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.
To calculate the amount of ethylene glycol needed to prevent water from freezing at -20°C, we need to use the concept of freezing point depression.
Freezing point depression refers to the lowering of the freezing point of a solvent (in this case, water) when a non-volatile solute (ethylene glycol) is added. The formula for freezing point depression is:
ΔTf = Kf * m * i
Where:
ΔTf = Change in freezing point
Kf = Cryoscopic constant
m = Molality (moles of solute per kilogram of solvent)
i = Van't Hoff factor (number of particles when the solute dissolves)
In this case, we want to calculate the molality (m) of the ethylene glycol needed to prevent freezing at -20°C. The change in freezing point (ΔTf) would be 20°C (since water's freezing point is 0°C), and the cryoscopic constant (Kf) for water is 1.86°C/m.
Rearranging the formula, we get:
m = ΔTf / (Kf * i)
Since ethylene glycol is a non-ionic compound, the Van't Hoff factor (i) is 1.
m = 20°C / (1.86°C/m * 1)
m ≈ 10.75 m
Now, we can calculate the amount of ethylene glycol needed in moles:
moles = m * kg of water
Considering that the density of water is 1 kg/L, the kg of water is 1 kg.
moles = 10.75 m * 1 kg = 10.75 moles
The molar mass of ethylene glycol is 62.07 g/mol. Therefore:
grams = moles * molar mass
grams = 10.75 moles * 62.07 g/mol ≈ 667.1 g
Therefore, approximately 667.1 grams of ethylene glycol must be added to 1L of water to keep it from freezing at -20°C.
delta T = Kf*m
You know delta T (it's 20C), and Kf, solve for m.
Then m = mols/kg solvent.
You know m and kg solvent (1 kg), solve for mols.
Then mols = grams/molar mass
You know mols and molar rmass, solve for grams.