How many grams of ethylene glycol (c2h6o2) must be added to 1.15 kg of water to produce a solution that freezes at -4.46â°c?

232.5 g C2H6O2

Explanation:

The equation you need to use here is ΔTf = i Kf m

Since pure water freezes at 0 C, your ΔTf is just 4.46 C

i = 1 (ethylene glycol is a weak electrolyte)

Kf = molal freezing constant, which for water is 1.86 C/m

m = molality = x mols C2H6O2 / 1.15 kg H2O (don't know the moles of ethylene glycol we're dissolving yet)

Than,

4.46 C = 1.86 C/m (x mol C2H6O2 / 1.15 kg H2O)

Solve for x, you should get x = 2.75 mol C2H6O2

3.75 mol C2H6O2 (62 g C2H6O2 / 1 mol C2H6O2) = 232.5 g C2H6O2

the answer is 75.0%.