The Mole & The Avogadro Constant

விளையாடிக் கற்றுக்கொள்ளுங்கள்

ஆற்றல் சம்பாதிக்க இந்த கேள்விகளுக்குப் பதிலளியுங்கள், பின்னர் மீன் பிடித்து ஆராயுங்கள். கணக்கு தேவையில்லை.

ஆசிரியர்களுக்கு: The Mole & The Avogadro Constant (Chemistry, CIE)-க்கான தயாரான பாட ஸ்லைடுகள், திருப்புதல் குறிப்புகள் — உங்கள் பாடத்தில் அவற்றைப் பயன்படுத்தவும், அல்லது மாணவர்கள் நேரலை விளையாட்டாக விளையாடும் ஊடாடும் வகுப்பு செயல்பாடாக தலைப்பை இயக்கவும்.

பாட குறிப்புகள்

The Mole & Avogadro Constant

The mole (mol) is the SI unit for amount of substance.
One mole contains 6.02 × 10²³ particles (atoms, molecules, or ions) – the Avogadro constant.
One mole of sodium (Na) contains 6.02 × 10²³ Na atoms; one mole of H₂ contains 6.02 × 10²³ H₂ molecules.
Molar mass of an element = relative atomic mass in grams; for a compound = relative formula mass in grams.
At RTP (20 °C, 1 atm), one mole of any gas occupies 24 dm³ (or 24 000 cm³) – the molar gas volume.
Gas volume (dm³) = moles × 24; moles = volume (dm³) ÷ 24.

Molar gas volume formula triangle

Linking Moles, Mass & Mr

Moles = mass (g) ÷ Mᵣ (or Aᵣ for elements).
Mass = moles × Mᵣ.
Example: 0.250 mol Zn (Aᵣ=65) has mass = 0.250 × 65 = 16.25 g.
Example: 2.64 g sucrose (Mᵣ=342) gives moles = 2.64 ÷ 342 = 7.72 × 10⁻³ mol.
Use the formula triangle: cover the quantity you want to find.

Moles, mass and molar mass formula triangle

Reacting Masses & Limiting Reactants

Use balanced equation to find molar ratio between reactants and products.
Convert given mass to moles, then use ratio to find moles of required substance, then convert back to mass.
Limiting reactant is the one used up first; it determines the amount of product.
To identify limiting reactant: calculate moles of each reactant, compare with the molar ratio from the equation.
Example: 2Mg + O₂ → 2MgO; 6.0 g Mg (0.25 mol) gives 0.25 mol MgO (mass = 10 g).

Calculating Concentration

Concentration in g/dm³ = mass of solute (g) ÷ volume (dm³).
Concentration in mol/dm³ = moles of solute ÷ volume (dm³).
To convert g/dm³ to mol/dm³, divide by molar mass; to convert mol/dm³ to g/dm³, multiply by molar mass.
Always convert cm³ to dm³ by dividing by 1000.
Example: 10 g NaOH in 2 dm³ gives 5 g/dm³.

Concentration, moles and volume formula triangle

Titration Calculations

Use the balanced equation to find the mole ratio between acid and base.
Calculate moles of known solution: moles = concentration × volume (dm³).
Use ratio to find moles of unknown, then concentration = moles ÷ volume (dm³).
Example: 25.0 cm³ HCl titrated with 12.1 cm³ 0.100 mol/dm³ NaOH (1:1 ratio) gives [HCl] = 0.0484 mol/dm³.

Concentration, moles and volume formula triangle

Empirical & Molecular Formula

Empirical formula is the simplest whole‑number ratio of atoms in a compound.
To find empirical formula: divide each element's mass (or % mass) by its Aᵣ, then divide by the smallest result to get a ratio; multiply to get whole numbers if needed.
Molecular formula = (empirical formula)ₙ, where n = Mᵣ(compound) ÷ Mᵣ(empirical formula).
Example: empirical formula CH₂O (Mᵣ=30); if Mᵣ=60, n=2, so molecular formula = C₂H₄O₂.
Water of crystallisation: use similar method to find x in hydrated salt formula (e.g., CuSO₄·xH₂O).

Percentage Yield & Purity

Percentage yield = (actual yield ÷ theoretical yield) × 100%.
Actual yield is the mass obtained; theoretical yield is calculated from the balanced equation.
Percentage purity = (mass of pure substance ÷ total mass of sample) × 100%.
Percentage composition by mass = (total mass of element in compound ÷ Mᵣ of compound) × 100%.
Example: Fe₂O₃ (Mᵣ=160) contains 112 g Fe → %Fe = (112÷160)×100 = 70%.

