Q.Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult's law?
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →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
Molality (m)=kilograms of solventmoles of solute
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 (C6H12O6, molar mass = 180 g/mol) is dissolved in 500 g of water. Calculate the molality of the solution.
Step 1: Find moles of solute
Moles of glucose=180 g/mol36 g=0.2 mol
Step 2: Convert solvent mass to kilograms
500 g=0.5 kg
Step 3: Apply the formula
m=0.5 kg0.2 mol=0.4 m
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
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=mass of solvent in kgmoles of solute
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 wsolute grams of solute with molar mass Msolute (g/mol):
moles of solute=Msolutewsolute
Step 2: Mass of Solvent in kg
If the solvent mass is Wsolvent grams:
mass of solvent in kg=1000Wsolvent
Step 3: Putting It Together
m=1000WsolventMsolutewsolute
Simplifying:
m=Msolute×Wsolventwsolute×1000
The Final Formula (Exam-Ready)
m=Msolute×Wsolventwsolute×1000
Where:
- wsolute = mass of solute in grams
- Msolute = molar mass of solute in g/mol
- Wsolvent = mass of solvent in grams
Why the ×1000 Factor? …
The key idea is that positive deviation from Raoult’s law occurs when the A–B interactions in the mixture are weaker than the A–A and B–B interactions in the pure components. This makes the vapour pressure higher than ideal.
Step 1 – Analyse pure components:
Methanol (CH₃OH) forms strong hydrogen bonds with itself. Acetone (CH₃COCH₃) has a polar carbonyl group but no –OH group; its self-interactions are weaker dipole–dipole forces.
Step 2 – Compare mixture interactions: …
Positive deviation from Raoult’s law occurs when A-B interactions are weaker than A-A and B-B interactions. In the given options, only methanol and acetone form weaker cross hydrogen bonds than the self-association in pure methanol, leading to higher vapour pressure. The correct option is (i).
Why This Approach Works
Raoult’s law states that the partial vapour pressure of each component in an ideal solution is proportional to its mole fraction. Real solutions deviate from this depending on how A-B interactions compare with A-A and B-B interactions.
Positive deviation: A-B interactions weaker than A-A/B-B - higher vapour pressure than ideal.
Negative deviation: A-B interactions stronger - lower vapour pressure than ideal.
Step-by-Step Reasoning
- (i) Methanol and acetone - Pure methanol has strong O-H...O self-association. The cross hydrogen bond with acetone's carbonyl oxygen is weaker than methanol-methanol bonds. A-B < A-A -> positive deviation. …
Concept: Deviation from Raoult’s Law and Hydrogen Bonding
Method: Intermolecular Force Comparison Method
Why this method works
Raoult’s law assumes ideal behaviour — the intermolecular forces between A–A, B–B, and A–B are all equal.
- Positive deviation occurs when A–B interactions are weaker than A–A and B–B.
- This makes the vapour pressure higher than predicted, because molecules escape more easily.
Steps to apply
- Identify the dominant intermolecular force in each pure component (usually hydrogen bonding here).
- Compare the strength of A–B hydrogen bonds with the pure component hydrogen bonds.
- If A–B bonds are weaker → positive deviation. If A–B bonds are stronger → negative deviation.
Applying to the options
(i) Methanol and acetone ✓
- Pure methanol: strong O–H···O hydrogen bonding.
- Pure acetone: only weak dipole-dipole (C=O is polar, but no –OH or –NH).
- In mixture: Methanol’s –OH can form H-bonds with acetone’s C=O, but acetone disrupts the strong methanol–methanol network.
- Result: A–B bonds are weaker than A–A bonds → positive deviation.
(ii) Chloroform and acetone ✗
- Chloroform (CHCl₃) has a slightly acidic H (C–H···O type H-bond possible).
- Acetone’s C=O accepts H-bonds.
- Result: A–B H-bonds are stronger than pure component interactions → negative deviation.
