We shall start by finding out the moles of the solute.
Moles of solute can be calculated as :
Moles= [tex] \frac{mass of the solute}{molar mass of the solute} [/tex]
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=[tex] \frac{Moles of the solute}{Mass of the solvent in kg} [/tex]
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
