Calculate the solubility of sodium trioxocarbonate (IV) in mol dm⁻³ given 30 g of it is dissolved in 40 g of water. Use appropriate atomic masses and the density of water.

The chemical name sodium trioxocarbonate (IV) refers to sodium carbonate, Na₂CO₃.
We need the atomic masses: Na = 23 g/mol, C = 12 g/mol, O = 16 g/mol.
Molar mass of Na₂CO₃ = (2 × Na) + C + (3 × O)
Molar mass = (2 × 23 g/mol) + 12 g/mol + (3 × 16 g/mol)
Molar mass = 46 g/mol + 12 g/mol + 48 g/mol
Molar mass of Na₂CO₃ = 106 g/mol

Moles = mass / molar mass
Moles of Na₂CO₃ = 30 g / 106 g/mol
Moles of Na₂CO₃ ≈ 0.283 mol

We assume the density of water is 1 g/mL (or 1 g/cm³).
Volume of water = mass / density
Volume of water = 40 g / (1 g/mL) = 40 mL
To convert mL to dm³: 1 dm³ = 1000 mL
Volume of water = 40 mL × (1 dm³ / 1000 mL)
Volume of water = 0.040 dm³

Solubility = moles of solute / volume of solvent (in dm³)
Solubility = 0.283 mol / 0.040 dm³
Solubility ≈ 7.075 mol/dm³
Rounding to two significant figures (based on the given masses of 30 g and 40 g):
Solubility = 7.1 mol/dm³