Question

How many grams of antifreeze C2H4(OH)2 would be required per 500 g of water to prevent the water from freezing at a temperature of -20.0 C

Answer 1

Answer:

333.7 g.

Explanation:

• The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: ΔTf = Kf.m.

Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).

Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).

m is the molality of the solution.

∴ m = ΔTf/Kf = (20.0°C)/(1.86 °C/m) = 10.75 m.

molaity (m) is the no. of moles of solute per kg of the solvent.

∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))

∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.

∵ no. of moles = mass/molar mass.

∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass = (5.376 mol)(62.07 g/mol) = 333.7 g.

Answer 2

The mass of antifreeze required is 345 g.

What are colligative properties?

The term colligative properties  refer to those properties that depend on the amount of substance. Freezing point is a colligative property.

We know that;

ΔT = k m i

ΔT = freezing point depression

k = freezing constant

m = molality

i = Van't Hoft factor

Hence;

0 - (-20) = 1.86 * x/0.5 * 1

Where x = number of moles of antifreeze

20 = 1.86x/0.5

x = 20 * 0.5/1.86

x = 5.56 moles

number of moles = mass/molar mass

Molar mass of antifreeze = 62 g/mol

Mass of antifreeze =  5.56 moles * 62 g/mol = 345 g

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