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CBSE · Class 11 · Chemistry

Chapter 1: Some Basic Concepts of Chemistry

6 concepts · 48 practice questions with answers · Free to read

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Properties of matter, SI units and measurement

Link to Std 9–10 : You already know that matter has mass and takes up space; now we learn to measure it in a way every scientist in the world agrees on.

From daily life

A tailor measures in inches, a vegetable seller in kilograms, a doctor in °F. If a recipe from another country says "250 mL", you need one common system to follow it. That system is the SI.

Simple definition

A physical property can be seen or measured without changing the substance (colour, melting point, density). A chemical property shows only when the substance changes into another (burning, rusting). The SI has 7 base units.

Temperature scalesIce meltsWater boils°C0100K273.15373.15°F32212K = °C + 273.15°F = (9 ÷ 5) × °C + 32

In the figure : The same two points (ice melting, water boiling) on the °C, K and °F scales.

Rule 1: SI base units and prefixes — length m, mass kg, time s, electric current A, temperature K, amount of substance mol, luminous intensity cd. Prefixes: kilo 10³, centi 10⁻², milli 10⁻³, micro 10⁻⁶, nano 10⁻⁹, pico 10⁻¹².

  • Easy : 3.5 km = 3500 m.
  • Medium : 1 L = 1000 mL = 1000 cm³ = 1 dm³.
  • Exam-type : 640 pm = 640 × 10⁻¹² m = 6.4 × 10⁻¹⁰ m = 0.64 nm.

Rule 2: Mass, weight and density — Mass is the amount of matter and does not change from place to place; weight is the pull of gravity on that mass and changes (less on the Moon). Density: ρ = m ÷ V (ρ = density, m = mass, V = volume).

  • Easy : 96 g of a metal occupies 40 cm³: ρ = 96 ÷ 40 = 2.4 g/cm³.
  • Medium : Mercury has ρ = 13.6 g/cm³; 25 cm³ of it has mass 13.6 × 25 = 340 g.
  • Exam-type : 7.8 g/cm³ = 7.8 × 1000 kg/m³ = 7800 kg/m³ (1 g/cm³ = 1000 kg/m³).

Rule 3: Temperature scales — K = °C + 273.15 and °F = (9 ÷ 5) × °C + 32. A kelvin and a degree Celsius are the same size.

  • Easy : 37 °C = 37 + 273.15 = 310.15 K.
  • Medium : 50 °C = (9 ÷ 5) × 50 + 32 = 122 °F.
  • Exam-type : Liquid nitrogen boils at 77 K, which is 77 − 273.15 = −196.15 °C.
Remember

Mass is measured in kg (SI); weight is a force measured in newton. Kelvin has no degree sign: write 310 K, not 310 °K.

Common mistake

Mixing units while using ρ = m ÷ V. If mass is in g and volume in cm³, density comes out in g/cm³ (not kg/m³).

Shortcut

Memory clue: "KILO is 1000, MILLI is one-thousandth"; and for temperature, "add 273 to get K".

Spot the difference

PointMassWeight
MeaningAmount of matterGravity's pull on matter
SI unitkgN
Changes with place?NoYes

Practice questions

Easy
  1. 1Name the SI unit of amount of substance.
  2. 2How many cm³ are there in 1 L?
  3. 3Is "melting point" a physical or a chemical property?
Medium
  1. 4A solid of mass 52 g has volume 20 cm³. Find its density.
  2. 5Convert 25 °C to kelvin.
  3. 6Give two differences between mass and weight.
Challenging
  1. 7What volume (in cm³) does 390 g of a liquid of density 1.3 g/cm³ occupy?
  2. 8Convert 98.6 °F to °C and then to K.
Show answers (try them yourself first)
  1. 1mole (mol).
  2. 21000 cm³.
  3. 3Physical property.
  4. 452 ÷ 20 = 2.6 g/cm³.
  5. 525 + 273.15 = 298.15 K.
  6. 6Mass is the amount of matter, measured in kg, and is the same everywhere; weight is a force measured in N and changes with gravity.
  7. 7V = m ÷ ρ = 390 ÷ 1.3 = 300 cm³.
  8. 8°C = (98.6 − 32) × 5 ÷ 9 = 37 °C; K = 37 + 273.15 = 310.15 K.

