Is the experimental Van't Hoff factor for ionic molecules always higher or lower than the theoretical Van't Hoff factor? Why?
Higher, because the ions have a greater kinetic energy than non-ionic compounds
The observed Van't Hoff factor is always the same as the calculated Van't Hoff factor
Lower, because all ionic solids prefer to stay in their crystal lattice and do not want to dissolve in water
Lower, because of the dynamic equilibrium between the ions dissolved in the solution and their lattice connected ion pairs
Higher, because many ionic solids contain a component that will react with the water
The correct answer is: Lower, because of the dynamic equilibrium between the ions dissolved in the solution and their lattice connected ion pairs.
The theoretical Van't Hoff factor is based on the assumption that all ionic compounds will completely dissociate into their individual ions when dissolved in water. However, in reality, there is a dynamic equilibrium between the ions that are dissolved in the solution and their lattice-connected ion pairs.
This means that not all of the ions will be free in the solution, contributing to a lower experimental Van't Hoff factor compared to the theoretical value. This is especially true for compounds with strong ionic bonds and higher lattice energies, as they are more likely to stay in their crystal lattice rather than fully dissociating in water.