An aqueous antifreeze solution is 60.0% ethylene glycol (HOCH2CH2OH) by mass and has a density of 1.06 g/mL. Calculate the molality of ethylene glycol. Enter your answer to 1 decimal place.

[HOCH₂CH₂OH] = 24.1 m

Explanation:

Ethylene glycol → HOCH₂CH₂OH

60% by mass means that 60 g of ethylene glycol are contained in 100 g of solution.

Solution mass = Solute mass + Solvent mass

100 g = 60 g + Solvent mass

Solvent mass = 40 g

Molality are the moles of solute contained in 1kg of solvent.

We determine the moles of solute → 60 g . 1mol/62 g = 0.967 moles

We convert the mass of solvent from g to kg → 40 g . 1kg/1000 g = 0.04 kg

Molality → 0.967 mol / 0.04 kg = 24.1 m

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