The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves in water is antifreeze (ethylene glycol). How many grams of antifreeze, CH2OHCH2OH (62.10 g/mol), must be dissolved in 280.0 grams of water to reduce the freezing point by 0.400 oC

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

Where ΔT is change in temperature = 0.400°C

Kf is freezing point constant of the solvent = 1.86°C/m

m is molality of the solution (Moles of solute / kg of solvent)

And i is Van't Hoff constant (1 for a nonelectrolyte)

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

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