Question

Calculate the freezing point of a 10% glucose(C6H12O6 ) in water if the freezing point of pure water is 273.15K. Kf of water is 1.86 K kg/mol.

Answer 1

Answer:

The number of moles of glucose = Molecular weight of glucoseMass of glucose = 180 g/mol  60 g =0.333 mol

The molality of glucose solution = Mass of solvent (in kg)Number of moles of glucose = 250 g ×1000 g1 kg0.333 mol=1.333 mol/kg

The depression in the freezing point of glucose solution :

ΔTf=Kfm=1.86K kg /mol×1.333 mol/kg=2.48 K

The freezing point of pure water is 0oC

The freezing point of glucose solution =0−2.48=−2.48oC