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State Board · Class 11 · Chemistry

Lesson 2: Introduction to Analytical Chemistry

7 concepts · 56 practice questions with answers · Free to read

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Qualitative and quantitative analysis

From daily life

Is milk adulterated (identity) and how much fat does it hold (amount)? Both are analysis questions.

Simple definition

Analytical chemistry finds out what a substance contains (qualitative) and how much (quantitative).

Qualitative and quantitativeAnalytical chemistrySampleQualitative: what is it?flame testprecipitation testLassaigneQuantitative: how much?titrationgravimetryspectrometryFirst identity (qualitative), then amount (quantitative)e.g. is Pb present in water ⇒ how many ppm?

In the figure : Qualitative and quantitative.

Rule 1: Qualitative analysis — Identifying ions or elements: flame test, precipitation, colour change, Lassaigne test.

  • Easy : Na⁺: golden-yellow flame.
  • Medium : Cl⁻ + AgNO₃ ⇒ white AgCl.
  • Exam-type : Fe³⁺ + SCN⁻ ⇒ blood-red.

Rule 2: Quantitative analysis — Measuring amounts: titration (volumetric), gravimetry (weighing precipitate), spectrometry (light absorption).

  • Easy : 20 mL of 0.1 M HCl neutralises 25 mL NaOH ⇒ 0.08 M.
  • Medium : SO₄²⁻ from the mass of BaSO₄.
  • Exam-type : Concentration from colour intensity.

Rule 3: Choosing a method — Based on sample size, expected amount (major/trace), accuracy needed and cost.

  • Easy : Large amount ⇒ titration.
  • Medium : Trace (ppm) ⇒ spectrometry.
  • Exam-type : Identity first, amount next.

Calculation: molarity of NaOH by titration

  • Step 1: M₁V₁ = M₂V₂ (1 : 1 reaction).
  • Step 2: 0.1 × 20 = M × 25.
  • Step 3: M = 2 ÷ 25 = 0.08 M.

CET level (MCQ)

  • CET 1) A quantitative method? 1) flame test 2) titration 3) precipitate colour 4) smell — Answer: 2).
  • CET 2) A qualitative question asks? 1) how much 2) what 3) where 4) when — Answer: 2).

Real-life use: Food adulteration, water testing, drug quality.

Remember

Qualitative = what; quantitative = how much; titration M₁V₁ = M₂V₂.

Common mistake

Thinking a flame test gives amounts; it only identifies.

CET trap

M₁V₁ = M₂V₂ in titration holds only for 1 : 1 reactions.

Shortcut

Memory clue: "quality = what, quantity = how much".

Spot the difference

ItemQualitativeQuantitative
Answerwhathow much
Exampleflame testtitration

Practice questions

Easy
  1. 1What is analytical chemistry?
  2. 2Two qualitative tests.
  3. 3Two quantitative methods.
Medium
  1. 4Test for Cl⁻?
  2. 520 mL 0.1 M HCl vs 25 mL NaOH ⇒ M?
  3. 6What is titration?
Challenging
  1. 7When use spectrometry?
  2. 8Which analysis comes first?
Show answers (try them yourself first)
  1. 1Finding what and how much.
  2. 2Flame and precipitation.
  3. 3Titration and gravimetry.
  4. 4AgNO₃ ⇒ white precipitate.
  5. 50.08 M.
  6. 6Finding an unknown with a known solution.
  7. 7For trace amounts.
  8. 8Qualitative.

Scientific notation, accuracy, precision and errors

From daily life

A scale showing 1.9 kg for 2 kg is an error; if it always shows 1.9 it is precise but not accurate.

Simple definition

Scientific notation: N × 10ⁿ (1 ≤ N < 10). Accuracy: closeness to the true value. Precision: closeness of repeated measurements.

Accuracy precisionAccuracy vs precisionaccurate + preciseprecise, not accurateaccurate, not preciseneither% error = (measured − true) ÷ true × 100

In the figure : Accuracy and precision.

Rule 1: Scientific notation — Decimal moved right ⇒ negative n; left ⇒ positive n. Multiply: add powers; divide: subtract.

  • Easy : 0.000345 = 3.45 × 10⁻⁴.
  • Medium : (3 × 10⁴)(2 × 10³) = 6 × 10⁷.
  • Exam-type : 0.5 mol = 3.011 × 10²³ particles.

Rule 2: Accuracy versus precision — Four cases: both; precise only (systematic error); accurate only (right average); neither.

