A solution has 30 g of ethylene glycol (molar mass=62 g/mol) in 1000 g of water. What is the molality?
Q1. A solution has 30 g of ethylene glycol (molar mass=62 g/mol) in 1000 g of water. What is the molality?
Answer: 0.48 m
Explanation: Moles = 30 / 62 ≈ 0.484 mol. Mass of solvent = 1 kg. Molality = 0.484 mol/kg. Option C uses 30 / 62 × 2.