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இலவச முன்னோட்டம் — 44-இல் 8 கேள்விகள். அனைத்தையும் பார்க்க பதிவு செய்யவும்.

1.Which row correctly describes the mole and the value of Avogadro's constant?
Easy
- Aone mole of a substance is equal to the substance's relative atomic or molecular mass in grams; one mole of a substance contains 6.02 × 10²³ atoms, molecules or formula units
- Bone mole of a substance is equal to the substance's atomic number in grams; one mole of a substance contains 6.02 × 10²³ atoms, molecules or formula units
- Cone mole of a substance is equal to the substance's relative atomic or molecular mass in grams; one mole of a substance contains 12.04 × 10²³ atoms, molecules or formula units
- Done mole of a substance is equal to the substance's atomic number in grams; one mole of a substance contains 12.04 × 10²³ atoms, molecules or formula units
2.What is the mass of one mole of carbon dioxide (CO₂)? (Ar: C = 12, O = 16)
Easy
- A28 g
- B44 g
- C32 g
- D12 g
3.How many molecules are there in one mole of hydrogen gas (H₂)?
Easy
- A6.02 × 10²³ molecules
- B1.204 × 10²⁴ molecules
- C3.01 × 10²³ molecules
- D12.04 × 10²³ molecules
4.The complete combustion of methane produces carbon dioxide and steam. CH₄(g) + 2O₂(g) → 2H₂O(g) + CO₂(g). Which statements about the reaction are correct? 1. The empirical formula of methane is CH₄ 2. The number of atoms in 1 mole of methane is 4 × Avogadro's constant 3. 1 mole of methane produces 72 dm³ of gaseous products at r.t.p. 4. 1 mole of methane occupies a volume of 12 dm³ at r.t.p.
Medium
- A1, 2 and 3
- B1 and 2
- C1 and 3
- D2 and 4
5.Magnesium carbonate and hydrochloric acid react: MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g). What is the volume of CO₂ produced when 21 g of magnesium carbonate (Mr = 84) reacts completely with excess hydrochloric acid? (One mole of gas occupies 24.0 dm³ at r.t.p.)
Medium
- A4 dm³
- B8 dm³
- C6 dm³
- D2 dm³
6.Calcium carbonate undergoes thermal decomposition: CaCO₃ → CaO + CO₂. What mass of calcium oxide (Mr = 56) is formed when 60 g of calcium carbonate (Mr = 100) is completely decomposed?
Medium
- A28 g
- B18.5 g
- C60 g
- D33.6 g
7.Substance X was analysed and found to contain 40.00% carbon, 6.67% hydrogen and 53.33% oxygen by mass. What is the empirical formula of X?
Medium
- AC₄H₈O₄
- BC₆H₁₂O₈
- CC₂H₄O₂
- DCH₂O
8.What is the concentration in mol dm⁻³ of a solution of sodium hydroxide that contains 16 g of NaOH in 200 cm³ of distilled water? (Mr NaOH = 40)
Medium
- A2 mol dm⁻³
- B2.5 mol cm⁻³
- C0.2 mol dm⁻³
- D1 mol cm⁻³