  • Easy : All 9.8 g (true 10.0) ⇒ precise, not accurate.
  • Medium : 9.7, 10.3, 10.0 ⇒ accurate on average.
  • Exam-type : 10.0, 10.0, 10.0 ⇒ both.

Rule 3: Types of errors — Systematic (one direction) and random. Absolute error = |measured − true|; % error = absolute ÷ true × 100.

  • Easy : True 10.0, measured 9.8 ⇒ 0.2, 2 %.
  • Medium : A wrong zero ⇒ systematic.
  • Exam-type : Reading variation ⇒ random.

Calculation: percent error

  • Step 1: Absolute error = |9.8 − 10.0| = 0.2.
  • Step 2: ÷ true value = 0.2 ÷ 10.0 = 0.02.
  • Step 3: × 100 = 2 %.

CET level (MCQ)

  • CET 1) 0.000345 = ? 1) 3.45×10⁻³ 2) 3.45×10⁻⁴ 3) 34.5×10⁻⁵ 4) 3.45×10⁴ — Answer: 2).
  • CET 2) A systematic error lowers? 1) precision 2) accuracy 3) both 4) neither — Answer: 2).

Real-life use: Weighing scales, lab measurement, quality control.

Remember

N × 10ⁿ; accurate = near true; precise = near each other; % error ÷ true.

Common mistake

Treating accuracy and precision as the same.

CET trap

Percent error is divided by the true value, not the measured one.

Shortcut

Memory clue: "precision = repeatability".

Spot the difference

ItemAccuracyPrecision
Meaningnear truenear each other
Errorsystematicrandom

Practice questions

Easy
  1. 10.000345 in scientific notation.
  2. 2(3×10⁴)(2×10³)?
  3. 3What is accuracy?
Medium
  1. 4What is precision?
  2. 5True 10.0, measured 9.8 ⇒ % error?
  3. 6Example of systematic error.
Challenging
  1. 79.8, 9.8, 9.8 (true 10.0) ⇒ ?
  2. 8Particles in 0.5 mol?
Show answers (try them yourself first)
  1. 13.45 × 10⁻⁴.
  2. 26 × 10⁷.
  3. 3Closeness to the true value.
  4. 4Closeness among measurements.
  5. 52 %.
  6. 6A wrong zero setting.
  7. 7Precise, not accurate.
  8. 83.011 × 10²³.

Significant figures and rounding off

From daily life

A scale saying 2.50 kg rather than 2.5 kg shows how finely it measured.

Simple definition

Significant figures: the certain digits plus the last uncertain one. An answer cannot be more precise than the measurement.

Significant figuresSignificant figures0.00345sig. figs: 32.0050sig. figs: 54500sig. figs: 24.500 × 10³sig. figs: 40.0300sig. figs: 3Rules: leading zeros don't count; middle zeros and trailing decimal zeros doMultiply/divide: fewest sig. figs; add/subtract: fewest decimals

In the figure : Significant figures.

Rule 1: Counting rules — Non-zero digits count; middle zeros count; leading zeros do not; trailing zeros after a decimal count; 4500 has 2 (4.500×10³ has 4).

  • Easy : 0.00345: 3.
  • Medium : 2.0050: 5.
  • Exam-type : 0.0300: 3.

Rule 2: Rules in arithmetic — Multiply/divide: fewest significant figures. Add/subtract: fewest decimal places. Exact counts (5 students) have no limit.

  • Easy : 2.5 × 3.42 = 8.55 ⇒ 8.6 (2 s.f.).
  • Medium : 12.11 + 18.0 + 1.012 = 31.122 ⇒ 31.1.
  • Exam-type : Exact numbers are unlimited.

Rule 3: Rounding off — First dropped digit less than 5 ⇒ the last kept digit stays; 5 or more ⇒ add 1 to it (textbook rule).

  • Easy : 3.142 ⇒ 3.14 (2 < 5).
  • Medium : 7.465 ⇒ 7.47 (5 ⇒ +1).
  • Exam-type : 6.78 ⇒ 6.8.

Calculation: 2.5 × 3.42

  • Step 1: 2.5 × 3.42 = 8.55.
  • Step 2: 2.5 has 2 s.f. ⇒ answer has 2 s.f.
  • Step 3: In 8.55 the dropped digit is 5 ⇒ add 1 ⇒ 8.6.

