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: …
- COMEDK 2025Set 2025-A1 markMCQQ.Lead storage battery contains 4.25MH2SO4 which has a density of 1.24 g/ml. Calculate the molality of aqueous solution of H2SO4. (A) 6.264 (B) 3.427 (C) 5.161 (D) 4.108
›Reveal solutionSolution
Molality is moles of solute per kilogram of solvent. Given molarity and density, we find the mass of 1 L of solution, subtract the mass of H₂SO₄ to get solvent mass, then compute molality. The result is 5.161 m, so option (C) is correct.
Concept & Intuition
Molality (m) depends on the mass of solvent, not the volume of solution. Molarity (M) gives moles per liter of solution, but the solvent mass is hidden inside the density. The trick: take exactly 1 liter of solution, find its total mass from density, subtract the mass of H₂SO₄ (from moles × molar mass), and you have the solvent mass in kg. Then molality = moles / kg solvent.
Step-by-step solution
-
Interpret the given data
- Molarity of H₂SO₄ = 4.25 M → 4.25 moles per liter of solution.
- Density of solution = 1.24 g/mL = 1240 g/L (since 1 mL = 1 g water equivalent, but here it’s the whole solution).
- Molar mass of H₂SO₄ = 2(1.008) + 32.06 + 4(16.00) = 98.08 g/mol (we’ll use 98.08).
-
Mass of 1 liter of solution
Mass of solution=1.24 mLg×1000 mL=1240 g
- Mass of H₂SO₄ in 1 liter
Moles of H₂SO₄=4.25 mol
Mass of H₂SO₄=4.25×98.08=416.84 g
- Mass of solvent (water) in 1 liter
-
- COMEDK 2025Set 2025-E1 markMCQQ.The mole fraction of an unknown solute in 1560 g of Benzene is 0.5 . What is the molality of the solution? (M. M of Benzene:78 amu) (A) 12.8 (B) 10.3 (C) 3.25 (D) 16.9
›Reveal solutionSolution
The key idea is to use the definition of mole fraction to find the moles of solute, then divide by the mass of solvent in kg. The molality is 12.8 m, so the correct option is (A).
Concept and Intuition
Mole fraction tells us the ratio of moles of one component to the total moles in the mixture. Here, the mole fraction of solute is 0.5, meaning the solute and solvent have equal moles. Since we know the mass of benzene (solvent) and its molar mass, we can find the moles of benzene, then the moles of solute, and finally the molality (moles of solute per kg of solvent). The trap is forgetting to convert grams to kilograms for molality.
Step-by-step solution
- Find moles of benzene (solvent) Mass of benzene = 1560 g Molar mass of benzene (C₆H₆) = 78 g/mol
nbenzene=78 g/mol1560 g=20 mol
- Use mole fraction to find moles of solute Mole fraction of solute, Xsolute=0.5 By definition:
Xsolute=nsolute+nbenzenensolute
Substitute known values:
0.5=nsolute+20nsolute
Multiply both sides by nsolute+20:
0.5(nsolute+20)=nsolute
0.5nsolute+10=nsolute
10=0.5nsolute⇒nsolute=20 mol
- Calculate molality …
- COMEDK 2024Set 2024-E1 markMCQQ.Sulphuric acid used in Lead Storage battery has a concentration of 4.5 M and a density of 1.28 g/ ml. The molality of the acid is __________. (A) 4.012 (B) 2.568 (C) 5.364 (D) 3.516
›Reveal solutionSolution
Take 1 L of solution: it holds 4.5 mol H2SO4 in (1280−441)=839 g water, giving molality ≈5.36 m.
Molar mass of H2SO4=98 g/mol. Consider 1 L of the 4.5 M solution.
Mass of solution:
1.28 g/mL×1000 mL=1280 g
Mass of H2SO4:
4.5 mol×98 g/mol=441 g …
- COMEDK 2023Set 2023-E1 markMCQQ.What is the mole fraction of solute in a 5 m aqueous solution? (A) 0.038 (B) 0.593 (C) 0.082 (D) 0.751
›Reveal solutionSolution
Mole fraction of solute: x_solute = 5 / (5 + 55.55) = 5 / 60.55 = 0.0826 ~ 0.082
Concept: molality m = moles of solute per 1 kg (1000 g) of solvent. Convert to mole fraction.
Basis: 1000 g of water.
moles of solute = 5 mol
moles of water = 1000 / 18 = 55.55 mol …
- COMEDK 2021Set 20211 markMCQQ.What would be the molarity of one litre solution of 22.2 g of CaCl2 ? (A) 0.2 M (B) 0.4 M (C) 0.6 M (D) 0.8 M
›Reveal solutionSolution
Molarity = 0.2 / 1 = 0.2 M
Concept: Molarity = moles of solute / volume of solution in litres.
Molar mass of CaCl2 = 40 + 2(35.5) = 111 g/mol
Moles = 22.2 / 111 = 0.2 mol
Volume = 1 L …
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