Practice question
Question
What is the molality of a solution prepared by dissolving 12 g of glucose (C₆H₁₂O₆) in 48 g of water, if the density of water is 1 g/mL? (Molar mass: C₆H₁₂O₆ = 180 g/mol)
Explanation
Moles of glucose = 12/180 ≈ 0.0667 mol. Mass of water = 48 g = 0.048 kg. Molality = 0.0667 / 0.048 ≈ 1.39 m.
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