Answer:
The volume of 8.625 g of sulphur dioxide at RTP is approximately 3,230.84 cm³
Explanation:
The question requires the determination of the volume occupied by the gas based on the molar volume of a gas at STP
The given parameters of the sulphur dioxide, SO₂, gas are;
The mass of the given SO₂ gas = 8.625 g
The molar mass of SO₂ = 64.07 g/mol
The number of moles, 'n', in the given sample of SO₂ gas = Mass of SO₂/(Molar Mass of SO₂)
∴ The number of moles of SO₂ in the gas sample = 8.625 g/(64.07 g/mol) ≈ 0.134618386 moles
The molar volume of a gas at RTP is approximately 24 dm³/mole
24 dm³ = 24,000 cm³
∴ The molar volume of a gas at RTP is approximately 24,000 cm³/mole
The volume occupied by a gas at RTP = (The number of moles of the gas) × (The Molar Volume of a gas at RTP)
∴ The volume occupied by the 8.625 g of SO₂ gas at RTP = 0.134618386 moles × 24,000 cm³/mole ≈ 3,230.84 cm³
The volume occupied by the 8.625 g of SO₂ gas at RTP ≈ 3,230.84 cm³.