CET level (MCQ)

  • CET 1) Significant figures in 0.0300? 1) 1 2) 2 3) 3 4) 4 — Answer: 3).
  • CET 2) 12.11 + 18.0 + 1.012 = ? 1) 31.122 2) 31.12 3) 31.1 4) 31 — Answer: 3).

Real-life use: Lab reports, calculations, instrument readings.

Remember

No leading zeros; add = decimals; multiply = sig. figs.

Common mistake

Counting sig. figs in addition; use decimal places there.

CET trap

4500 has no decimal point ⇒ its end zeros are not significant (2 s.f.); write 4.500 × 10³ for 4.

Shortcut

Memory clue: "× ÷ = figures, + − = decimals".

Spot the difference

OperationRule
× ÷fewest sig. figs
+ −fewest decimals

Practice questions

Easy
  1. 1Sig. figs of 0.00345?
  2. 2Sig. figs of 2.0050?
  3. 32.5 × 3.42?
Medium
  1. 412.11 + 18.0 + 1.012?
  2. 5Round 2.345 to 3 s.f.
  3. 6Significant figures in 4500?
Challenging
  1. 7Rule for addition?
  2. 8S.f. of exact numbers?
Show answers (try them yourself first)
  1. 13.
  2. 25.
  3. 38.6.
  4. 431.1.
  5. 52.35 (5 ⇒ +1).
  6. 62 (no decimal point, so end zeros do not count).
  7. 7Fewest decimal places.
  8. 8Unlimited.

Percent composition, empirical and molecular formula

From daily life

'Carbs 70 %' on a packet is a composition in percent.

Simple definition

Percent composition = mass of element ÷ total × 100. Empirical formula (EF): simplest whole-number ratio of atoms. Molecular formula (MF) = (EF) × n.

Formula flowFrom % composition to empirical and molecular formula%take 100 gg ÷ Mmoles÷ min÷ smallestEFempirical× nmoleculare.g. 40 % C, 6.7 % H, 53.3 % O; M = 180C : 40/12 = 3.33 · H : 6.7/1 = 6.7 · O : 53.3/16 = 3.33÷ 3.33 ⇒ 1 : 2 : 1 ⇒ EF = CH₂O (M = 30)n = 180 ÷ 30 = 6 ⇒ MF = C₆H₁₂O₆If a ratio ends in .5 multiply by 2; in .33 by 3MF = (EF) × n, n = M ÷ M(EF)

In the figure : % ⇒ EF ⇒ MF.

Rule 1: Percent composition — In Fe₂O₃ the Fe is (2 × 56) ÷ 160 × 100 = 70 %.

  • Easy : H in H₂O: 2/18 = 11.1 %.
  • Medium : Na in NaCl: 23/58.5 = 39.3 %.
  • Exam-type : Fe₂O₃: Fe 70 %, O 30 %.

Rule 2: Steps for EF — Take 100 g ⇒ g ÷ M ⇒ moles ⇒ ÷ smallest ⇒ whole numbers (.5 ⇒ ×2; .33 ⇒ ×3).

  • Easy : 40 % C, 6.7 % H, 53.3 % O ⇒ CH₂O.
  • Medium : 85.7 % C, 14.3 % H ⇒ CH₂.
  • Exam-type : 52.9 % Al, 47.1 % O ⇒ Al₂O₃.

Rule 3: Getting MF — n = M ÷ M(EF); MF = (EF)n. CH₂O with M = 180 ⇒ n = 6 ⇒ C₆H₁₂O₆.

  • Easy : CH₂, M = 84 ⇒ C₆H₁₂.
  • Medium : CH, M = 78 ⇒ C₆H₆.
  • Exam-type : HO, M = 34 ⇒ H₂O₂.

Calculation: 85.7 % C, 14.3 % H, M = 84

  • Step 1: C = 85.7/12 = 7.14; H = 14.3/1 = 14.3.
  • Step 2: ÷ 7.14 ⇒ C : H = 1 : 2 ⇒ EF = CH₂ (14).
  • Step 3: n = 84 ÷ 14 = 6 ⇒ C₆H₁₂.

CET level (MCQ)

  • CET 1) CH₂O with M = 60 ⇒ MF? 1) C₂H₄O₂ 2) CH₂O 3) C₃H₆O₃ 4) C₆H₁₂O₆ — Answer: 1).
  • CET 2) % Fe in Fe₂O₃? 1) 30 2) 56 3) 70 4) 72 — Answer: 3).

Real-life use: Drug purity, mineral analysis.

