Question

A solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. What is the freezing point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

Answer 1

We shall start by finding out the moles of the solute.

Moles of solute can be calculated as :

Moles=
`(mass of the solute)/(molar mass of the solute)`

Now,molar mass of the solute (C₆H₁₂O₆) will be = 6×12.0107+12×1.008+6×16

=180.73 grams/mol

Therefore moles of glucose will be:

Moles=21.5 grams/180.73 grams mol⁻¹

Moles=0.119 moles.

Now in order to calculate the freezing point depression ΔTf , we need to find the molality.

Molality=
`(Moles of the solute)/(Mass of the solvent in kg)`

m=0.119 moles/0.255 kg

m=0.46 molal.

The freezing point depression can be calculated as under :

∴ ΔTf=Kf × m

ΔTf = -1.86 °C/m × 0.46 m = -0.8556 °C