Colligative properties

Question

Solvent water freezing point 0 degrees Celsius, Kf 1.86 boiling point 100 degrees Celsius, Kb 0.512, vapor pressure 2.33kpa

1) 5.7 grams of calcium bromide are dissolved in 50.0 mL of water. Remember that calcium bromide is ionic, and dissociates in solution!

1. Determine the molality of all particles dissolved in this solution.
2. What will be the freezing point of this solution?
3. What will be the boiling point of this solution?
4. Determine the mole fraction of solvent in this solution.
5. What is the vapor pressure of this solution?

Answer 1

Calcium bromide = CaBr₂; molar mass = 200.0 g / mol

(1). molality = moles solute / kg solvent

moles solute = 5.7 g CaBr₂ x 1 mol / 200. g = 0.0285 mols CaBr₂

kg solvent = 50.0 ml H2O x 1 g / ml = 50 g H2O = 0.050 kg H2O. (interesting that they give all the information about the water, but left out that the density is 1 g / ml. So, we have to assume this or else there is no way to determine the mass of H2O).

molality = 0.0285 / 0.050 kg = 0.57 molal

(2). Freezing point determination:

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 3 for CaBr₂ (CaBr₂ ==> Ca²⁺ + 2Br^-, so there are 3 ions)

m = molality = 0.57 m

K = freezing constant = 1.86º/m (given in the problem)

∆T = (3)(0.57)(1.86) = 3.2ºC .. this is the change in freezing point.

New freezing point of solution = 0º - 3.2º = -3.2ºC

(3). Boiling point determination:

∆T = imK

∆T = ?

i = van't Hoff factor = 3 for CaBr₂ (CaBr₂ ==> Ca²⁺ + 2Br^-, so there are 3 ions)

m = molality = 0.57 m

K = boiling constant = 0.512º/m (given in the problem)

∆T = (3)(0.57)(0.512) = 0.88º ... this is the change in boiling point

Boiling point of the solution = 100º + 0.88º = 100.9º

(4). The vapor pressure of the solution = X_H2O x Pº_H2O (see Raoult's law)

X_H2O = mole fraction of H2O = moles H2O / moles H2O + moles CaBr₂

moles H2O = 50.0 mls H2O x 1 g / ml x 1 mol / 18 g = 2.78 moles H2O

moles CaBr₂ = 0.0285 mols (see calculation in #1 above)

total moles = 2.78 + 0.0285 = 2.81 mols

X_H2O = 2.78 / 2.81 = 0.989

Pº_H2O = original vapor pressure of H2O = 2.33 kPa

Vapor pressure of solution = X_H2O x Pº_H2O = 0.989 x 2.33 kPa = 2.30 kPa