Volume of 8.625 g of sulphur di oxide at RTP in cm3

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.

that's what i got you can count the number of atoms