50 g of ethylene glycol (CH₂OH)₂ dissolved in 200 g water; cooled to −9.3°C. Find mass of ice that separates. (Kf = 1.86 K·kg/mol)

Molar mass of ethylene glycol (C₂H₆O₂) = 62 g/mol

Step 1 — Moles of solute:

n = 50 / 62 = 0.8065 mol

Step 2 — Freezing point depression formula: ΔTf = Kf × m

9.3 = 1.86 × (0.8065 / W_solvent_kg)

Step 3 — Mass of liquid water remaining:

W_solvent = (1.86 × 0.8065) / 9.3 = 0.1613 kg = 161.3 g

Step 4 — Ice separated = total water − liquid water:

Ice = 200 − 161.3 = 38.7 g

Answer: ≈38.71 g of ice separates out

When a solution cools, only pure water freezes. As ice forms, the remaining solution becomes more concentrated, lowering its freezing point further.