ChemistryISC Class 11 · Some Basic Concepts of Chemistry

Concentration Terms, Dilution and the Limiting Reagent

Updated 17 October 2026 · Fact-checked

Molarity M = moles of solute per litre of solution; molality m = moles per kg of solvent; mole fraction x = moles of one component/total moles. Dilution follows M₁V₁ = M₂V₂. In a reaction, the limiting reagent is the one that gives the least product, and it fixes the theoretical yield.

Divide each reactant's moles by its coefficient; the smallest answer (O₂ here) is the limiting reagent and fixes the product, while H₂ is left over.

Illustration drawn for QuestionBuddy.

Understand Concentration Terms and Limiting Reagent

Concentration terms. Mass percentage = (mass of solute/mass of solution) × 100. Molarity M = n/V (mol L⁻¹) depends on temperature because volume changes; molality m = n/mass of solvent in kg is independent of temperature. Mole fraction xₐ = nₐ/(nₐ + n_b) is a pure number, and the mole fractions of all components add to 1. Parts per million (ppm) = (mass of solute/mass of solution) × 10⁶.

Conversion. Molarity from percentage and density: M = (% × 10 × d)/M₀, where d is the density in g mL⁻¹ and M₀ the molar mass of the solute. For example, 36.5% HCl of density 1.2 g mL⁻¹ has M = (36.5 × 10 × 1.2)/36.5 = 12 mol L⁻¹.

Dilution and mixing. Moles of solute do not change on dilution: M₁V₁ = M₂V₂. On mixing two solutions of the same solute, M = (M₁V₁ + M₂V₂)/(V₁ + V₂).

Stoichiometry. Use the balanced equation as a mole ratio. The limiting reagent is the reactant that is used up first; divide the available moles of each reactant by its coefficient, and the smallest quotient identifies it. The product formed, and the amount of the other reactant left over, follow from the limiting reagent.

Yield. Percentage yield = (actual yield/theoretical yield) × 100. Theoretical yield is computed from the limiting reagent.

Concentration Terms and Limiting Reagent: formulas and key facts

Concentrations: $M = n / V (L) ∣ m = n / ma sso f so l v e n t (k g) ∣ x_{a} = n_{a} / (n_{a} + n_{b})$; Molality does not change with temperature.
Conversion: $M = ($; d in g mL⁻¹.
Dilution: $M_{1} V_{1} = M_{2} V_{2}$; Amount of solute is constant.
Yield: Theoretical yield from the limiting reagent.

Concentration Terms and Limiting Reagent in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Making a dilute acid from stock

Suppose a laboratory stock of hydrochloric acid is 36.5% by mass with density 1.2 g mL⁻¹ (HCl = 36.5 g mol⁻¹).

Find its molarity and the volume needed to make 250 mL of 0.5 M acid.

Try it first, then show the solutionSolution

M = (36.5 × 10 × 1.2)/36.5 = 12 mol L⁻¹.
M₁V₁ = M₂V₂: 12 × V₁ = 0.5 × 250.
V₁ = 125/12 = 10.4 mL, made up to 250 mL with water (add acid to water).

Answer: 12 M; 10.4 mL.

Memory hook: Moles of HCl are the same before and after dilution.

Tricks to remember Concentration Terms and Limiting Reagent

Short memory aids. Say each one aloud once, then use it on a question the same day.

Divide by the coefficient
The limiting reagent has the smallest (moles/coefficient).
Use it when: Finding the limiting reagent quickly.
Molality is the odd one
Molarity and normality use solution volume; molality uses solvent mass.
Use it when: Temperature-dependence MCQs.

Concentration Terms and Limiting Reagent: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Temperature-independent concentration · 1 mark · MCQBoard level

Which concentration term does not change with temperature?
(a) molarity
(b) normality
(c) molality
(d) formality

Try it first, then show the solutionSolution

Molality uses the mass of the solvent, which does not change with temperature; the others use volume.

Answer: (c) molality

Example 2: Molarity and molality · 3 marks · Short answerBoard level

A solution contains 4 g of NaOH (Na = 23, O = 16, H = 1) in 500 mL of solution. Find its molarity. If the same 4 g is dissolved in 250 g of water, find the molality.

Try it first, then show the solutionSolution

M₀ = 40 g mol⁻¹, so n = 4/40 = 0.1 mol. (1 mark)
Molarity = 0.1/0.5 = 0.2 mol L⁻¹. (1 mark)
Molality = 0.1/0.250 = 0.4 mol kg⁻¹. (1 mark)

Answer: Molarity 0.2 M; molality 0.4 m.

Example 3: Mole fractions · 4 marks · Short answerHarder board level

A solution contains 36 g of water and 46 g of ethanol (C₂H₅OH). Find the mole fraction of each component and the molality of ethanol. (C = 12, H = 1, O = 16)

Try it first, then show the solutionSolution

n(H₂O) = 36/18 = 2 mol; n(C₂H₅OH) = 46/46 = 1 mol. (1 mark)
x(water) = 2/3 = 0.667; x(ethanol) = 1/3 = 0.333. (1 mark)
The mole fractions add to 1. (0.5 mark)
Molality of ethanol = 1/0.036 = 27.8 mol kg⁻¹. (1.5 marks)

Answer: x(water) = 0.667, x(ethanol) = 0.333; molality 27.8 m.

Example 4: Preparing a dilute solution · 4 marks · Short answerHarder board level

Concentrated HCl is 36.5% by mass with density 1.2 g mL⁻¹. Find its molarity and the volume required to prepare 250 mL of 0.5 M HCl.

