The Mole And The Avogadro Constant
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लेसन नोट्स
The Mole and Avogadro Constant
- The mole (mol) is the SI unit for amount of substance.
- One mole of any substance contains 6.02 × 10²³ particles (Avogadro constant).
- Particles can be atoms, molecules, ions, or formula units.
- Example: 1 mol Na = 6.02 × 10²³ Na atoms; 1 mol H₂ = 6.02 × 10²³ H₂ molecules.
- Molar mass is the mass of one mole of a substance (g/mol).
- For elements, molar mass = relative atomic mass in grams.
- For compounds, molar mass = relative molecular/formula mass in grams.
Mole–Mass–Mr Relationship
- Use the formula triangle: moles = mass / Mr.
- Mass = moles × Mr; Mr = mass / moles.
- Always show working to avoid errors and gain credit.
- Example: 0.250 mol Zn 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.
Formula triangle for moles, mass and molar mass
Mole and Volume of Gas
- Avogadro's Law: equal moles of any gas occupy equal volume at same T and P.
- At RTP (20 °C, 1 atm), molar gas volume = 24 dm³ (24 000 cm³).
- Volume = moles × 24 dm³ (or × 24 000 cm³).
- Moles = volume (dm³) / 24 (or volume (cm³) / 24 000).
- Example: 3 mol H₂ occupies 3 × 24 = 72 dm³.
- Example: 1200 cm³ O₂ gives moles = 1200 / 24 000 = 0.05 mol.
Molar gas volume (dm³) formula triangle
Calculating Number of Particles
- Number of particles = moles × Avogadro constant (6.02 × 10²³).
- For compounds, count atoms per formula unit.
- Example: 1 mol MgCl₂ contains 6.02 × 10²³ molecules, 3 × 6.02 × 10²³ atoms.
- Example: 15.7 g H₂O (Mr = 18) → moles = 15.7/18 = 0.872 mol → molecules = 0.872 × 6.02 × 10²³ = 5.25 × 10²³.
Reacting Masses
- Use balanced equation to find mole ratio between reactants/products.
- Convert given mass to moles using moles = mass / Mr.
- Use ratio to find moles of required substance.
- Convert back to mass using mass = moles × Mr.
- Example: 6.0 g Mg (Mr = 24) → 0.25 mol Mg → 1:1 ratio → 0.25 mol MgO → mass = 0.25 × 40 = 10 g.
Limiting Reactant
- Limiting reactant is used up first and determines amount of product.
- Excess reactant remains after reaction stops.
- To find limiting reactant: calculate moles of each, compare with balanced equation ratio.
- Example: 9.2 g Na (0.40 mol) + 8.0 g S (0.25 mol); equation 2Na + S → Na₂S requires 0.40 mol Na to react with 0.20 mol S; S (0.25 mol) is excess, Na is limiting.
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प्रैक्टिस सवाल
फ्री प्रीव्यू — 60 में से 8 सवाल। सभी देखने के लिए साइन अप करें।
1.What is the Avogadro constant?
Easy- A6.02 × 1023
- B6.02 × 1022
- C6.02 × 1024
- D6.02 × 1021
2.How many moles are there in 2.64 g of sucrose (C12H22O11, Mr = 342)? Give your answer in standard form to 3 significant figures.
Medium- A7.72 × 10⁻³
- B7.72 × 10⁻²
- C1.30 × 10²
- D2.72 × 10⁻³
3.State the volume, in dm³, occupied by one mole of any gas at room temperature and pressure.
Easy4.One mole of any substance contains 6.02 × 1023 particles.
EasyTrue or false?
5.Complete the sentence.
EasyThe mass of one mole of a substance is called its ____ mass.
6.How many moles of carbon dioxide gas are present in 960 dm³ at RTP? (Molar volume = 24 dm³/mol)
Easy- A40
- B36
- C48
- D44
7.Match each quantity to its correct value.
Medium- Number of atoms in 1 mole of MgCl2
- Number of chloride ions in 1 mole of MgCl2
- Number of molecules in 1 mole of MgCl2
- 6.02 × 1023
- 1.204 × 1024
- 1.806 × 1024
8.Arrange the following steps in the correct order to calculate the mass of product from a given mass of reactant in a chemical reaction.
Medium- Calculate moles of reactant using mass ÷ Mr
- Use the balanced equation to find the mole ratio
- Calculate mass of product using moles × Mr
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