3.0 g of a non-electrolyte dissolved in 50 g of water raises the boiling point by 0.26 K. Kb for water is 0.52 K kg mol−1. What is the molar mass of the solute?
- Molality = ΔTb ÷ Kb = 0.26 ÷ 0.52 = 0.50 mol kg−1
- Moles of solute in 50 g (0.050 kg) of water = 0.50 × 0.050 = 0.025 mol
- Molar mass = mass ÷ moles = 3.0 ÷ 0.025 = 120 g mol−1
Answer: 120 g mol−1