Remember

Take 100 g; moles; ÷ smallest; MF = EF × n.

Common mistake

Treating percentages as the ratio; convert to moles first.

CET trap

1.33 ⇒ ×3 (not 4); 1.5 ⇒ ×2.

Shortcut

Memory clue: "% → g → mol → ratio".

Spot the difference

ItemEFMF
GlucoseCH₂OC₆H₁₂O₆
BenzeneCHC₆H₆
H₂O₂HOH₂O₂

Practice questions

Easy
  1. 1% Fe in Fe₂O₃?
  2. 2% H in H₂O?
  3. 3What is EF?
Medium
  1. 4EF for 40 % C, 6.7 % H, 53.3 % O?
  2. 5MF: EF CH₂O, M = 180.
  3. 685.7 % C, 14.3 % H, M = 84 ⇒ MF?
Challenging
  1. 7What if a ratio is 1.5?
  2. 8CH, M = 78 ⇒ ?
Show answers (try them yourself first)
  1. 170 %.
  2. 211.1 %.
  3. 3Simplest whole-number ratio.
  4. 4CH₂O.
  5. 5C₆H₁₂O₆.
  6. 6C₆H₁₂.
  7. 7Multiply by 2.
  8. 8C₆H₆.

Stoichiometric calculations

From daily life

A cake recipe fixes eggs-to-flour ratios; a chemical equation fixes mole ratios.

Simple definition

Stoichiometry: using mole ratios from a balanced equation to find reactant and product amounts.

Mole mapThe mole mapmolenmass g× 1 ÷ Mparticles× 6.022×10²³STP volume× 22.4 LsolutionM × V (L)10 g CaCO₃ ⇒ 0.1 mol ⇒ 4.4 g CO₂ ⇒ 2.24 L (STP)CaCO₃ → CaO + CO₂ (1 : 1)

In the figure : Mole map.

Rule 1: Mole map — n = m ÷ M = particles ÷ 6.022×10²³ = V(STP) ÷ 22.4 = M × V(L).

  • Easy : 10 g CaCO₃ = 0.1 mol.
  • Medium : 0.1 mol CO₂ = 2.24 L (STP).
  • Exam-type : 0.2 M in 0.5 L = 0.1 mol.

Rule 2: Mass–mass problems — Balance ⇒ moles of given ⇒ ratio ⇒ moles of wanted ⇒ mass.

  • Easy : CaCO₃ → CaO + CO₂: 10 g ⇒ 4.4 g CO₂.
  • Medium : 2Al + 6HCl: 5.4 g Al ⇒ 0.3 mol H₂.
  • Exam-type : N₂ + 3H₂: 56 g N₂ ⇒ 68 g NH₃.

Rule 3: Volume and particle problems — For gases at STP 1 mol = 22.4 L; Avogadro gives particles. At equal T, P gas volume ∝ moles.

  • Easy : 5.4 g Al ⇒ 6.72 L H₂.
  • Medium : 2H₂ + O₂: 4 L H₂ needs 2 L O₂.
  • Exam-type : 0.1 mol ⇒ 6.022×10²² molecules.

Calculation: H₂ (STP) from 5.4 g Al + HCl

  • Step 1: 2Al + 6HCl → 2AlCl₃ + 3H₂; n(Al) = 5.4 ÷ 27 = 0.2 mol.
  • Step 2: 2 : 3 ⇒ H₂ = 0.2 × 3/2 = 0.3 mol.
  • Step 3: V = 0.3 × 22.4 = 6.72 L.

CET level (MCQ)

  • CET 1) CO₂ (STP) from 10 g CaCO₃? 1) 1.12 L 2) 2.24 L 3) 4.48 L 4) 22.4 L — Answer: 2).
  • CET 2) In 2Al : 3H₂, 0.2 mol Al gives H₂? 1) 0.2 2) 0.3 3) 0.4 4) 0.6 — Answer: 2).

Real-life use: Industrial yield, fertiliser amounts, fuel combustion.

Remember

Balanced equation ⇒ mole ratio; 22.4 L; 6.022×10²³.

Common mistake

Using grams directly in the ratio; use moles first.

CET trap

22.4 L is only at STP (0 °C, 1 atm).

Shortcut

Memory clue: "grams → mol → mol → grams".

