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What is the normal boiling point in ∘C of an antifreeze solution prepared by dissolving 766.9 g of ethylene glycol (C2H6O2) in 500.0 g of water? The molal boiling-point-elevation constant for water is 0.51 (∘C⋅kg)/mol.

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Answer:

Boiling T° of solution is 112.6°C

Explanation:

Boiling point elevation → ΔT =  Kb . m

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Let's determine m (molality)

Moles of solute / 1kg of solvent. We determine moles of solute:

766.9 . 1mol / 62 g = 12.37 moles

Now, we conver the mass of solvent from g to kg → 500 g. 1kg/1000g = 0.5kg

Molal → 12.37 mol / 0.5kg = 24.74 m

Boiling T° of solution - 100°C = 0.51 °C /m . 24.74 m

Boiling T° of solution = 0.51 °C /m . 24.74 m + 100°C → 112.6°C

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