Molality and Mole Fraction

1. Molality and Mole Fraction • In Chapter 3 we introduced two important concentration units. • % by mass of solute Molarity

2. Molality is a concentration unit based on the number of moles of solute per kilogram of solvent. Molality and Mole Fraction

3. Molality and Mole Fraction • Calculate the molarity and the molality of an aqueous solution that is 10.0% glucose, C6H12O6. The density of the solution is 1.04 g/mL. 10.0% glucose solution has several medical uses. 1 mol C6H12O6 = 180 g

4. Molality and Mole Fraction • Calculate the molality of a solution that contains 7.25 g of benzoic acid C6H5COOH, in 2.00 x 102 mL of benzene, C6H6. The density of benzene is 0.879 g/mL. 1 mol C6H5COOH = 122 g

5. Molality and Mole Fraction • Mole fraction is the number of moles of one component divided by the moles of all the components of the solution • Mole fraction is literally a fraction using moles of one component as the numerator and moles of all the components as the denominator. • In a two component solution, the mole fraction of one component, A, has the symbol XA.

6. Molality and Mole Fraction • The mole fraction of component B - XB

7. Molality and Mole Fraction • What are the mole fractions of glucose and water in a 10.0% glucose solution?

8. Colligative Properties of Solutions • Colligative properties are properties of solutions that depend solely on the number of particles dissolved in the solution. • Colligative properties do not depend on the kinds of particles dissolved. • Colligative properties are a physical property of solutions. • There are four common types of colligative properties: • Vapor pressure lowering • Freezing point depression • Boiling point elevation • Osmotic pressure • Vapor pressure lowering is the key to all four of the colligative properties.

9. Lowering of Vapor Pressure and Raoult’s Law • Addition of a nonvolatile solute to a solution lowers the vapor pressure of the solution. • The effect is simply due to fewer solvent molecules at the solution’s surface. • The solute molecules occupy some of the spaces that would normally be occupied by solvent. • Raoult’s Law models this effect in ideal solutions.

10. Lowering of Vapor Pressure and Raoult’s Law • This graph shows how the solution’s vapor pressure is changed by the mole fraction of the solute, which is Raoult’s law.

11. Fractional Distillation • Distillation is a technique used to separate solutions that have two or more volatile components with differing boiling points. • A simple distillation has a single distilling column. • Simple distillations give reasonable separations. • A fractional distillation gives increased separations because of the increased surface area. • Commonly, glass beads or steel wool are inserted into the distilling column.

12. Boiling Point Elevation • Addition of a nonvolatile solute to a solution raises the boiling point of the solution above that of the pure solvent. • This effect is because the solution’s vapor pressure is lowered as described by Raoult’s law. • The solution’s temperature must be raised to make the solution’s vapor pressure equal to the atmospheric pressure. • The amount that the temperature is elevated is determined by the number of moles of solute dissolved in the solution.

13. Boiling point elevation relationship is: Boiling Point Elevation

14. Freezing Point Depression • Relationship for freezing point depression is:

15. Fundamentally, freezing point depression and boiling point elevation are the same phenomenon. The only differences are the size of the effect which is reflected in the sizes of the constants, Kf & Kb. This is easily seen on a phase diagram for a solution. Freezing Point Depression • Notice the similarity of the two relationships for freezing point depression and boiling point elevation.

16. Freezing Point Depression

17. Boiling Point Elevation • What is the normal boiling point of a 2.50 m glucose, C6H12O6, solution?

18. Freezing Point Depression • Calculate the freezing point of a solution that contains 8.50 g of benzoic acid (C6H5COOH, MW = 122) in 75.0 g of benzene, C6H6.

19. Determination of Molecular Weight by Freezing Point Depression • The size of the freezing point depression depends on two things: • The size of the Kf for a given solvent, which are well known. • And the molal concentration of the solution which depends on the number of moles of solute and the kg of solvent. • If Kf and kg of solvent are known, as is often the case in an experiment, then we can determine # of moles of solute and use it to determine the molecular weight.

20. Determination of Molecular Weight by Freezing Point Depression • A 37.0 g sample of a new covalent compound, a nonelectrolyte, was dissolved in 2.00 x 102 g of water. The resulting solution froze at -5.58oC. What is the molecular weight of the compound?

21. Colligative Properties and Dissociation of Electrolytes • Electrolytes have larger effects on boiling point elevation and freezing point depression than nonelectrolytes. • This is because the number of particles released in solution is greater for electrolytes • One mole of sugar dissolves in water to produce one mole of aqueous sugar molecules. • One mole of NaCl dissolves in water to produce two moles of aqueous ions: • 1 mole of Na+ and 1 mole of Cl- ions

22. Osmotic Pressure • Osmosis is the net flow of a solvent between two solutions separated by a semipermeable membrane. • The solvent passes from the lower concentration solution into the higher concentration solution. • Examples of semipermeable membranes include: • cellophane and saran wrap • skin • cell membranes

23. Osmotic Pressure • Osmosis is a rate controlled phenomenon. • The solvent is passing from the dilute solution into the concentrated solution at a faster rate than in opposite direction, i.e. establishing an equilibrium. • The osmotic pressure is the pressure exerted by a column of the solvent in an osmosis experiment.

24. Osmotic Pressure • For very dilute aqueous solutions, molarity and molality are nearly equal. • M m

25. Osmotic Pressure • Osmotic pressures can be very large. • For example, a 1 M sugar solution has an osmotic pressure of 22.4 atm or 330 p.s.i. • Since this is a large effect, the osmotic pressure measurements can be used to determine the molar masses of very large molecules such as: • Polymers • Biomolecules like • proteins • ribonucleotides

26. Osmotic Pressure • A 1.00 g sample of a biological material was dissolved in enough water to give 1.00 x 102 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25oC. Calculate the molarity and approximate molecular weight of the material.