From the diagram above, find the amount of solute deposited when 200 cm³ of the solution is cooled from 55 °C to 40 °C.

Step 1. Read values from the solubility curve:

• At 55 °C, the solubility ≈ 6.0 moles of solute per dm³ (from the graph).
• At 40 °C, the solubility ≈ 5.0 moles of solute per dm³.

Step 2. Find the difference in solubility:

\(\Delta n = 6.0 - 5.0 = 1.0 \ \text{mol per dm}^3\)

Step 3. Adjust for the actual volume (200 cm³ = 0.200 dm³):

Moles deposited = \(1.0 \times 0.200 = 0.20 \ \text{mol}\)

Final Answer: 0.20 mol