Uncertainty in measurement: significant figures and dimensional analysis

Link to Std 9–10 : You used standard form for very big and very small numbers; Std 11 adds rules for how many digits of a measured number are meaningful.

From daily life

A kitchen scale shows 250 g, a lab balance shows 250.00 g. The second reading carries more information, because it claims to be right up to the second decimal place.

Simple definition

Significant figures are the digits that are known with certainty plus one uncertain digit. Accuracy is closeness to the true value; precision is closeness of repeated readings to each other.

Accuracy and precisionAccurateand precisePrecise,not accurateAccurate,not preciseNeitherCentre = true value

In the figure : Four dart boards: accurate and precise, precise but not accurate, accurate but not precise, neither.

Rule 1: Counting significant figures — (a) all non-zero digits count; (b) zeros between non-zero digits count; (c) leading zeros do not count; (d) trailing zeros count only if there is a decimal point; (e) exact numbers (like 12 students) have unlimited significant figures.

  • Easy : 4 605 has 4 and 0.0070 has 2 significant figures.
  • Medium : 305.0 has 4; 1200 (no decimal point) is taken to have 2.
  • Exam-type : 4.50 × 10³ has 3 significant figures; scientific notation removes the doubt about trailing zeros.

Rule 2: Calculations — In addition/subtraction keep as many decimal places as the number with the fewest. In multiplication/division keep as many significant figures as the number with the fewest. Round up when the dropped digit is 5 or more.

  • Easy : 12.11 + 0.3 + 1.234 = 13.644 → 13.6 (one decimal place, from 0.3).
  • Medium : 2.5 × 4.12 = 10.3 → 10 (two significant figures, from 2.5).
  • Exam-type : 0.0072 ÷ 3.600 = 0.002 → 0.0020 (two significant figures).

Rule 3: Dimensional analysis (factor-label method) — Multiply by conversion factors written as fractions equal to 1, so that unwanted units cancel.

  • Easy : 2 h = 2 × 60 × 60 = 7200 s.
  • Medium : 72 km/h = 72 × 1000 m ÷ 3600 s = 20 m/s.
  • Exam-type : 2.4 days = 2.4 × 24 × 60 × 60 = 207 360 s = 2.0736 × 10⁵ s.
Remember

Write the final answer in scientific notation when trailing zeros matter: 8.50 × 10⁴ shows exactly three significant figures.

Common mistake

Rounding in the middle of a calculation. Round only at the last step.

Shortcut

Memory clue: "Add → decimals, Multiply → digits" (addition looks at decimal places, multiplication at significant digits).

Spot the difference

PointAccuracyPrecision
Compares withTrue valueOther readings
Improved byCalibrating the instrumentCareful repeated measurement
Can exist without the other?YesYes

Practice questions

Easy
  1. 1How many significant figures are there in 0.0305?
  2. 2Write 85 300 in scientific notation.
  3. 3Convert 3 h 20 min into minutes.
Medium
  1. 4The true mass is 5.00 g. Student A reads 4.98, 4.99, 5.01 g; student B reads 4.20, 4.21, 4.19 g. Who is accurate, who is precise?
  2. 5Add 6.75 + 0.082 + 12.1 with the correct number of decimal places.
  3. 6Convert 90 km/h into m/s.
Challenging
  1. 7Multiply 3.14 × 2.0 and give the answer with correct significant figures.
  2. 8A cyclist covers 15.0 km in 40 minutes. Find the speed in m/s.
Show answers (try them yourself first)
  1. 1Three (3, 0, 5).
  2. 28.53 × 10⁴.
  3. 3200 min.
  4. 4A is both accurate and precise; B is precise but not accurate.
  5. 5Sum = 18.932 → 18.9 (one decimal place, from 12.1).
  6. 690 × 1000 ÷ 3600 = 25 m/s.
  7. 76.28 → 6.3 (two significant figures, from 2.0).
  8. 815 000 m ÷ (40 × 60 s) = 15 000 ÷ 2400 = 6.25 m/s.

