Answer: - 1.86°C
Step-by-step explanation:
The depression of freezing points of solutions is a colligative property.
That means that the depression of freezing points of solutions depends on the number of molecules or particles dissolved and not the nature of the solute.
To solve the problem follow these steps:
Data:
Tf = ?
solute = glucosa (this implies i factor is 1)
mass of solue = 36.0 g
mass of water = 500 g
Kf = 1.86 °/m
mm glucose = 180.0 g / mol
2) Formulas
Tf = Normal Tf - ΔTf
ΔTf = i * kf * m
m = number of moles of solute / kg of solvent
number of moles of solute = mass in grams / molar mass
3) Solution
number of moles of solute = 36.0 g / 180.0 g/mol = 0.2 mol
m = 0.2 mol / 0.5 kg = 1.0 m
ΔTf = i * Kb * m = 1 * 1.86 °C/m * 1 m = 1.86°C
Tf = 0°C - 1.86°C = - 1.86°C
Answer: - 1.86 °C