Question

How much ethylene glycol (C2H6O2, the major component of antifreeze) must be added to 1 L of water to keep it from freezing at -20 oC?

Kf = 1.86 oC/m.

Answer 1

To prevent water from freezing at -20°C, you need to add about 667.25 grams of ethylene glycol to 1L of water, based on the freezing point depression formula and the molar mass of ethylene glycol.

Step-by-step explanation:

To calculate how much ethylene glycol (C2H6O2) must be added to 1L of water to prevent freezing at -20℃, we use the freezing point depression formula ΔTf = i * Kf * m, where ΔTf is the freezing point depression, i is the van't Hoff factor (which is 1 for ethylene glycol because it does not ionize in solution), Kf is the freezing point depression constant for water (1.86℃/m), and m is the molality of the solution. In order to prevent water from freezing at -20℃, the freezing point must be lowered by 20 degrees since pure water freezes at 0℃.

ΔTf = 20℃ = (1) * (1.86℃/m) * m

We solve for m:

m = ΔTf / Kf

m = 20℃ / 1.86℃/m = 10.75 m

Next, we calculate the mass of ethylene glycol needed using its molar mass (62.07 g/mol). The molality (m) is the moles of solute per kilogram of solvent. Therefore:

Moles of ethylene glycol = 10.75 moles/kg * 1 kg water = 10.75 moles

Mass of ethylene glycol = Moles * Molar mass

Mass of ethylene glycol = 10.75 moles * 62.07 g/mol = 667.2525 g

So, to keep water from freezing at -20℃, you would need to add approximately 667.25 grams of ethylene glycol to 1L of water.

Answer 2

Colligative properties calculations are used for this type of problem. Calculations are as follows:

ΔT(freezing point) = -(Kf)m
-20.0 = -1.86 °C kg / mol (m)
m = 10.75 mol/kg

10.75 mol C2H6O2 / kg water ( 1 L water ) ( 1 kg/L ) (62.07 g/mol )= 667.25 g

Hope this answers the question. Have a nice day.