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Molality and Mole Fraction Modified from: Chem%20102%20week%202.ppt Molality is a concentration unit based.

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Presentation on theme: "Molality and Mole Fraction Modified from: Chem%20102%20week%202.ppt Molality is a concentration unit based."— Presentation transcript:

1 Molality and Mole Fraction Modified from: http://www.chem.tamu.edu/class/fyp/stone/notes/ Chem%20102%20week%202.ppt Molality is a concentration unit based on the number of moles of solute per kilogram of solvent.

2 Molality and Mole Fraction Calculate the molality of an aqueous solution that is 10.0% glucose, C 6 H 12 O 6. The density of the solution is 1.04 g/mL. 10.0% glucose solution has several medical uses. 1 mol C 6 H 12 O 6 = 180 g Hint: Since molality is moles per kilogram of solvent, grams must be converted to moles and mL must be converted to kg. 1 mL of water = 0.001 kg water

3 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: 1. Vapor pressure lowering 2. Freezing point depression 3. Boiling point elevation 4. Osmotic pressure  Vapor pressure lowering is the key to all four of the colligative properties.

4 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.

5 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.

6 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.

7 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.

8 Boiling Point Elevation Boiling point elevation relationship is:

9 Freezing Point Depression Relationship for freezing point depression is:

10 Freezing Point Depression Notice the similarity of the two relationships for freezing point depression and boiling point elevation. 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, K f & K b. This is easily seen on a phase diagram for a solution.

11 Freezing Point Depression

12 Boiling Point Elevation What is the normal boiling point of a 2.50 m glucose, C 6 H 12 O 6, solution?

13 Freezing Point Depression Calculate the freezing point of a solution that contains 8.50 g of benzoic acid (C 6 H 5 COOH, MW = 122) in 75.0 g of benzene, C 6 H 6.

14 Determination of Molecular Weight by Freezing Point Depression The size of the freezing point depression depends on two things: 1.The size of the K f for a given solvent, which are well known. 2.And the molal concentration of the solution which depends on the number of moles of solute and the kg of solvent. If K f 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.

15 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 10 2 g of water. The resulting solution froze at -5.58 o C. What is the molecular weight of the compound?

16 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

17 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: 1.cellophane and saran wrap 2.skin 3.cell membranes

18 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.

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

20 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: 1.Polymers 2.Biomolecules like proteins ribonucleotides

21 Osmotic Pressure A 1.00 g sample of a biological material was dissolved in enough water to give 1.00 x 10 2 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25 o C. Calculate the molarity and approximate molecular weight of the material.


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