Laws of chemical combination and Dalton's atomic theory

Link to Std 9–10 : In Std 9 you met the laws of conservation of mass and constant proportions; here we add multiple proportions, the gas laws of combination and the theory that explains them.

From daily life

When a candle burns, it seems to vanish. But if all the gases are collected, their total mass equals the candle plus the oxygen used. Matter only changes form.

Simple definition

Law of conservation of mass: mass is neither created nor destroyed in a chemical reaction. Law of definite proportions: a compound always has the same elements in the same mass ratio. Law of multiple proportions: if two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.

Conservation of massC 6 g+ O₂ 16 gCO₂22 gReactants6 + 16226 g + 16 g = 22 g : the beam stays level

In the figure : 6 g of carbon + 16 g of oxygen give 22 g of carbon dioxide; the two sides weigh the same.

Rule 1: Conservation of mass — Total mass of reactants = total mass of products (in a closed system).

  • Easy : 6 g C + 16 g O₂ → 22 g CO₂.
  • Medium : 10 g of calcium carbonate gives 4.4 g CO₂ on heating; the solid residue is 10 − 4.4 = 5.6 g.
  • Exam-type : If a mixture loses mass on heating in an open dish, the missing mass has gone out as gas; in a sealed vessel the mass stays constant.

Rule 2: Definite proportions — Pure copper oxide, from any source, has copper and oxygen in the mass ratio 4 : 1.

  • Easy : 10 g of the oxide has 8 g Cu and 2 g O.
  • Medium : 25 g of the same oxide has 20 g Cu and 5 g O, ratio again 4 : 1.
  • Exam-type : A sample with 9 g Cu and 3 g O is not this oxide, since 9 : 3 = 3 : 1.

Rule 3: Multiple proportions — Hydrogen forms H₂O (2 g H : 16 g O) and H₂O₂ (2 g H : 32 g O).

  • Easy : For 2 g of hydrogen the oxygen masses are 16 g and 32 g.
  • Medium : Their ratio 16 : 32 = 1 : 2, a simple whole-number ratio.
  • Exam-type : Carbon forms CO and CO₂; for 12 g of carbon the oxygen is 16 g and 32 g, again 1 : 2.

Rule 4: Gay-Lussac's law and Avogadro's law — Gases react in simple whole-number ratios of volumes (same T and P). Avogadro: equal volumes of all gases at the same T and P contain equal numbers of molecules.

  • Easy : 2 L H₂ + 1 L O₂ → 2 L steam.
  • Medium : N₂ + 3H₂ → 2NH₃: 10 L N₂ needs 30 L H₂ and gives 20 L NH₃.
  • Exam-type : 1 L of H₂ and 1 L of CO₂ at the same T and P have the same number of molecules, though their masses differ.

Rule 5: Dalton's postulates — Matter is made of indivisible atoms; atoms of one element are identical; atoms of different elements differ; compounds form when atoms combine in small whole-number ratios; atoms are neither created nor destroyed in a reaction. Today we know atoms are divisible and isotopes exist, but the theory still explains the three laws.

Remember

Dalton's idea of atoms combining in whole-number ratios is exactly why the law of multiple proportions holds.

Common mistake

Gay-Lussac's law is about volumes of gases, not masses. Do not use it for solids.

Shortcut

Memory clue: "CDM-G-A": Conservation, Definite, Multiple, Gay-Lussac, Avogadro, the order of the laws.

