Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
ΔTf = (Kf)(m),
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
So, the mass of 575 mL is 575 g = 0.575 kg.
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
∴ ΔTf = (Kf)(m) = (-1.86 °C/m)(2.36 m) = - 4.39 °C.
∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.
∴ The freezing point of the solution is - 4.39 °C.
