What approximate proportions by volume of water and ethylene glycol (d=1.12g/cc) must be mixed to ensure protection of an automobile cooling system to -10*C?

According to depression of freezong point,T = Kf m m = T1.86W2M × W1 =101.86 W262 × 1 =101.86W2 = 10 × 62 × 11.86 = 333.33 gSo for 1000 g of water 333.33 g of ethylene glycol is neededFor 1000 ml of water 333.331.12 = 297.62 g neededSo the amount of ethylene glycol is 297.62 g for 1000 ml of water

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