Spot the difference

QuantityFormula
Molesm ÷ M
Particlesn × 6.022×10²³
Volumen × 22.4 L

Practice questions

Easy
  1. 1Moles in 10 g CaCO₃?
  2. 2Mass of CO₂ from it?
  3. 3H₂ in L from 5.4 g Al?
Medium
  1. 4Molecules in 0.1 mol?
  2. 5NH₃ from 56 g N₂?
  3. 6O₂ for 4 L H₂?
Challenging
  1. 7What is STP?
  2. 8Why not grams directly?
Show answers (try them yourself first)
  1. 10.1 mol.
  2. 24.4 g.
  3. 36.72 L.
  4. 46.022×10²².
  5. 568 g.
  6. 62 L.
  7. 70 °C and 1 atm.
  8. 8Equations are in moles.

Limiting reagent and percent yield

From daily life

With 10 bread slices and 3 cheese slices you make only 3 sandwiches; cheese is limiting.

Simple definition

Limiting reagent is used up first and fixes the product. Percent yield = actual ÷ theoretical × 100.

Limiting reagentLimiting reagent: 2H₂ + O₂ → 2H₂OStartAfterH₂ 2 mol(4 g)O₂ 0.5 mol(16 g)H₂ left1 mol, 2 gO₂ 0H₂O 1 mol(18 g)0.5 mol O₂ needs just 1 mol H₂ ⇒ O₂ is limitingYield is always from the limiting reagent

In the figure : Limiting reagent.

Live figure

Limiting reagent and percent yield

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Rule 1: Finding it — Moles ÷ coefficient for each; the smaller value is limiting.

  • Easy : 2H₂ + O₂: 2/2 = 1; 0.5/1 = 0.5 ⇒ O₂.
  • Medium : N₂ 0.1, H₂ 0.5: 0.1/1 vs 0.5/3 = 0.167 ⇒ N₂.
  • Exam-type : Equal values ⇒ both fully used.

Rule 2: Product and leftover — Product from the limiting reagent; leftover = initial − used.

  • Easy : 0.5 mol O₂ ⇒ 1 mol H₂O = 18 g.
  • Medium : H₂ left 2 − 1 = 1 mol = 2 g.
  • Exam-type : 0.1 mol N₂ ⇒ 0.2 mol NH₃ = 3.4 g.

Rule 3: Percent yield — % yield = (actual ÷ theoretical) × 100; always ≤ 100 %.

  • Easy : 3.0 g / 3.4 g = 88.2 %.
  • Medium : Theoretical 20 g, actual 15 g = 75 %.
  • Exam-type : Above 100 % = wet or impure product.

Calculation: 4 g H₂ and 16 g O₂

  • Step 1: n(H₂) = 2 mol; n(O₂) = 0.5 mol.
  • Step 2: 2/2 = 1; 0.5/1 = 0.5 ⇒ O₂ is limiting.
  • Step 3: H₂O = 2 × 0.5 = 1 mol = 18 g; 1 mol H₂ is left.

CET level (MCQ)

  • CET 1) 2H₂ + O₂: 4 g H₂ + 16 g O₂ ⇒ H₂O? 1) 9 g 2) 18 g 3) 20 g 4) 36 g — Answer: 2).
  • CET 2) Theoretical 20 g, actual 15 g ⇒ % yield? 1) 25 2) 60 3) 75 4) 133 — Answer: 3).

Real-life use: Cost control in factories, drug synthesis.

Remember

Moles ÷ coefficient ⇒ smaller = limiting; % yield ≤ 100.

Common mistake

Calling the reagent with fewer grams limiting; check moles.

CET trap

Forgetting to divide by the coefficient gives the wrong limiting reagent.

Shortcut

Memory clue: "smallest share = limiting".

Spot the difference

ItemLimitingExcess
Stateused upleft over
Decidesproductnothing

Practice questions

Easy
  1. 1What is a limiting reagent?
  2. 24 g H₂ + 16 g O₂ ⇒ limiting?
  3. 3H₂O formed?
Medium
  1. 4H₂ left over?
  2. 5NH₃ from 2.8 g N₂ + 1 g H₂?
  3. 6Actual 3.0 g, theoretical 3.4 g ⇒ %?
Challenging
  1. 7Why not above 100 %?
  2. 8Why divide by the coefficient?
Show answers (try them yourself first)
  1. 1The reactant used up first.
  2. 2O₂.
  3. 318 g.
  4. 41 mol (2 g).
  5. 53.4 g.
  6. 688.2 %.
  7. 7Actual ≤ theoretical.
  8. 8To match the equation ratio.

