Asolution of water (kf=1.86 ∘c/m) and glucose freezes at − 2.15 ∘c. what is the molal concentration of glucose in this solution? assume that the freezing point of pure water is 0.00 ∘c.

1.16 moles/liter The equation for freezing point depression in an ideal solution is ΔTF = KF * b * i where ΔTF = depression in freezing point, defined as TF (pure) â’ TF (solution). So in this case ΔTF = 2.15 KF = cryoscopic constant of the solvent (given as 1.86 âc/m) b = molality of solute i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1. Solving for b, we get ΔTF = KF * b * i ΔTF/KF = b * i ΔTF/(KF*i) = b And substuting known values. ΔTF/(KF*i) = b 2.15âc/(1.86âc/m * 1) = b 2.15/(1.86 1/m) = b 1.155913978 m = b So the molarity of the solution is 1.16 moles/liter to 3 significant figures.

The mass has escaped into a gas which you can not see so the law stays true.