மேலும் கேள்விகள்

9.A student titrated 25 cm³ of hydrochloric acid against a standard solution of sodium hydroxide of 0.1 mol dm⁻³. The average titre of sodium hydroxide was 30 cm³. What is the concentration of the acid? (HCl + NaOH → NaCl + H₂O)
Medium
- A0.50 mol dm⁻³
- B0.12 mol dm⁻³
- C1.25 mol dm⁻³
- D1.0 mol dm⁻³
10.How many oxygen atoms are there in 10.0 g of magnesium nitrate, Mg(NO₃)₂? (Avogadro constant L = 6.02 × 10²³ mol⁻¹, Ar: Mg=24, N=14, O=16)
Hard
- A4.07 × 10²²
- B8.14 × 10²²
- C1.22 × 10²³
- D2.44 × 10²³
11.What is the volume of hydrogen gas produced, measured at r.t.p., when 5.0 g of magnesium reacts with an excess of dilute hydrochloric acid? (Mg + 2HCl → MgCl₂ + H₂, Ar Mg = 24, molar gas volume = 24 dm³ at r.t.p.)
Medium
- A0.21 dm³
- B2.5 dm³
- C5.0 dm³
- D120 dm³
12.A sample of chalk (limestone) with a mass of 1.70 g was reacted with excess HCl and 0.36 dm³ of CO₂ gas was collected at r.t.p. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. What is the percentage purity of the chalk? (Mr CaCO₃ = 100, molar gas volume = 24 dm³ at r.t.p.)
Hard
- A75%
- B88%
- C90%
- D98%
13.A student investigates the reaction of zinc with hydrochloric acid, using a gas syringe to collect the hydrogen gas produced. Several volume readings are taken as the reaction proceeds, and one reading falls noticeably off the general trend. Which of the following is the most likely reason for that one inaccurate reading?
Medium
- AThe zinc was not added all at once
- BThe temperature of the acid was not constant
- CThe gas syringe had a leak
- DThe concentration of acid was too high
14.Calcium carbide and water react: CaC₂(s) + 2H₂O(l) → C₂H₂(g) + Ca(OH)₂(s). What volume of ethyne (C₂H₂) is produced at r.t.p. when 45 g of water reacts completely? (Molar volume at r.t.p. = 24 dm³ mol⁻¹; Ar: H=1, O=16, Ca=40, C=12)
Medium
- A72 dm³
- B30 dm³
- C24 dm³
- D6 dm³
15.A student thermally decomposes 24.8 g of copper(II) carbonate and obtains 11.2 g of copper(II) oxide. CuCO₃ → CuO + CO₂. What is the percentage yield of CuO? (Ar: Cu=63.5, C=12, O=16)
Medium
- A55%
- B70%
- C82%
- D90%
16.A student was asked to find the concentration in g dm⁻³ of a solution of sulfuric acid. She transferred 25 cm³ of the acid to a 250 cm³ volumetric flask and made it up to the mark. She titrated 25 cm³ of the diluted acid against 0.100 mol dm⁻³ NaOH. The average titre of NaOH was 18.0 cm³. What is the concentration of the original acid in g dm⁻³? (Mr H₂SO₄ = 98)
Hard
- A9.85 g dm⁻³
- B12.25 g dm⁻³
- C98 g dm⁻³
- D35.28 g dm⁻³
17.A chemist reacted 0.6 g of magnesium with 0.6 dm³ of oxygen in a sealed container to produce magnesium oxide: 2Mg + O₂ → 2MgO. Which row correctly describes the limiting reactant? (Ar: Mg=24, molar volume = 24 dm³ mol⁻¹)
Medium
- AMg is the limiting reactant because it is used up first.
- BO₂ is the limiting reactant because it is used up first.
- CMg is in excess, so O₂ is limiting.
- DO₂ is in excess, so Mg is limiting.
18.What is the mass of one mole of calcium nitrate, Ca(NO₃)₂? (Ar: Ca=40, N=14, O=16)
Easy
- A102 g
- B132 g
- C164 g
- D196 g
19.How many atoms are there in 0.500 moles of phosphorus pentoxide, P₄O₁₀? (Avogadro constant L = 6.02 × 10²³ mol⁻¹)
Medium
- A4.21 × 10²⁴
- B7.53 × 10²³
- C3.01 × 10²³
- D1.20 × 10²⁴
20.What is the concentration in mol dm⁻³ of a solution containing 4.0 g of sodium hydroxide, NaOH, dissolved in 250 cm³ of water? (Mr NaOH = 40)
Medium
- A0.10 mol dm⁻³
- B0.20 mol dm⁻³
- C0.40 mol dm⁻³
- D1.00 mol dm⁻³
21.A compound contains 43.4% sodium, 11.3% carbon, and 45.3% oxygen by mass. What is its empirical formula? (Ar: Na=23, C=12, O=16)
Medium
- ANa₂CO₃
- BNaCO₂
- CNa₂C₂O₅
- DNaCO₃
22.The reaction between hydrogen and iodine is H₂(g) + I₂(g) → 2HI(g). How many moles of hydrogen iodide are produced when 2.0 moles of hydrogen react completely with excess iodine?
Easy
- A1.0 mol
- B2.0 mol
- C3.0 mol
- D4.0 mol
