Question

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 ΔTf 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 ΔTf 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.

Part A

At what temperature would a 1.85 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86 ∘C/m .

Express your answer with the appropriate units.

Answer 1

We will use below formula,

Delta Tf = i * m * Kf

Where Delta Tf is depression in freezing point, i is van't Hoff factor , m is molality of solution and Kf is freezing point constant.

Delta Tf = 1.9 * 1.85 * 1.86 = 6.538° C

So final freezing point = 0 - 6.538 = - 6.538 °C .... Answer