Ethylene glycol (c2h6o2) is used as an antifreeze in cars. if 400 g of ethylene glycol is added to 4.00 kg of water, what is the molality? calculate how much the freezing point of water will be lowered. the freezing-point depression constant for water is kf = –1.86°c/m. show your work.

∴ The freezing point of water will be lowered by 3.0 °C.

Explanation:

ΔTf = Kf.m,

Where, ΔTf is the depression of freezing point,

Kf is the freezing point depression constant (Kf for water = -1.86 °C/m),

m is the molality of the solution.

m = (mass / molar mass) solute x (1000 / mass of solvent)

m = (400.0 g / 62.07 g/mol) x (1000 / 4000.0 g) = 1.61 m.

∴ ΔTf = Kf.m = (-1.86 °C/m) (1.61 m) = -2.99 °C ≅ -3.0°C.

∴ The freezing point of water will be lowered by 3.0 °C.

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