What is the change in the freezing point of water when 68.5 g of sucrose is dissolved in 500.0 g of water?Kf of water = -1.86°C/molmolar mass sucrose = 342.30 g/moli value of sugar = 1A.0.0186°CB.0.254°CC.-0.744°CD.-0.215°C

To solve this problem, we need to use the formula for freezing point depression:

ΔTf = Kf * m * i

where: ΔTf is the change in freezing point, Kf is the cryoscopic constant of the solvent (water in this case), m is the molality of the solution, and i is the van't Hoff factor (which is 1 for sucrose because it doesn't ionize in solution).

First, we need to calculate the molality of the solution. Molality is defined as the number of moles of solute (sucrose) divided by the mass of the solvent (water) in kilograms.

The number of moles of sucrose can be calculated by dividing the mass of the sucrose by its molar mass:

moles of sucrose = 68.5 g / 342.30 g/mol = 0.2 mol

The mass of the water is 500.0 g, which is 0.5 kg.

So, the molality of the solution is:

m = 0.2 mol / 0.5 kg = 0.4 mol/kg

Now we can substitute the values into the freezing point depression formula:

ΔTf = -1.86°C/mol * 0.4 mol/kg * 1 = -0.744°C

So, the change in the freezing point of the water when 68.5 g of sucrose is dissolved in it is -0.744°C. Therefore, the correct answer is C. -0.744°C.