(iii) Nitric acid and water ✗ …
Common Mistakes & How to Avoid Them
This question tests your understanding of Raoult's law deviations in liquid-liquid solutions, specifically linked to hydrogen bonding between components.
✗ Mistake 1: Confusing "positive deviation" with "stronger intermolecular forces"
The error:
Students think that if a mixture forms strong hydrogen bonds, it will show positive deviation.
Actually, positive deviation occurs when A–B interactions are weaker than A–A and B–B interactions. This makes the mixture easier to vaporise (higher vapour pressure than ideal).
How to avoid:
Remember the rule:
Positive deviation → weaker A–B forces → vapour pressure higher than ideal
Negative deviation → stronger A–B forces → vapour pressure lower than ideal
So for positive deviation, the mixture must break strong self-association (like in pure methanol or water) and form weaker cross hydrogen bonds.
✗ Mistake 2: Assuming all hydrogen-bonding mixtures show negative deviation
The error:
Many students think "hydrogen bond = negative deviation". But that’s only true if the cross hydrogen bond is stronger than the self-hydrogen bonds in the pure components.
Example:
- Methanol + acetone: Methanol has strong self-association. Acetone can only accept H-bonds (C=O). The cross bond is weaker than methanol–methanol bonds → positive deviation.
- Chloroform + acetone: Chloroform’s H is acidic, acetone’s O is basic → stronger cross bond → negative deviation.
How to avoid:
Compare the strength of the cross bond vs. the self bonds. If the pure component has strong self-association (like alcohols, water, carboxylic acids), and the other component is a weak H-bond acceptor (like acetone, ether), expect positive deviation.
✗ Mistake 3: Not recognising when both components form strong self-association
The error:
Students sometimes pick options like (iii) Nitric acid + water or (iv) Phenol + aniline, thinking they show positive deviation because both have H-bonding.
Correction:
- Nitric acid + water: Both form very strong self-association, but the cross H-bond (acid–water) is even stronger → negative deviation.
- Phenol + aniline: Both have strong self-association, but phenol (acidic H) and aniline (basic N) form a very strong cross H-bond → negative deviation.
How to avoid: …
- KEAM 2026Set eng-2026-04174 marksMCQQ.If 'c' is the molarity of a solution, 'm' the molality, M2 the molecular weight of the solute in a binary solution and 'ρ', is the density of the solution in g/cm3, then the relationship between molality 'm' and molarity 'c' is given by (A) m = c/[ρ-cM2/1000] mol kg−1 (B) m = 1000c/[ρ-cM2] mol kg−1 (C) m = cρ/[1+cM2/100] mol kg−1 (D) m = 1000c/ρmol kg−1 (E) m = 1000c/[1000ρ+cM2]
›Reveal solutionSolution
Convert molarity to molality by finding the mass of solvent in 1 L of solution; the result is m=1000ρ−cM21000c, identical to option (A).
Take exactly 1 litre (1000 cm3) of solution.
Mass of solution =volume×density=1000ρ g.
Moles of solute in 1 L =c (definition of molarity), so mass of solute =cM2 g.
Mass of solvent =(1000ρ−cM2) g =10001000ρ−cM2 kg. …
- KEAM 2026Set eng-2026-04194 marksMCQQ.What is the mass of ethanoic acid required to prepare 0.5 m solution containing 100 g of be water? (Molar mass of ethanoic acid = 60 g mol−1). (A) 3 g (B) 6 g (C) 0.3 g (D) 7.5 g (E) 2 g
›Reveal solutionSolution
A 0.5m solution in 100g water needs 0.05mol (3g) of ethanoic acid.
