If 50 cm$^{3}$ of a saturated solution of KNO$_{3}$ at 40 °C contained 5.05 g of the salt, its solubility at the same temperature would be [KNO$_{3}$ = 101]

Analyze the question: Solubility here is expressed in mol dm⁻³, so we need the concentration of KNO₃ in the saturated solution.

Concept:

• Moles = mass ÷ molar mass
• Concentration (mol dm⁻³) = moles ÷ volume in dm³

Solution:

1. Moles of KNO₃ = 5.05 g ÷ 101 g/mol = 0.05 mol

2. Volume = 50 cm³ = 50 ÷ 1000 = 0.05 dm³

3. Concentration = 0.05 mol ÷ 0.05 dm³ = 1.0 mol dm⁻³

Tip: Always convert cm³ to dm³ by dividing by 1000 before calculating concentration. A quick shortcut is that concentration = (mass ÷ molar mass) × (1000 ÷ volume in cm³). Here that is 0.05 × 20 = 1.0. Also notice the numbers were chosen to be neat (5.05 is exactly 101 ÷ 20), which is a good sign you're on the right track.