Spot the difference

PointDefinite proportionsMultiple proportions
Number of compoundsOne compoundTwo or more compounds of the same elements
StatementFixed ratio in one compoundSimple ratio between different compounds

Practice questions

Easy
  1. 1State the law of conservation of mass.
  2. 23 L of H₂ reacts completely with N₂. How many litres of NH₃ can form at most if enough N₂ is present (N₂ + 3H₂ → 2NH₃)?
  3. 3Who proposed the atomic theory of matter described here?
Medium
  1. 415 g of reactant A and 7 g of reactant B react completely to give one product. What is its mass?
  2. 5Pure CuO from two sources has 16 g Cu and 4 g O in one and 12 g Cu and ? g O in the other. Find ?.
  3. 6State two Dalton postulates that are no longer fully accepted.
Challenging
  1. 7Oxides of sulphur: SO₂ has 32 g S with 32 g O; SO₃ has 32 g S with 48 g O. Show that multiple proportions apply.
  2. 820 L of N₂ and 50 L of H₂ react (N₂ + 3H₂ → 2NH₃). Which gas is left, and how much?
Show answers (try them yourself first)
  1. 1Mass is neither created nor destroyed in a chemical reaction.
  2. 23 L H₂ gives (2 ÷ 3) × 3 = 2 L NH₃.
  3. 3John Dalton.
  4. 415 + 7 = 22 g.
  5. 5Ratio 4 : 1, so 12 g Cu has 3 g O.
  6. 6Atoms are indivisible; all atoms of an element are identical (isotopes differ in mass).
  7. 7For 32 g S, oxygen is 32 g and 48 g; ratio 32 : 48 = 2 : 3, a simple whole-number ratio.
  8. 820 L N₂ needs 60 L H₂, but only 50 L is available, so H₂ is used up; N₂ used = 50 ÷ 3 ≈ 16.7 L, N₂ left ≈ 3.3 L.

Atomic and molecular masses, the mole and molar mass

Link to Std 9–10 : You calculated molecular masses by adding atomic masses; now we count atoms in bulk using the mole.

From daily life

Eggs are sold by the dozen, paper by the ream. Atoms are so tiny that chemists need a far bigger counting unit; that unit is the mole.

Simple definition

Atomic mass is the mass of an atom in u, relative to 1/12 of the mass of a ¹²C atom. A mole is the amount of substance that contains 6.022 × 10²³ particles (Avogadro constant). Molar mass is the mass of one mole in g/mol.

The mole as a bridgeMoles nMass (g)ParticlesGas volume (L)÷ M× M× Nₐ÷ Nₐ× 22.4÷ 22.4Nₐ = 6.022 × 10²³ per mol22.4 L per mol at 0 °C, 1 atm

In the figure : The mole is a bridge between mass (g), number of particles and volume of a gas at STP.

Rule 1: Average atomic mass — Mass of each isotope × its fraction, added up.

  • Easy : Isotopes of an element, 50 % of mass 10 and 50 % of mass 12, average = 11.
  • Medium : Copper: ⁶³Cu 69 % and ⁶⁵Cu 31 %: 0.69 × 63 + 0.31 × 65 = 63.62 u.
  • Exam-type : The average atomic mass need not be a whole number, though each isotope's mass number is.

Rule 2: Molecular and formula mass — Add the atomic masses of all atoms in the formula (H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Cl = 35.5).

  • Easy : H₂O = 2 + 16 = 18 u.
  • Medium : Glucose C₆H₁₂O₆ = 72 + 12 + 96 = 180 u.
  • Exam-type : For ionic compounds like NaCl we say formula mass (58.5 u), as there are no separate molecules.

Rule 3: Mole relations — n = m ÷ M (n = moles, m = mass in g, M = molar mass); N = n × Nₐ (N = number of particles, Nₐ = Avogadro constant); at STP (0 °C, 1 bar) one mole of any gas occupies about 22.7 L; the older value at 0 °C and 1 atm is 22.4 L.

  • Easy : 36 g water = 36 ÷ 18 = 2 mol.
  • Medium : 2 mol of water contain 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules.
  • Exam-type : 11 g CO₂ = 0.25 mol; each molecule has 3 atoms, so atoms = 0.25 × 3 × 6.022 × 10²³ = 4.5165 × 10²³.