Molality =kg of solventmoles of solute. For 0.5m in 100g=0.100kg water:
n=0.5×0.100=0.05mol. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.3.0 g of a salt of molecular weight 30 is dissolved in 250 mL of water. The molarity of the solution is (A) 0.1 M (B) 0.2 M (C) 0.3 M (D) 0.4 M (E) 0.5 M
›Reveal solutionSolution
Moles = mass/MW =3.0/30=0.1 mol in 0.25 L, so molarity =0.4 M.
n=30 g mol−13.0 g=0.1 mol …
- KEAM 2026Set pha-2026-0418F4 marksMCQQ.The molarity of an aqueous solution containing 0.4g of NaOH (molar mass=40g/mol) in 250mL of a solution is (A) 0.04M (B) 0.02M (C) 0.20M (D) 0.40M (E) 0.08M
›Reveal solutionSolution
0.4 g NaOH is 0.01 mol; in 0.250 L that is 0.04 M.
Reasoning
Moles of NaOH =40 g mol−10.4 g=0.01 mol.
Volume =250 mL=0.250 L. …
- KEAM 2025Set eng-2025-04234 marksMCQQ.The density of 3 M aqueous solution of a solute 'X' is 1.86 g mL−1. The molality of the solution is (Molar mass of solute 'X' is 120 g mol−1) (A) 3 m (B) 4 m (C) 2 m (D) 5 m (E) 1 m
›Reveal solutionSolution
The molality of the 3 M solution is 2 m.
Concept and Intuition
Molarity is per litre of solution; molality is per kilogram of solvent. Using the density to get the mass of 1 L of solution, subtracting the solute mass gives the solvent mass, from which molality follows.
Step-by-Step Solution
- Mass of 1 L solution = 1000 mL * 1.86 g/mL = 1860 g.
- Solute in 1 L (3 mol) = 3 * 120 = 360 g.
- Solvent mass = 1860 - 360 = 1500 g = 1.5 kg. …
- KEAM 2024Set eng-2024-06084 marksMCQQ.260 g of an aqueous solution contains 60 g of urea (Molar mass = 60 g mol−1). The molality of the solution is (A) 2m (B) 3m (C) 4m (D) 5m (E) 6m
›Reveal solutionSolution
Molality = moles of solute per kg of solvent. Here 1 mol urea in 0.2 kg water gives 5 m.
The number of moles of urea is
n=60 g mol−160 g=1 mol.
The solvent (water) mass is the total solution mass minus the solute mass:
260−60=200 g=0.2 kg. …
- KEAM 2024Set eng-2024-06084 marksMCQQ.The molarity of a solution containing 8 g of NaOH (Molar mass = 40 g mol−1) in 250 mL solution is (A) 0.8M (B) 0.4M (C) 0.2M (D) 0.5M (E) 0.6M
›Reveal solutionSolution
Molarity = moles of solute per litre of solution: 0.2 mol in 0.25 L gives 0.8 M.
The number of moles of NaOH is
n=40 g mol−18 g=0.2 mol.
The solution volume is 250 mL=0.25 L, so the molarity is …
- KEAM 2024Set eng-2024-06094 marksMCQQ.The molarity of sodium hydroxide in the solution prepared by dissolving 6 g in 600 mL of water is (molar mass of NaOH = 40 g mol−1) (A) 0.5 M (B) 0.4 M (C) 0.25 M (D) 0.1 M (E) 0.2 M
›Reveal solutionSolution
6 g NaOH is 0.15 mol; in 0.6 L this is 0.25 M.
Moles of NaOH:
n=406=0.15 mol.
Molarity: …
- KEAM 2024Set eng-2024-06094 marksMCQQ.The volume of ethanol required to prepare 3 L of 0.25 M aqueous solution is (density of ethanol= 0.36 kg L−1, molar mass = 60 g mol−1) (A) 125 mL (B) 25mL (C) 75mL (D) 50mL (E) 12.5mL
›Reveal solutionSolution
3 L of 0.25 M needs 0.75 mol=45 g ethanol; at density 360 gL−1 that is 125 mL.
Moles required:
n=M×V=0.25×3=0.75 mol.
Mass:
m=n×Mmolar=0.75×60=45 g. …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.