If 24.4 g of lead(II) trioxonitrate(V) were dissolved in 42 g of distilled water at 20 °C, calculate the solubility of the solute in g dm⁻³.

Correct answer: 581.000 g dm⁻³

Explanation: Solubility in g per 100 g of water is
\(\displaystyle \frac{24.4}{42}\times 100 = 58.1\) g per 100 g H₂O.

At 20 °C, take density of water ≈ 1 g cm⁻³, so 100 g H₂O ≈ 100 cm³ = 0.1 dm³. Therefore, solubility per 1 dm³ is:

58.1 × 10 = 581 g dm⁻³

• 581.000 — Correct.
• 0.581 — Misses the “per 100 g → per dm³” conversion (×1000 factor).
• 5.810 — Off by a factor of 100.
• 58.100 — This is g per 100 g H₂O, not per dm³.