What is the freezing point of a solution that contains 36.0 g of glucose ( ) in 500.0 g of water? ( for water is 1.86°c/m. the molar mass of glucose is 180.0 g per

- 1.86°C

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

- 1.86 °C

Exothermic,  physical change after it is already baked