What is the boiling point of a 0.200 m solution of glucose in water?
It’s based on the increase of boiling point. You’ll find the next equation: DT = i * Ke * m, where DT is the change in temperatures from the boiling point of the pure solvent to the boiling point of the solution, i is the number of species generated because of the disolution (because glucose is a molecular solid, it doesn’t generate any ions, so i = 1), Ke is the ebulloscopic constant for the solvent (0.51°C/m for water) and m is the molality of the solution. DT = 1 * 0.51 °C/m * 0.200 m = 0.102 °C The normal boiling point of pure water is 100.000°C, so the boiling point of the solution increases in 0.102 °C, so it’s 100.102 °C. I hope it can help you.
