On the basis of information given below mark the correct option. Information: (A) In bromoethane and chloroeth

NCERT Exemplar · Q21

Q. On the basis of information given below mark the correct option. Information: (A) In bromoethane and chloroethane mixture intermolecular interactions of A-A and B-B type are nearly same as A-B type interactions. (B) In ethanol and acetone mixture A-A or B-B type intermolecular interactions are stronger than A-B type interactions. (C) In chloroform and acetone mixture A-A or B-B type intermolecular interactions are weaker than A-B type interactions.

(i) Solution

(ii) and

(iii) will follow Raoult's law.

(ii) Solution

(i) will follow Raoult's law.

(iii) Solution

(ii) will show negative deviation from Raoult's law.

(iv) Solution

(iii) will show positive deviation from Raoult's law.

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Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

Imagine you're making a cup of sweet tea. You add sugar to hot water, stir, and taste. If you let the tea cool to room temperature, the amount of sugar hasn't changed — but the volume of the liquid has shrunk slightly. If you measured concentration as "grams of sugar per litre of solution," that number would change just because the temperature changed. That's annoying if you're a chemist who needs a reliable, temperature-independent way to describe how much solute is present.

Molality was invented to solve exactly this problem.

The Intuition

Instead of measuring the volume of the solution (which expands and contracts with temperature), molality measures the mass of the solvent. Mass doesn't change with temperature. So molality gives you a concentration that stays the same whether your solution is hot or cold.

Think of it this way:

Molarity = moles of solute per litre of solution (temperature-sensitive)
Molality = moles of solute per kilogram of solvent (temperature-independent)

The solvent is the substance doing the dissolving — usually water. The solute is what gets dissolved — sugar, salt, etc.

The Precise Definition

\text{Molality } (m) = \frac{\text{moles of solute}}{\text{kilograms of solvent}}

The symbol for molality is a lowercase m (not to be confused with M for molarity).

Key points to remember:

The denominator is solvent mass, not solution mass
The unit is mol/kg (often written as simply "m")
It is independent of temperature because mass doesn't change with temperature

Worked Example

Problem: 36 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180 g/mol) is dissolved in 500 g of water. Calculate the molality of the solution.

Step 1: Find moles of solute

\text{Moles of glucose} = \frac{36 \text{ g}}{180 \text{ g/mol}} = 0.2 \text{ mol}

Step 2: Convert solvent mass to kilograms

500 \text{ g} = 0.5 \text{ kg}

Step 3: Apply the formula

m = \frac{0.2 \text{ mol}}{0.5 \text{ kg}} = 0.4 \text{ m}

Note

The answer is 0.4 m (or 0.4 mol/kg). Notice we used the mass of water (500 g), not the mass of the solution (which would be 536 g).

Common Mistake to Avoid

Watch out

Do not use the mass of the solution in the denominator. The formula specifically asks for the mass of the solvent alone. If the problem gives you the total mass of the solution, subtract the mass of the solute to find the solvent mass.

When Do You Use Molality?

Molality is the star in two important situations: …

Why this formula?

Molality Calculation: Why the Formula Works

Molality is a measure of concentration that is temperature-independent — this is its key advantage over molarity. Let's understand why the formula takes the form it does.

The Definition First

Molality ( $m$ ) is defined as:

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

The unit is mol/kg, often written as m (e.g., 0.5 m glucose solution).

Why Mass of Solvent, Not Solution?

This is the critical conceptual point.

The Reasoning

Molarity uses volume of solution → volume changes with temperature (expansion/contraction). So molarity changes with temperature.
Molality uses mass of solvent → mass is invariant with temperature. So molality remains constant regardless of temperature changes.

Key insight: By using the solvent's mass (not the solution's volume), we eliminate temperature dependence. This is why molality is preferred for colligative properties (boiling point elevation, freezing point depression) — these properties depend on the number of solute particles, not on temperature.

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

If you have $w_{\text{solute}}$ grams of solute with molar mass $M_{\text{solute}}$ (g/mol):

\text{moles of solute} = \frac{w_{\text{solute}}}{M_{\text{solute}}}

Step 2: Mass of Solvent in kg

If the solvent mass is $W_{\text{solvent}}$ grams:

\text{mass of solvent in kg} = \frac{W_{\text{solvent}}}{1000}

Step 3: Putting It Together

m = \frac{\frac{w_{\text{solute}}}{M_{\text{solute}}}}{\frac{W_{\text{solvent}}}{1000}}

Simplifying:

m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}

The Final Formula (Exam-Ready)

\boxed{m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}}

Where:

$w_{\text{solute}}$ = mass of solute in grams
$M_{\text{solute}}$ = molar mass of solute in g/mol
$W_{\text{solvent}}$ = mass of solvent in grams

Why the ×1000 Factor? …

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The type of deviation from Raoult's Law depends on the relative strengths of intermolecular forces; solution (i) has similar A-A, B-B, and A-B interactions, making it an ideal solution that follows Raoult's Law.

When two volatile liquids, A and B, are mixed to form a solution, the vapor pressure of the solution is influenced by the intermolecular forces between the molecules. Raoult's Law describes the behavior of an ideal solution, where these interactions are perfectly balanced. Deviations from Raoult's Law occur when the interactions in the mixture differ significantly from those in the pure components.

