What is the molality of a solution prepared by dissolving 18.0 g of glucose (C6H12O6, molar mass = 180 g/mol) in 250 g of pure water?
Answer and explanation
Correct answer: A. 0.40 mol/kg
- Step 1: Moles of glucose = 18.0 g / 180.0 g/mol = 0.10 mol. Step 2. Mass of solvent in kilograms = 250 g / 1000 = 0.250 kg. Step 3. Molality (m) = moles of solute / mass of solvent (kg) = 0.10 mol / 0.250 kg = 0.40 mol/kg. Answer: 0.40 mol/kg. Why not the others: 0.10 is total moles; 0.04 results from dividing by 2500.
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