Colligative properties, such as boiling point elevation, depend on the number of dissolved particles in solution. For nonelectrolytes, no dissociation occurs, and so you can use the number of moles of solute to calculate both molality and molarity. In contrast, electrolytes dissociate and therefore the molality and molarity must be calculated based on the number of moles of dissociated particles or ions. There are two ions per formula unit of NaCl Therefore, we would expect the freezing-point depression AT; of a NaCl solution to be twice that of a sugar solution of the same concentration. However, it has been experimentally determined that in the typical unsaturated solution AT; for the salt solution is only 1.9 times that of the sugar solution. This indicates that not all ion pairs in the NaCl solution are dissociated. The number 1.9 is called the van't Hoff factor (symbolized by i) and can be thought of as the number of dissociated particles per NaCl formula unit. This factor changes based on the concentration of the solution, and each salt will have a series of experimentally determined values. How to express the van't Hoff factor Here are three different methods of expressing the van't Hoff factor: Part A At what temperature would a 1.45 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? K for water is 1.86 °C/m. Express your answer with the appropriate units. ▾ View Available Hint(s) μA Value Submit Hint 1. How to approach the problem A van't Hoff factor of 1.9 indicates that the actual particle concentration in the NaCl solution is 1.9 times that of the given concentration. Therefore, the actual freezing-point depression will be 1.9 times greater than the freezing point calculated using the given molality. 5 C Units Request Answer i apparent moles of particles in solution moles of solute apparent particle concentration solute concentration observed colligative property colligative property assuming no dissociation i= ? i i= Review | Constants | Periodic Table