1. States of Matter Lesson 4.8
CHEMISTRY 2 HONORS
Jeff Venables Northwestern High School

2. Ways of Expressing Concentration Mass Percentage, ppm, and ppb
All methods involve quantifying amount of solute per amount of solvent (or solution).
Generally amounts or measures are masses, moles or liters.
Qualitatively solutions are dilute or concentrated.
Definitions:

3. Parts per billion (ppb) are 1 g of solute per kilogram of solution.
Parts per million (ppm) can be expressed as 1 mg of solute per kilogram of solution.
If the density of the solution is 1g/mL, then 1 ppm = 1 mg solute per liter of solution.
Parts per billion (ppb) are 1 g of solute per kilogram of solution.

4. Mole Fraction, Molarity, and Molality
Recall mass can be converted to moles using the molar mass.

5. Mole Fraction, Molarity, and Molality
We define
Converting between molarity (M) and molality (m) requires density.

7. Example - 1. 00 g of NaCl is dissolved in 50
Example g of NaCl is dissolved in 50.0 g of H2O, to make a solution with a total volume of 50.7 mL. Calculate the molarity, mass percent, mole fraction, molality, ppm, and ppb of NaCl in this solution.
0.338 M %
X = m
19600 ppm ppb

8. Colligative Properties Lowering Vapor Pressure
Colligative properties depend on quantity of solute molecules. (E.g. freezing point depression and boiling point elevation.)
Lowering Vapor Pressure
Non-volatile solvents reduce the ability of the surface solvent molecules to escape the liquid.
Therefore, vapor pressure is lowered.
The amount of vapor pressure lowering depends on the amount of solute.

9. Colligative Properties Lowering Vapor Pressure

10. Lowering Vapor Pressure
Raoult’s Law: PA is the vapor pressure with solute, PA is the vapor pressure without solute, and A is the mole fraction of A, then
Recall Dalton’s Law:
Ptotal = P1 + P2 = c1Po1 + c2Po2

11. Ideal solution: one that obeys Raoult’s law.
Raoult’s law breaks down when the solvent-solvent and solute-solute intermolecular forces are greater than solute-solvent intermolecular forces.
Example - Calculate the Vapor Pressure of a solution prepared by dissolving 50.0 g of sugar (C12H22O11) in 200. g of H2O. (Vapor Pressure of pure H2O = torr)
23.45 torr

12. Example
Calculate the vapor pressure of a solution of 25.0 g of H2O and 30.0 g of C2H5OH at 25o C.
Vapor pressures:
H2O = torr
C2H5OH = torr
36.9 torr

13. Deviations From Raoult’s Law
If observed vapor pressure is lower than predicted by Raoult’s Law:
NEGATIVE deviation
A-B > A-A and B-B
If observed vapor pressure is higher than predicted by Raoult’s Law:
POSITIVE deviation
A-B < A-A and B-B

14. Vapor Pressure
Positive Deviation
Negative Deviation
Pure A Pure B

15. Deviations from Raoult’s Law
If Solvent has a strong affinity for solute (H bonding).
Lowers solvent’s ability to escape.
Lower vapor pressure than expected.
Negative deviation from Raoult’s law.
Hsoln is large and negative exothermic.
Endothermic Hsoln indicates positive deviation.

16. Boiling-Point Elevation
Goal: interpret the phase diagram for a solution.
Non-volatile solute lowers the vapor pressure.
Therefore the triple point - critical point curve is lowered.
At 1 atm (normal boiling point of pure liquid) there is a lower vapor pressure of the solution. Therefore, a higher temperature is required to teach a vapor pressure of 1 atm for the solution (Tb).
Molal boiling-point-elevation constant, Kb, expresses how much Tb changes with molality, m:

18. i = vant Hoff factor
= number of particles produced when a substance dissolves
For nonionic substances, i=1
For ionic substances, i is the number of ions produced per formula unit that dissolves.
NaCl, i = 2 Na3PO4 i = 4
Na3PO4  3 Na+ + PO43-

19. Example
What mass of NaCl is needed to raise the boiling point of 500. g of H2O to 101.0˚ C? (Kb = 0.51 ˚C∙kg/mol)
29 g NaCl

20. Freezing Point Depression
At 1 atm (normal boiling point of pure liquid) there is no depression by definition
When a solution freezes, almost pure solvent is formed first.
Therefore, the sublimation curve for the pure solvent is the same as for the solution.
Therefore, the triple point occurs at a lower temperature because of the lower vapor pressure for the solution.

21. The melting-point (freezing-point) curve is a near-vertical line from the triple point.
The solution freezes at a lower temperature (Tf) than the pure solvent.
Decrease in freezing point (Tf) is directly proportional to molality (Kf is the molal freezing-point-depression constant):

22. Example
A solution was prepared by dissolving 25.0 g of CaCl2 in 150. g of H2O. What is the freezing point of this solution? (Kf = 1.86 ˚C/m)
-8.38ºC