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Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24 is kg (Molar mass in g mol-1 for ethylene glycol 62, Kf of water = 1.86 K kg mol-1)
15
Tf = Kf m = Kf nsoluteWwater (kg) (ethylene glycol: non-electrolyte, i = 1)
Tf = 0 − (−24) = 24 K
⇒ 24 = 1.86 w/6218.6
⇒ w = 24 62 18.61.86 = 24 × 62 × 10 = 14880 g
∴ w = 14.88 kg ≈ 15 kg
Answer: 15
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