Question

The boiling point of an aqueous 3.0M ethylene glycol (C₂H₆O₂) solution is 102^oC and the freezing point is-10.8^oC. What is the experimental van't Hoff factor (i) and how does it compare to the theoretical van't Hoff factor?

Answer 1

The density of solution is necessary to solve this problem.It can also be solved by assuming the density.Given: A 3.0 M solution.It contains 3.0 moles of ethylene glycol in 1 liter of solution.Assuming volume of solvent alone is 1 Liter and Density of solvent is 1.0 g/ mL, we get mass of solvent = 1000 gSo molality = 3.0 moles * 1000 / 1000 g = 3.0 mKf for water = 1.86 C/ mFreezing point depression, ΔTf = 0 - (-10.8) = 10.8 CΔTf = i * Kf * mSo i = ΔTf / Kf * m= 10.8 C / 1.86 C/m * 3.0 m= 1.94But the theoretical vant Hoff factor for ethylene glycol is 1 since it is assumed to not dissociate in solution.