Concentration of solutions

From daily life

One spoon versus three spoons of sugar in tea is a difference in concentration.

Simple definition

Molarity M = mol ÷ L of solution. Molality m = mol ÷ kg of solvent. Mole fraction x = n₁ ÷ (n₁ + n₂). ppm = parts per 10⁶.

ConcentrationConcentration of solutionsMolarity Mmol ÷ L ( द्रावण )Molality mmol ÷ kg ( solvent )Mole fraction xn₁ ÷ (n₁ + n₂)ppmg ÷ 10⁶ gMass %g ÷ 100 g4 g NaOH in 500 mL ⇒ 0.2 M · in 100 g water ⇒ 1 mDilution: M₁V₁ = M₂V₂ (2 M, 100 mL → 0.5 M = 400 mL)M depends on temperature; m does not

In the figure : Concentration units.

Live figure

Concentration of solutions

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Rule 1: Molarity and dilution — M = n ÷ V(L). In dilution moles stay constant: M₁V₁ = M₂V₂.

  • Easy : 4 g NaOH in 500 mL ⇒ 0.2 M.
  • Medium : 2 M, 100 mL ⇒ 0.5 M needs 400 mL.
  • Exam-type : 0.5 L of 0.04 M NaOH ⇒ 0.8 g.

Rule 2: Molality and mole fraction — Molality is temperature-independent; molarity is not. Mole fractions add to 1.

  • Easy : 4 g NaOH in 100 g water ⇒ 1 m.
  • Medium : x(NaOH) = 0.1 ÷ 5.656 = 0.0177.
  • Exam-type : x₁ + x₂ = 1.

Rule 3: Percent and ppm — Mass % = g/100 g; ppm = g/10⁶ g ≈ mg/L (in water).

  • Easy : 5 mg/L ≈ 5 ppm.
  • Medium : 10 g salt in 90 g water ⇒ 10 %.
  • Exam-type : 0.2 ppm Cl₂ ⇒ 0.2 g/m³.

Calculation: dilute 100 mL of 2 M to 0.5 M

  • Step 1: M₁V₁ = M₂V₂ ⇒ 2 × 100 = 0.5 × V₂.
  • Step 2: V₂ = 200 ÷ 0.5 = 400 mL.
  • Step 3: Water to add = 400 − 100 = 300 mL.

CET level (MCQ)

  • CET 1) 4 g NaOH in 500 mL ⇒ M? 1) 0.1 2) 0.2 3) 0.4 4) 2 — Answer: 2).
  • CET 2) Independent of temperature? 1) M 2) N 3) m 4) V — Answer: 3).

Real-life use: Drug solutions, fertiliser mixes, water testing.

Remember

M = mol/L; m = mol/kg solvent; M₁V₁ = M₂V₂; ppm = mg/L.

Common mistake

Using kg of solution in molality; it is kg of solvent.

CET trap

In M, V is litres of solution, not of water.

Shortcut

Memory clue: "M = Litre of solution, m = kg of solvent".

Spot the difference

UnitFormulaTemperature
Mmol/Ldepends
mmol/kgindependent
xn₁/Σnindependent

Practice questions

Easy
  1. 14 g NaOH in 500 mL ⇒ M?
  2. 24 g NaOH in 100 g water ⇒ m?
  3. 3Mole fraction of NaOH?
Medium
  1. 4How to make 0.5 M from 100 mL of 2 M?
  2. 55 mg/L = ? ppm
  3. 6Why does M depend on temperature?
Challenging
  1. 7Sum of mole fractions?
  2. 8NaOH in 0.5 L of 0.04 M?
Show answers (try them yourself first)
  1. 10.2 M.
  2. 21 m.
  3. 30.0177.
  4. 4Add 300 mL of water.
  5. 55 ppm.
  6. 6Volume changes.
  7. 71.
  8. 80.8 g.

Quick revision of the whole lesson

ConceptKey formula / fact
Qualitative/quantitativewhat / how much
% errormeasured − true÷ true × 100
Sig. figs× ÷ figures; + − decimals
EF → MFMF = EF × n
Molem ÷ M = V ÷ 22.4
Limitingsmallest mol ÷ coeff.
% yieldactual ÷ theoretical × 100
DilutionM₁V₁ = M₂V₂
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🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Qualitative/quantitative.
  • Accuracy, errors.
  • Significant figures.
  • EF and MF.
  • Stoichiometry.
  • Limiting reagent.
  • Concentration.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.