Question

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.

Answer 1

Answer:

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

Explanation:

• We can solve this problem using the relation:

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.

• We can get the molality of the solution using the relation:

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.

Answer 2

The freezing point lowering is depicted in the equation,
                                F = Fs + Kf x m
where F and Fs are the freezing point of the solution and water, respectively. Kf and m are the freezing point depression and molality. Substituting the given values,
                                 F = 0°C + (-1.86 °C/m) x ((400/ 64) / 4) 
The, the freezing point is approximately equal to -3.0°C.