1) The depression of the freezing point is a colligative property.
For non ionic compounds it may be calculated with the formula:
ΔTf = Kf * m
Where Kf is the cryoscopic constant of the water = 1.86°C/mol
And m is the molality, m = number of moles of solute / kg of solvent
You have 500.0 g of solvent = 0.5000 kg
And you can calculate the moles in 36.0 g of glucosa dividing by its molar mass:
number of moles = mass in grams / molar mass = 36.0 g / 180.0 g/mol = 0.2 mol
=> m = 0.2 mol / 0.5000 kg = 0.4 m
=> ΔT = 1.86°C/m * 0.4m = 0.744 °C.
Given that the freezing point of pure water is 0.0°C, the new freezing point is 0°C - 0.744 °C = - 0.744°C
Answer: - 0.744 °C
2) Elevation of boiling point is another colligative property.
It is calculate as:
ΔTb = i * Kb * m.
Whre i is van't Hoff constant, which accounts for the number of ions generated in ionic compounds.
Here, you do not have the i constant, but you can tell that each molecule of KCl dissociates into 2 ions (one K+ and one Cl-) which is to say that i = 2.
Kb is the boiling constant of the solvent, which here is water => Kb = 0.512 °C/m.
You also know the boiling point of the solution as 103.07°C. Given that the normal boiling point of pure water is 100.0°C, ΔT is 103.07°C - 100.0 °C = 3.07°C.
Then you can determine the molality, m = ΔTb / (i*Kb) = 3.07°C / (2*0.512°C/m) = 3.00 °C.
Answer: 3.0 °C
