Q.On the basis of information given below mark the correct option. Information: On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.

The key idea is the relationship between intermolecular forces, deviation from Raoult's law, and azeotrope formation.

1. Methanol molecules are associated through strong hydrogen bonds. When acetone is added, these strong methanol-methanol hydrogen bonds are disrupted.
2. This disruption means the new intermolecular forces between methanol and acetone molecules (A-B interactions) are weaker than the original methanol-methanol hydrogen bonds (A-A interactions). …

The question asks us to determine the behavior of a methanol-acetone mixture based on the information that "On adding acetone to methanol some of the hydrogen bonds between methanol molecules break." To answer this, we need to understand how the disruption of hydrogen bonds affects intermolecular forces, vapor pressure, deviation from Raoult's law, and azeotrope formation.

Concept and Intuition: Molality Calculation

Methanol ($\text{CH}_3\text{OH}$) is a highly polar molecule capable of forming strong intermolecular hydrogen bonds. The hydrogen atom attached to the electronegative oxygen atom in one methanol molecule can form a strong attractive interaction with the oxygen atom of another methanol molecule. These hydrogen bonds are significant intermolecular forces that hold methanol molecules together in the liquid phase.

Acetone ($\text{CH}_3\text{COCH}_3$) is also a polar molecule due to its carbonyl group ($\text{C=O}$). However, acetone molecules primarily interact through dipole-dipole forces and London dispersion forces. While acetone can act as a hydrogen bond acceptor (via its oxygen atom), it cannot donate a hydrogen bond because it lacks a hydrogen atom directly bonded to an oxygen, nitrogen, or fluorine atom.

When acetone is added to methanol, the acetone molecules insert themselves between the methanol molecules. The new interactions formed between methanol and acetone molecules are primarily dipole-dipole interactions and some weaker hydrogen bonds (where methanol acts as donor and acetone as acceptor). Crucially, these new methanol-acetone interactions are weaker than the strong hydrogen bonds that existed between pure methanol molecules. The overall effect is a reduction in the average intermolecular forces within the mixture compared to pure methanol.

Weaker intermolecular forces mean that molecules require less energy to escape from the liquid phase into the vapor phase. This directly translates to a higher tendency for the solution to vaporize, resulting in a higher vapor pressure than would be expected for an ideal solution. This higher vapor pressure is the hallmark of a positive deviation from Raoult's law.

A solution that exhibits a positive deviation from Raoult's law has a higher vapor pressure than predicted. If the vapor pressure is sufficiently high at a specific composition, the solution will boil at a temperature lower than the boiling point of either pure component. Such a mixture is called a minimum boiling azeotrope.

Step-by-step Reasoning

1. Analyze Intermolecular Forces in Pure Methanol:

   Methanol ($\text{CH}_3\text{OH}$) molecules are extensively hydrogen-bonded to each other. The hydrogen atom attached to the oxygen in one methanol molecule forms a strong attractive force with the oxygen atom of an adjacent methanol molecule. These strong hydrogen bonds contribute significantly to methanol's relatively high boiling point and low vapor pressure compared to non-hydrogen-bonded compounds of similar molar mass.

2. Analyze the Effect of Adding Acetone:

   The problem states that "On adding acetone to methanol some of the hydrogen bonds between methanol molecules break." This is the key piece of information. When acetone molecules are introduced, they disrupt the strong methanol-methanol hydrogen bonds. While new interactions form between methanol and acetone, these interactions (primarily dipole-dipole and weaker hydrogen bonds where acetone is the acceptor) are not as strong as the original methanol-methanol hydrogen bonds. Therefore, the overall intermolecular attractive forces in the methanol-acetone mixture are weaker than the average of the intermolecular forces in the pure components.

3. Relate Intermolecular Forces to Vapor Pressure: …

Concept: Azeotropes and Deviations from Raoult’s Law

Method: Intermolecular Forces Analysis

Step 1: Identify the type of interaction in the pure liquids.

• Pure methanol ($CH_3OH$) molecules are held together by strong hydrogen bonds.
• Pure acetone ($CH_3COCH_3$) molecules have dipole-dipole interactions (no hydrogen bonding between themselves).

Step 2: Predict what happens when they are mixed.

• When acetone is added to methanol, acetone molecules break the hydrogen bonds between methanol molecules.
• The new interactions (methanol–acetone) are weaker than the original methanol–methanol hydrogen bonds.

Step 3: Relate to Raoult’s law.

• Weaker A–B interactions compared to A–A and B–B interactions mean:
  • Molecules escape more easily into vapour phase.
  • Vapour pressure of the mixture is higher than predicted by Raoult’s law. …

Here’s a breakdown of the common mistakes students make on this concept and how to avoid each.

Mistake 1: Confusing the direction of deviation with the type of azeotrope

The error:

Students often think that positive deviation always means a maximum boiling azeotrope, or that negative deviation always means a minimum boiling azeotrope. This is exactly backwards.

Why it happens:

They memorise the words “positive” and “maximum” as if they go together, but the logic is the opposite:

• Positive deviation → vapour pressure is higher than ideal → mixture boils at a lower temperature → minimum boiling azeotrope.
• Negative deviation → vapour pressure is lower than ideal → mixture boils at a higher temperature → maximum boiling azeotrope.

How to avoid:

Always connect the physical reason to the outcome:

• If A–B interactions are weaker than A–A and B–B (like here: H-bonds break), molecules escape more easily → higher vapour pressure → positive deviation → boils at lower T → minimum boiling azeotrope.
• If A–B interactions are stronger, molecules are held tighter → lower vapour pressure → negative deviation → boils at higher T → maximum boiling azeotrope.

Key link:

Weaker interactions → positive deviation → minimum boiling azeotrope.

Stronger interactions → negative deviation → maximum boiling azeotrope.

Mistake 2: Misinterpreting the given information about hydrogen bonds

The error:

Students read “some of the hydrogen bonds between methanol molecules break” and think this means the mixture is more stable or more attracted to each other. They then incorrectly choose negative deviation.

Why it happens:

Breaking bonds sounds like “energy is released” or “something is formed”, but here it means the methanol–methanol attraction is lost without a strong enough acetone–methanol attraction to replace it. The net interaction becomes weaker.

How to avoid:

Always ask: Are the new A–B interactions stronger or weaker than the original A–A and B–B interactions?

• Methanol has strong H-bonds with itself.
• Acetone can only form weaker dipole-dipole or weak H-bonds (C=O with O–H).
• When you add acetone, you break strong H-bonds and replace them with weaker interactions.
• Result: molecules escape more easily → positive deviation.

Rule of thumb:

If the information says “bonds break” or “interactions weaken”, it’s almost always positive deviation.

Mistake 3: Forgetting that azeotrope formation requires a specific composition

The error:

Students think that any mixture showing deviation will form an azeotrope. They don’t check whether the deviation is large enough to cause a minimum or maximum in the boiling point curve.

Why it happens:

Many textbooks state that “mixtures showing positive deviation form minimum boiling azeotropes” without emphasising that this happens only at a specific composition where the vapour and liquid compositions become equal.

How to avoid: …