C)

The Mistake:

Students compare osmotic pressure ($\Pi$) using only molarity ($C$). They see all are 0.01 M and think they are equal, or they rank them incorrectly (e.g., putting $KCl$ above $BaCl_2$).

Why it happens:

The formula $\Pi = iCRT$ is often remembered as $\Pi = CRT$ (Statement B). Students forget that $i$ (the number of particles the solute dissociates into) is the most important factor for comparing solutions of the same molarity.

How to Avoid:

• Always calculate $i$ first. For 0.01 M solutions:
  • $BaCl_2$ $\rightarrow Ba^{2+} + 2Cl^-$ (3 ions) $\Rightarrow i \approx 3$
  • $KCl$ $\rightarrow K^+ + Cl^-$ (2 ions) $\Rightarrow i \approx 2$
  • $CH_3COOH$ (weak acid) $\rightarrow$ partial dissociation $\Rightarrow i \approx 1 + \alpha$ (slightly > 1)
  • Sucrose (non-electrolyte) $\rightarrow$ no dissociation $\Rightarrow i = 1$
• Rank by $i$: Higher $i$ means higher $\Pi$. The order $3 > 2 > \sim 1.1 > 1$ is correct. So statement (iii) is true.

Mistake 3: Assuming “Same Molality” Means “Same Solvent” (Statement A)

The Mistake:

Students think that if two solutions have the same molality ($m$), they must have the same freezing point depression ($\Delta T_f$). They ignore the solvent’s identity.

Why it happens:

The formula is $\Delta T_f = i \cdot K_f \cdot m$. Students focus on $m$ and forget that $K_f$ (cryoscopic constant) is a property of the solvent. Water has $K_f = 1.86 \text{ K kg mol}^{-1}$, while benzene has $K_f = 5.12 \text{ K kg mol}^{-1}$.

How to Avoid:

• Read the question carefully: It explicitly says “prepared in different solvents.”
• Apply the formula: If $K_f$ is different, $\Delta T_f$ will be different even if $m$ is the same. Therefore, statement (i) is false.

Mistake 4: Misinterpreting the “False” Statement (Statement B)

The Mistake: …