Q.One mole of any substance contains 6.022×1023 atoms/molecules. Number of molecules of H2SO4 present in 100 mL of 0.02M H2SO4 solution is ______.
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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? …
Concept: Relating molarity, volume, and Avogadro's number to find the number of molecules.
The molarity tells us moles of solute per liter of solution. For a 0.02 M H2SO4 solution, we have 0.02 moles in 1000 mL.
Step 1: Find moles in 100 mL.
Moles of H2SO4=0.02×1000100=0.002 mol
Step 2: Convert moles to molecules using Avogadro's number (NA=6.022×1023). …
Use n=M×V to find moles in the solution, then multiply by Avogadro's number to convert moles to molecules. The answer is 12.044×1020 molecules.
The question asks for the number of molecules, not the amount in moles. Whenever you need to count individual particles—atoms, molecules, ions—you bridge from the macroscopic world (moles, concentration, volume) to the microscopic one using Avogadro's number, NA=6.022×1023mol−1.
The strategy is straightforward: first calculate how many moles of H2SO4 are present in the given volume of solution, then convert that amount to molecules.
Step-by-step calculation
-
Identify what you know.
The solution is 0.02 M (molarity), meaning there are 0.02 moles of H2SO4 per litre of solution. The volume given is 100 mL, which is 1000100=0.1 L.
-
Calculate the number of moles.
Molarity is defined as moles per litre:
M=Vn⇒n=M×V
Substituting the values:
n=0.02mol/L×0.1L=0.002mol
So the solution contains 2×10−3 moles of H2SO4.
- Convert moles to molecules. …
Concept: Molarity and Avogadro's Number
Molarity (M) = number of moles of solute per litre of solution.
Avogadro's Number = 6.022×1023 particles per mole.
Method: Mole–Molarity–Particle Conversion
Steps
-
Write the given data
- Volume of solution = 100 mL=0.1 L
- Molarity = 0.02 M
- Avogadro’s number = 6.022×1023 molecules/mol
-
Find moles of H2SO4
Moles=Molarity×Volume (in L)
Moles=0.02×0.1=0.002 mol
- Convert moles to molecules …
🧠 Concept Recap
We are given:
- Volume = 100 mL = 0.1 L
- Molarity = 0.02 M = 0.02 mol/L
- Avogadro’s number = 6.022×1023 molecules/mol
Correct approach:
- Find moles of H2SO4:
Moles=Molarity×Volume (in L)=0.02×0.1=0.002 mol
- Find number of molecules:
Molecules=0.002×6.022×1023=1.2044×1021
Which equals 12.044×1020 molecules → Option (i).
✗ Common Mistake #1: Forgetting to convert mL to L
What students do:
They plug in 100 mL directly:
0.02×100=2 moles
Then multiply by Avogadro’s number → huge wrong answer like 12.044×1023 (option (iv)).
Why it’s wrong:
Molarity is mol/L, so volume must be in litres.
✓ How to avoid:
Always write the unit:
Volume in L=1000Volume in mL
Make it a habit — never skip this step.
✗ Common Mistake #2: Confusing atoms with molecules
What students do:
They calculate molecules correctly but then multiply by the number of atoms per molecule (e.g., 7 atoms in H2SO4) and give an answer for atoms, not molecules.
Why it’s wrong:
The question explicitly asks for number of molecules, not atoms.
✓ How to avoid:
Read the question carefully — underline “molecules” or “atoms”. If it says molecules, stop after multiplying moles by Avogadro’s number.
✗ Common Mistake #3: Misplacing the decimal in scientific notation
What students do:
They compute 0.002×6.022×1023 and write 1.2044×1021 but then match it to 12.044×1020 incorrectly, thinking it’s a different number.
Why it’s wrong:
12.044×1020 is exactly equal to 1.2044×1021 — just a different notation.
✓ How to avoid:
Practice converting between forms: …
- 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. …
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