What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?

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.