Question

Answer these please.

Answer 1

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