Try it first, then show the solutionSolution

M = (% × 10 × d)/M₀ = (36.5 × 10 × 1.2)/36.5. (1 mark)
= 12 mol L⁻¹. (1 mark)
M₁V₁ = M₂V₂ → V₁ = 0.5 × 250/12. (1 mark)
= 10.4 mL. (1 mark)

Answer: 12 M; 10.4 mL.

Example 5: Limiting reagent · 5 marks · Long answerHOTS / competency

In the reaction 2Al + 3Cl₂ → 2AlCl₃, 5.4 g of aluminium reacts with 10.65 g of chlorine (Al = 27, Cl = 35.5). Find the limiting reagent, the mass of AlCl₃ formed and the mass of the excess reagent left.

Try it first, then show the solutionSolution

n(Al) = 5.4/27 = 0.20 mol; n(Cl₂) = 10.65/71 = 0.15 mol. (1 mark)
Moles ÷ coefficient: Al 0.20/2 = 0.10; Cl₂ 0.15/3 = 0.05. Cl₂ is limiting. (1 mark)
AlCl₃ formed = 0.15 × 2/3 = 0.10 mol. (1 mark)
Mass = 0.10 × 133.5 = 13.35 g. (1 mark)
Al used = 0.10 mol, so Al left = 0.10 mol = 2.7 g. (1 mark)

Answer: Cl₂ limiting; 13.35 g AlCl₃; 2.7 g Al left.

Where students lose marks in Concentration Terms and Limiting Reagent

The slips examiners see most often, and what to write instead.

Mistake: Using volume of solvent for molarity
Using the volume of water instead of the volume of the solution.
Do this: Molarity uses the volume of the solution.
Usually costs: 1 to 2 marks.
Mistake: Choosing the limiting reagent by mass
Taking the reagent with the smaller mass as limiting.
Do this: Compare moles ÷ coefficient.
Usually costs: 2 marks.
Mistake: Unbalanced equation
Using 1 : 1 ratios without balancing.
Do this: Balance the equation before using the mole ratio.
Usually costs: 2 marks.

Exam tips

Convert everything to moles before comparing reactants.
Write the balanced equation first.
State units: mol L⁻¹, mol kg⁻¹, g.
Check that mole fractions add to 1.

Practice questions on Concentration Terms and Limiting Reagent

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If the temperature of a solution is raised, which of these changes?
(a) molality
(b) mole fraction
(c) molarity
(d) mass percentage
Hint
The volume of the solution changes with temperature.
Answer
(c) molarity
Q2 · 2 marks · Short answerBoard level
What mass of NaCl (Na = 23, Cl = 35.5) is needed to prepare 250 mL of 0.1 M solution?
Hint
n = MV; m = nM₀.
Answer
n = 0.1 × 0.250 = 0.025 mol; m = 0.025 × 58.5 = 1.46 g.
Q3 · 3 marks · Short answerHarder board level
100 mL of 0.5 M HCl is mixed with 200 mL of 0.2 M HCl. Find the molarity of the mixture.
Hint
Add the moles; add the volumes.
Answer
n = 0.05 + 0.04 = 0.09 mol; V = 0.3 L; M = 0.30 mol L⁻¹.
Q4 · 3 marks · Short answerHarder board level
A water sample contains 2.5 mg of lead in 500 g of water. Express the lead content in ppm.
Hint
ppm = (mass of solute/mass of solution) × 10⁶.
Answer
(2.5 × 10⁻³ g/500 g) × 10⁶ = 5 ppm.
Q5 · 3 marks · Short answerHarder board level
Heating 6.0 g of CaCO₃ (Ca = 40, C = 12, O = 16) gives 2.2 g of CO₂. Find the percentage yield. (CaCO₃ → CaO + CO₂)
Hint
Theoretical yield from the balanced equation.
Answer
Theoretical CO₂ = 0.06 mol × 44 = 2.64 g. Yield = 2.2/2.64 × 100 = 83.3%.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a fertiliser plant reacts 28 g of N₂ with 10 g of H₂ (N = 14, H = 1): N₂ + 3H₂ → 2NH₃.
(i) Find the limiting reagent. (1 mark)
(ii) Find the mass of NH₃ formed. (1 mark)
(iii) Find the mass of the excess reactant left over. (1 mark)
(iv) If the actual yield is 80%, find the mass of NH₃ obtained. (1 mark)
Hint
n(N₂) = 1 mol; n(H₂) = 5 mol.
Answer
(i) N₂ needs 3 mol H₂ but 5 mol are present, so N₂ is limiting. (ii) 2 mol NH₃ = 34 g. (iii) H₂ left = 5 − 3 = 2 mol = 4 g. (iv) 0.80 × 34 = 27.2 g.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define molality.Show answer
Moles of solute per kg of solvent.
2.Dilution formula?Show answer
M₁V₁ = M₂V₂.
3.How do you find the limiting reagent?Show answer
The reactant with the smallest (moles ÷ coefficient).
4.Formula for percentage yield?Show answer
(actual/theoretical) × 100.

Concentration Terms and Limiting Reagent: frequently asked questions

Why do we add acid to water and not water to acid?

Diluting a concentrated acid releases a lot of heat; adding the acid slowly to water spreads the heat and avoids spattering.

Can molarity and molality be equal?

Only approximately for dilute aqueous solutions, where 1 L of solution has nearly 1 kg of water.

What is the excess reagent?

The reactant left over when the limiting reagent has been used up completely.