23.A student investigated the reaction of zinc powder with dilute hydrochloric acid. The same mass of zinc was added to different volumes of acid and the total volume of hydrogen gas produced was measured. The results are shown in the table. | Volume of acid / cm³ | 5 | 10 | 15 | 20 | 25 | 30 | 35 | |---|---|---|---|---|---|---|---| | Volume of hydrogen / cm³ | 30 | 60 | 90 | 120 | 150 | 180 | 180 | Which volume of acid gives the maximum volume of hydrogen?
Medium
- A25 cm³
- B30 cm³
- C35 cm³
- D20 cm³
24.What is the SI unit of amount of substance?
Easy
- Amole
- Bgram
- Clitre
- DAvogadro
25.One mole of any substance contains how many particles?
Easy
- A6.02 × 10²³
- B12.04 × 10²³
- C6.02 × 10²²
- D1.00 × 10²⁴
26.What is the mass of one mole of helium atoms? (Ar: He = 4)
Easy
- A4 g
- B2 g
- C6 g
- D8 g
27.Which of the following statements about the mole is correct?
Easy
- AOne mole of a substance contains the same number of particles as the number of atoms in 12 g of carbon-12.
- BOne mole of a substance contains 6.02 × 10²² particles.
- COne mole of a substance has a mass equal to its atomic number in grams.
- DOne mole of any gas occupies 24 dm³ at any temperature and pressure.
28.How many molecules are there in 0.5 moles of carbon dioxide? (Avogadro constant L = 6.02 × 10²³ mol⁻¹)
Easy
- A3.01 × 10²³
- B6.02 × 10²³
- C1.20 × 10²⁴
- D1.51 × 10²³
29.What is the volume of 2 moles of oxygen gas at room temperature and pressure (molar volume = 24 dm³/mol)?
Easy
- A48 dm³
- B12 dm³
- C24 dm³
- D96 dm³
30.One mole of any substance contains 6.02 × 10²³ particles.
Easy
True or false?
31.The mass of one mole of carbon atoms is 12 g.
Easy
True or false?
32.Match each quantity to its correct value for one mole of oxygen gas (O₂). (Avogadro constant L = 6.02 × 10²³ mol⁻¹, Ar: O=16)
Medium
- Number of molecules
- Number of atoms
- Mass
- 32 g
- 6.02 × 10²³
- 1.204 × 10²⁴
33.At room temperature and pressure (RTP), what volume does one mole of any gas occupy?
Easy
- A24 dm³
- B24 cm³
- C22.4 dm³
- D2400 cm³
34.Arrange the following steps in the correct order for calculating the mass of product from a given mass of reactant using reacting masses.
Easy
- Write the balanced chemical equation.
- Convert mass of reactant to moles.
- Use the mole ratio to find moles of product.
- Convert moles of product to mass.
35.Concentration can be expressed in units of g/dm³ or mol/dm³.
Easy
True or false?
36.Which of the following statements about the mole is correct? (Select all that apply)
Easy
- AOne mole of any substance contains 6.02 × 10²³ particles.
- BThe mole is the SI unit for amount of substance.
- COne mole of carbon atoms has a mass of 6 g.
- DThe Avogadro constant is 6.02 × 10²³ mol⁻¹.
37.At room temperature and pressure (RTP), one mole of any gas occupies a volume of 24 dm³.
Easy
True or false?
38.A student heats 8.4 g of sodium hydrogencarbonate (NaHCO₃, Mr = 84) until it decomposes completely: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂. What mass of sodium carbonate (Na₂CO₃, Mr = 106) is produced?
Medium
- A5.3 g
- B10.6 g
- C2.65 g
- D8.4 g
39.What volume, in dm³, is occupied by 0.5 moles of carbon dioxide gas at RTP? (molar volume = 24 dm³/mol)
Easy
40.Define the term 'limiting reactant' in a chemical reaction.
Easy
41.A solution of sodium hydroxide contains 4.0 g of NaOH in 500 cm³ of water. What is the concentration in mol/dm³? (Ar: Na=23, O=16, H=1)
Medium
42.A student titrated 25.0 cm³ of hydrochloric acid against a standard solution of sodium hydroxide of concentration 0.100 mol/dm³. The average titre of sodium hydroxide was 30.0 cm³. Calculate the concentration of the hydrochloric acid in mol/dm³. The equation for the reaction is: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l).
Medium
43.What volume, in dm³, is occupied by 0.5 moles of carbon dioxide gas at RTP? (molar volume = 24 dm³/mol)
Easy
- A11 dm³
- B14 dm³
- C13 dm³
- D12 dm³
44.A solution of sodium hydroxide contains 4.0 g of NaOH in 500 cm³ of water. What is the concentration in mol/dm³? (Ar: Na=23, O=16, H=1)
Medium
- A2.2 mol/dm³
- B0.2 mol/dm³
- C1.2 mol/dm³
- D-0.8 mol/dm³

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