Rule 4: Gas volume (using 22.4 L/mol at 0 °C and 1 atm) — 5.6 L of N₂ = 5.6 ÷ 22.4 = 0.25 mol = 0.25 × 28 = 7 g.

  • Easy : 22.4 L of any gas = 1 mol.
  • Medium : 44.8 L of O₂ = 2 mol = 64 g.
  • Exam-type : 7 g N₂ and 7 g CO have equal molar mass (28), so equal moles and equal volumes.
Live figure

Atomic and molecular masses, the mole and molar mass

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Remember

One mole of different substances has different masses but the same number of particles.

Common mistake

Using "mole of oxygen" without saying atoms or molecules. 1 mol of O atoms is 16 g; 1 mol of O₂ molecules is 32 g.

Shortcut

Memory clue: "Grams → ÷M → Moles → ×Nₐ → Particles".

Spot the difference

PointAtomic massMolar mass
Unitug/mol
Refers toOne atomOne mole of atoms
Value for carbon12 u12 g/mol

Practice questions

Easy
  1. 1Find the molecular mass of CO₂ (C = 12, O = 16).
  2. 2How many moles are there in 90 g of water?
  3. 3What is the value of the Avogadro constant?
Medium
  1. 4How many molecules are in 0.5 mol of H₂?
  2. 5What is the mass of 3 mol of NaCl (58.5 g/mol)?
  3. 6Boron has isotopes of mass 10 (20 %) and 11 (80 %). Find the average atomic mass.
Challenging
  1. 7How many atoms are present in 8.8 g of CO₂?
  2. 8What volume does 3.2 g of O₂ occupy at 0 °C and 1 atm (22.4 L/mol)?
Show answers (try them yourself first)
  1. 112 + 32 = 44 u.
  2. 290 ÷ 18 = 5 mol.
  3. 36.022 × 10²³ mol⁻¹.
  4. 40.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules.
  5. 53 × 58.5 = 175.5 g.
  6. 60.20 × 10 + 0.80 × 11 = 10.8 u.
  7. 78.8 ÷ 44 = 0.2 mol; atoms = 0.2 × 3 × 6.022 × 10²³ = 3.613 × 10²³.
  8. 83.2 ÷ 32 = 0.1 mol; volume = 0.1 × 22.4 = 2.24 L.

Percentage composition, empirical and molecular formula

Link to Std 9–10 : You wrote formulas from valencies; now we work backwards from the data to find the formula.

From daily life

A nutrition label tells you the percentage of fat, protein and sugar in a packet of biscuits. In the same way a chemist reports the percentage of each element in a compound.

Simple definition

Mass percent of an element = (mass of the element in 1 mole of compound ÷ molar mass) × 100. The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of atoms in a molecule.

From percentages to the formula1. Take % as grams (100 g sample)2. ÷ atomic mass → moles3. ÷ smallest → ratio4. Whole numbers → empirical formula5. × n → molecular formula

In the figure : Steps from percentages to the empirical formula and then to the molecular formula.

Rule 1: Percentage composition — Mass % = (part ÷ whole) × 100.

  • Easy : CaCO₃ (100 g/mol): Ca = 40 %, C = 12 %, O = 48 %.
  • Medium : In H₂O: H = 2 ÷ 18 × 100 = 11.11 %.
  • Exam-type : In Na₂SO₄ (142 g/mol): Na = 46 ÷ 142 × 100 = 32.4 %, S = 22.5 %, O = 45.1 %.

Rule 2: Finding the empirical formula — (1) treat the percentages as grams; (2) divide by atomic mass to get moles; (3) divide all by the smallest; (4) if a ratio is like 1.5, multiply all by 2.

