Answer:
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
