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Answer:

  • a)  1.20 × 10²⁴ atoms
  • b) 10 g
  • c) 0.7 mol
  • d) 1,000 g
  • e) 0.5 mol
  • f) 3.0 × 10²² atoms
  • g) 0.20 mol
  • h) 2.0 mol/dm³
  • i) 2.8 g

Explanation:

a)

1. Data:

  • Cl: 71g
  • Ar: Cl 35.5

2. Solution:

i) Formulae:

  • number of moles = mass in grams/Ar
  • number of atoms = number of moles × Avogadro constant
  • Avogadro constant = 6.022 × 10²³ atoms/mol

ii) Calculations:

  • number of moles = 71g / 35.5(g/mol) = 2 mol
  • number of atoms = 2mol × 6.022 × 10²³ atoms/mol = 1.20 × 10²⁴ atoms

b)

1.  Data:

  • 0.2 mol KOH
  • Ar: H = 1, O = 16, k = 39

2. Solution

i)Formula:

  • mass = number of moles × molar mass

ii) Molar mass:

  • 1×1g/mol + 1×39g/mol + 1×16g/mol = 56g/mol

iii) Calculations:

  • mass = 0.2 mol × 56g/mol = 11.2 g/mol ≈ 10g/mol (rounded to 1 significant figure)

c)

1. Data:

  • 1.4g Li
  • Ar: Li = 7

2. Solution

i) Formula:

  • number of moles = mass in grams / Ar

ii) Calculations:

  • number of moles = 1.4g / 7 gmol = 0.7 mol

d)

1. Data:

  • S₈
  • Ar: S = 32

2. Solution

i) Formula:

  • mass = number of moles × molar mass

ii) Calculations:

  • molar mass = 8 × 32g/mol = 256 g/mol
  • mass = 4 mol × 256 g/mol = 1,024 g ≈ 1,000 g (rounded to 1 significant figure)

e)

1. Data:

  • mass: 50 g
  • Ca(NO₃)₂ . 2H₂O
  • Mr: Ca(NO₃)₂ . 2H₂O = 200

2. Solution

i) Formulae:

  • number of moles = mass in grams / molar mass

ii) Calculations:

  • number of moles = 50 g / 200 g/mol = 0.25 mol of Ca(NO₃)₂ . 2H₂O

The number of moles of water molecules is 2 times the number of moles of Ca(NO₃)₂ . 2H₂O.

  • number of moles of water molecules = 2 × 0.25 mol = 0.5 mol

f)

1. Data:

  • 4.9g
  • Mr: H₂SO₄ = 98

2. Solution:

i) Formulae:

  • number of atoms = number of moles × Avogadro constant
  • number of moles = mass in grams / molar mass

ii) Calculations

  • number of moles = 4.9g / 98g/mol = 0.050 mol
  • number of atoms = 0.050 mol × 6.022 × 10²³ atoms/mol = 3.0×10²² atoms

g)

1. Data:

  • V = 24 dm³ (air)
  • 20% oxygen
  • r.t.p ⇒ T = 298K, p = 1 atm

2. Solution

i) Formula:

  • pV = nRT

ii) Calculation:

  • n = (pV)/(RT)
  • R = 0.08206 (atm.dm³/K.mol)
  • n = [ 1 atm × 24 dm³] / (0.08206atm.dm³/K.mol × 298K) = 0.98mol of air

  • moles of oxygen = 20% × 0.98 mol ≈ 0.20 mol

h)

1. Data:

  • V = 125cm³
  • 0.25 mol

2. Solution

i) Formula:

  • Molarity  = moles of solute / volume of solution in dm³

ii) Calculations:

  • Convert 125 cm³ to liter

       25cm³ × (0.1cm/dm)³ = 0.125 dm³

  • Molarity = 0.25 mol / 0.125 dm³ = 2.0 mol/dm³

i)

1. Data:

  • V = 35cm³
  • M = 2 mol/dm³
  • Mr NaOH = 40

2. Solution

i) Formulae:

  • number of moles = M × V (in dm³)
  • mass = number of moles × mola rmass

ii) Calculations:

  • V = 35cm³ × (0.1dm/cm)³ = 0.035dm³
  • number of moles = 2 mol/dm³ × 0.035 dm³ = 0.070 mol
  • mass = 0.070 mol × 40 g/mol = 2.8 g

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