  • Easy : 85.7 % C and 14.3 % H: moles 7.14 and 14.3, ratio 1 : 2, empirical formula CH₂.
  • Medium : 40 % C, 6.7 % H, 53.3 % O: moles 3.33 : 6.7 : 3.33 = 1 : 2 : 1, empirical formula CH₂O.
  • Exam-type : 29.1 % Na, 40.5 % S, 30.4 % O: moles 1.265 : 1.266 : 1.9 = 1 : 1 : 1.5, multiply by 2 → Na₂S₂O₃.

Rule 3: Molecular formula — Molecular formula = (empirical formula) × n, where n = molar mass ÷ empirical formula mass.

  • Easy : CH₂ (14) with molar mass 56 gives n = 4, so C₄H₈.
  • Medium : CH₂O (30) with molar mass 90 gives n = 3, so C₃H₆O₃.
  • Exam-type : Empirical and molecular formula are equal when n = 1 (for example H₂O).
Remember

Percentages that add to 100 are treated as grams in a 100 g sample.

Common mistake

Writing the mole ratio without making it whole numbers. A ratio like 1 : 1 : 1.5 must be doubled.

Shortcut

Memory clue: "% → grams → moles → ratio → whole numbers".

Spot the difference

PointEmpirical formulaMolecular formula
ShowsSimplest ratioActual atoms
Example for glucoseCH₂OC₆H₁₂O₆
Needs molar mass?NoYes

Practice questions

Easy
  1. 1Find the mass percent of oxygen in CO₂.
  2. 2What is the empirical formula of C₆H₆?
  3. 3What is the empirical formula of C₂H₄O₂?
Medium
  1. 4Find the mass percent of nitrogen in NH₃.
  2. 5A compound has 80 % C and 20 % H. Find its empirical formula.
  3. 6Find n if the empirical formula is CH₂O (30) and molar mass is 180.
Challenging
  1. 7A compound has 52.2 % C, 13.0 % H, 34.8 % O and molar mass 46. Find its molecular formula.
  2. 8Find the mass percent of water in CuSO₄·5H₂O (CuSO₄ = 159.5, H₂O = 18).
Show answers (try them yourself first)
  1. 132 ÷ 44 × 100 = 72.73 %.
  2. 2CH.
  3. 3CH₂O.
  4. 414 ÷ 17 × 100 = 82.35 %.
  5. 5C: 80 ÷ 12 = 6.67; H: 20 ÷ 1 = 20; ratio 1 : 3, so CH₃.
  6. 6n = 180 ÷ 30 = 6 (molecular formula C₆H₁₂O₆).
  7. 7Moles: C 4.35, H 13.0, O 2.175 → 2 : 6 : 1, empirical C₂H₆O (46); n = 1, so C₂H₆O.
  8. 8Molar mass 159.5 + 90 = 249.5; water = 90 ÷ 249.5 × 100 = 36.07 %.

Stoichiometry, limiting reagent and concentration of solutions

Link to Std 9–10 : You balanced equations; now we use them to find how much substance is needed or formed.

From daily life

To make 10 sandwiches with 2 slices each you need 20 slices and 10 fillings. If you have only 15 slices, bread runs out first and decides how many sandwiches you can make.

Simple definition

Stoichiometry uses the coefficients of a balanced equation as mole ratios. The limiting reagent is the reactant that is used up first and decides the amount of product. Molarity (M) = moles of solute per litre of solution.

Limiting reagent: N₂ + 3H₂ → 2NH₃N₂0.5 mol : used up, limitingH₂1.5 mol used0.5 leftH₂ needed = 3 × 0.5 = 1.5 molAvailable: 0.5 mol N₂ and 2 mol H₂ → 1 mol NH₃

In the figure : Three molecules of H₂ are needed for each N₂; the shorter bar runs out first.

Rule 1: Mass–mass calculations — (1) balance the equation; (2) convert given mass to moles; (3) use the mole ratio; (4) convert to the mass wanted.

  • Easy : 2Mg + O₂ → 2MgO: 2 mol Mg (48 g) gives 2 mol MgO (80 g).
  • Medium : 12 g of Mg gives 12 ÷ 24 = 0.5 mol, so 0.5 mol MgO = 20 g.
  • Exam-type : For 40 g MgO, Mg needed = 40 ÷ 40 = 1 mol = 24 g.

