Question

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezing point depression of the solvent if the freezing point constant is -1.86 °C/m?

Answer 1

Freezing point depression for dilute solution can be calculated using the formula: ΔT_f = K_f · c · i where K_f = cryoscopic constant of the solvent c = molality of solution, i..e. number os moles solute per kilogram solvent i = van't Hoff factor 
The cryscopic constant for the solvent water is K_f = 1.853 °C/m 
Glucose has a molar mass of: M = (6·12 + 12·1 + 6·16) g/mol = 180 g/mol So you dissolved n = 15.5 g / 180 mol = 8.61×10⁻² mol Hence molaltity of solution is c = 8.61×10⁻² mol / 0.245 kg = 0.3515 m 
The van' Hoff factor gives you the number of particles in solution per particle dissolved. Since glucose does not dissociate, when it is dissolved in water, the van't Hoff factor for this solution is i = 1 
So the freezing point depression for this solution is: ΔT_f = 1.853 °C/m · 0.3515 m · 1 = 0.65 °C 
The freezing point of the solution equals freezing point of pure solvent minus freezing point depression: T_f = T_f° - ΔT_f = 0 °C - 0.65 °C = - 0.65 °C