Let's consider the intermolecular forces:

A-A interactions: Forces between molecules of pure liquid A.
B-B interactions: Forces between molecules of pure liquid B.
A-B interactions: Forces between molecules of A and B in the solution.

Ideal Solutions and Raoult's Law:

An ideal solution is one where the intermolecular attractive forces between A-A, B-B, and A-B are all very similar in magnitude. In such a case, the tendency of molecules A and B to escape into the vapor phase from the solution is the same as from their pure states.

P_A = x_A P_A^\circ

Deviations from Raoult's Law:

Positive Deviation:

This occurs when the A-B intermolecular interactions are weaker than the average of A-A and B-B interactions.
- Intuition: If A and B molecules don't attract each other as strongly as they attract their own kind, they find it easier to escape from the solution into the vapor phase.
- Effect: This leads to a higher vapor pressure than predicted by Raoult's Law ( $P_A > x_A P_A^\circ$ and $P_B > x_B P_B^\circ$ ).
- Examples: Ethanol and acetone, carbon disulfide and acetone.
Negative Deviation:

This occurs when the A-B intermolecular interactions are stronger than the average of A-A and B-B interactions.
- Intuition: If A and B molecules attract each other more strongly, they are held more tightly within the solution, making it harder for them to escape into the vapor phase.
- Effect: This leads to a lower vapor pressure than predicted by Raoult's Law ( $P_A < x_A P_A^\circ$ and $P_B < x_B P_B^\circ$ ).
- Examples: Chloroform and acetone, nitric acid and water.

Now, let's analyze each piece of information provided:

Analyze Information (i): "In bromoethane and chloroethane mixture intermolecular interactions of A-A and B-B type are nearly same as A-B type interactions."
- This statement directly describes the condition for an ideal solution. When the intermolecular forces between like molecules (A-A, B-B) are similar to those between unlike molecules (A-B), the solution behaves ideally.
- Therefore, the bromoethane and chloroethane mixture will follow Raoult's Law.
Analyze Information (ii): "In ethanol and acetone mixture A-A or B-B type intermolecular interactions are stronger than A-B type interactions."
- This means that the forces holding ethanol molecules together (A-A) or acetone molecules together (B-B) are stronger than the forces between ethanol and acetone molecules (A-B). In other words, the A-B interactions are weaker. …

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This problem tests Raoult’s law and its deviations. Below are the common mistakes students make on this type of question and how to avoid each.

🔍 Concept Recap (Why this matters)

Raoult’s law says:

For an ideal solution,

$P_{\text{total}} = P_A^0 x_A + P_B^0 x_B$

Deviations happen when intermolecular forces differ:

Ideal solution → A–A ≈ B–B ≈ A–B interactions → obeys Raoult’s law
Positive deviation → A–B weaker than A–A / B–B → vapour pressure higher than expected
Negative deviation → A–B stronger than A–A / B–B → vapour pressure lower than expected

✗ Mistake #1: Confusing “stronger” with “positive deviation”

What students do wrong:

They see “A–B interactions are stronger” and think → “stronger attraction → more vapour → positive deviation.”

This is backwards.

Correct logic:

Stronger A–B → molecules hold on to each other → less escaping tendency → lower vapour pressure → negative deviation
Weaker A–B → molecules escape easily → higher vapour pressure → positive deviation

How to avoid:

Memorise this:

Stronger A–B → negative deviation

Weaker A–B → positive deviation

✗ Mistake #2: Thinking “nearly same” means “exactly equal”

What students do wrong:

They think “nearly same” is not good enough for Raoult’s law — they look for “exactly equal.”

Correct logic:

In exam language, “nearly same” means ideal behaviour.

So case (i) → ideal solution → follows Raoult’s law.

How to avoid:

Treat “nearly same” as ideal in such MCQs.

Only “stronger” or “weaker” cause deviations.

✗ Mistake #3: Mixing up which case is positive vs negative

What students do wrong:

They match:

(ii) ethanol + acetone → stronger A–A / B–B → they call it negative deviation ✗

Correct logic:

(ii) A–A and B–B are stronger than A–B → A–B is weaker → positive deviation
(iii) A–A and B–B are weaker than A–B → A–B is stronger → negative deviation

How to avoid:

Draw a quick table:

Case	Interaction comparison	Deviation
(ii)	A–A / B–B stronger	Positive
(iii)	A–A / B–B weaker	Negative

✗ Mistake #4: Not reading the options carefully …

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Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

The Intuition

The Precise Definition

Worked Example

Common Mistake to Avoid

When Do You Use Molality?

Why this formula?

Molality Calculation: Why the Formula Works

The Definition First

Why Mass of Solvent, Not Solution?

The Reasoning

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

Step 2: Mass of Solvent in kg

Step 3: Putting It Together

The Final Formula (Exam-Ready)

Why the ×1000 Factor? …

Concept: Deviations from Raoult’s Law

Method: Compare Intermolecular Interaction Strengths

Steps:

🔍 Concept Recap (Why this matters)

✗ Mistake #1: Confusing “stronger” with “positive deviation”

✗ Mistake #2: Thinking “nearly same” means “exactly equal”

✗ Mistake #3: Mixing up which case is positive vs negative

✗ Mistake #4: Not reading the options carefully …

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