Rule 2: Limiting reagent — Divide moles of each reactant by its coefficient; the smaller value is the limiting reagent.

  • Easy : N₂ + 3H₂ → 2NH₃ with 14 g N₂ (0.5 mol) and 4 g H₂ (2 mol): 0.5 ÷ 1 = 0.5 and 2 ÷ 3 = 0.67, so N₂ is limiting.
  • Medium : NH₃ formed = 2 × 0.5 = 1 mol = 17 g.
  • Exam-type : H₂ used = 3 × 0.5 = 1.5 mol, so 0.5 mol (1 g) of H₂ is left.

Rule 3: Concentration terms — Mass % = (mass of solute ÷ mass of solution) × 100; mole fraction x = n(component) ÷ total moles; molarity M = moles of solute ÷ volume of solution in L; molality m = moles of solute ÷ mass of solvent in kg.

  • Easy : 4 g NaOH (40 g/mol) in 250 mL: M = 0.1 ÷ 0.25 = 0.4 M.
  • Medium : 9 g glucose in 200 g water: moles = 0.05; molality = 0.05 ÷ 0.2 = 0.25 m.
  • Exam-type : 36 g water (2 mol) + 46 g ethanol (1 mol): mole fraction of ethanol = 1 ÷ 3 = 0.333.

Rule 4: Dilution — M₁V₁ = M₂V₂ (the moles of solute stay the same).

  • Easy : To make 200 mL of 0.5 M from 2 M stock, V₁ = 0.5 × 200 ÷ 2 = 50 mL.
  • Medium : 100 mL of 1 M HCl diluted to 500 mL becomes 0.2 M.
  • Exam-type : Molality does not change with temperature, but molarity does, because volume changes.
Live figure

Stoichiometry, limiting reagent and concentration of solutions

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Remember

Always compare mole ratios, not masses, to find the limiting reagent.

Common mistake

Using the volume of the solvent instead of the volume of the whole solution in molarity.

Shortcut

Memory clue: "Moles ÷ coefficient: smallest is limiting".

Spot the difference

PointMolarityMolality
DenominatorLitres of solutionkg of solvent
Changes with temperature?YesNo

Practice questions

Easy
  1. 1Define limiting reagent.
  2. 2What is the molarity of 0.2 mol of solute in 500 mL of solution?
  3. 3Mole fraction of A in a mixture of 1 mol A and 3 mol B?
Medium
  1. 42H₂ + O₂ → 2H₂O: how many grams of water form from 4 g of H₂ (excess O₂)?
  2. 5How many grams of NaCl (58.5 g/mol) are in 250 mL of 0.2 M solution?
  3. 6How much water (mL) must be added to 100 mL of 1 M solution to get 0.25 M?
Challenging
  1. 75.6 g of N₂ and 3 g of H₂ react (N₂ + 3H₂ → 2NH₃). Find the limiting reagent and the mass of NH₃ formed.
  2. 8Calculate the molality of a solution of 11.7 g NaCl in 500 g water.
Show answers (try them yourself first)
  1. 1The reactant that is completely used up first and limits the amount of product.
  2. 20.2 ÷ 0.5 = 0.4 M.
  3. 31 ÷ (1 + 3) = 0.25.
  4. 44 g H₂ = 2 mol; water formed = 2 mol = 36 g.
  5. 5Moles = 0.2 × 0.25 = 0.05; mass = 0.05 × 58.5 = 2.925 g.
  6. 6Final volume = 1 × 100 ÷ 0.25 = 400 mL, so add 300 mL.
  7. 7N₂ = 0.2 mol, H₂ = 1.5 mol; 0.2 ÷ 1 < 1.5 ÷ 3, so N₂ is limiting; NH₃ = 0.4 mol = 6.8 g.
  8. 8NaCl = 0.2 mol; molality = 0.2 ÷ 0.5 = 0.4 m.