Step-by-step explanation:
The freezing point depression is a colligative property. We have to find the mass of ethylene glycol that we have to add to 100.0 g of water to change its freezing point from 0.0 °C to -4.3 °C.
The freezing point depression for a solution can be calculated using the following equation:
ΔTf = kf * molality * i
Where ΔTf is the freezing point depression, kf is the freezing point depression constant (it depends on the solvent and for water is 1.86 °C/m), molality is the molality of the solution and i is the Van't Hoff factor.
The Van't Hoff factor represents the number of particles formed when that compound dissolves. In our case the solute is ethylene glycol, a covalent compound, so it won't form ions when dissolved in the water. Then i is equal to 1.
The temperature must change from 0.0 °C to -4.3 °C, then the freezing point depression is 4.3 °C. So we know that:
i = 1 kf = 1.86 °C/m ΔTf = 4.3 °C
We can replace those values in the formula and find the molality of the solution.
ΔTf = kf * molality * i
4.3 °C = 1.86 °C/m * molality * 1
molality = 4.3 °C/(1.86 °C/m)
molality = 2.31 m
Now we can get the moles of ethylene glycol from the definition of the molal concentration. Molality are the moles of solute per kg of solvent. The mass of water is 100 g.
mass of solvent = 100.0 g * 1 kg/(1000 g)
mass of solvent = 0.100 kg
molality = moles of solute/(mass of solvent in kg)
moles of solute = molality * mass of solvent in kg
moles of solute = 2.31 m * 0.100 kg
moles of solute = 0.231 moles
And finally we can convert the moles of ethylene glycol to grams using its molar mass.
molar mass of C₂H₆O₂ = 62.07 g/mol
mass of C₂H₆O₂ = 0.231 moles * 62.07 g/mol
mass of C₂H₆O₂ = 14.4 g
Answer: 14.4 g of ethylene glycol must be mixed.