Question

In a molecular weight determination by freezing point depression experiment:

Why is it not necessary to wait for the entire sample to freeze in order to determine its freezig point??

Answer 1

:In a molecular weight determination by freezing point depression experiment:

1. 1) To determine the apparent molecular weight of the solute from the lowering of freezing point. 2) To determine the extent of dimerization and dissociation of acetic acid in benzene
2. 3. Introduction: The properties of a solution differ from those of a pure solvent due to interactions that take place between the solute and solvent molecules. The properties that exhibit such changes are called the colligative properties and include vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure change. These properties are dependent only upon the number of particles (ions or molecules) which are dissolved in the solvent and not on the identity of the particles. Freezing-point depression, one of the colligative properties of solution, is used in this experiment to determine the molecular weight of an unknown compound A. The temperature at which a solution freezes will be lower than the temperature at which the pure solvent freezes.
3. 4. The difference between the freezing point of a solvent (ΔTf) and the freezing point of the solution (Tf) is referred to as the freezing point depression (ΔT). A solution’s freezing point is related to the molality (m) of the solution. The equation that describes the relationship between freezing point depression and molality is: ΔTf = (Kf)(m) ΔT represents the freezing point depression: ΔT = Tf- Ti Tf is the freezing point of the pure solvent. Tf is the freezing point of the solution. Kf is the molal freezing point depression constant. The value of Kf is characteristic for a given solvent. The units of Kf are given in oC/m. m is the molality of the solution. m = moles of solute/ kg of solvent.
4. 5. Insert the thermometer and stirring rod inside the tube and place it in the ice bath. Pipet out about 25.00 mL benzene into the freezing point test tube. Record the its initial temperature. Fill the small Dewar flask with ice-water mixture Pre-chill the benzene in an ice bath.
5. 6. Add four 0.05 mL glacial acetic acid successively repeating steps 4 to 7 with each addition. Add 0.10 mL glacial acetic acid and repeat the previous steps of taking temperatures. Get the tube from the icebath and warm it with hand until the crystals formed by supercooling have melted. Continue taking temperature readings at 30 sec-intervals for another 5 mins or when the temperature becomes constant. Stir steadily up and down and take temperature readings at 30-second interval until the temperature remains constant
6. 7. Data and Calculations: TEMPERATURE READINGS BENZENE After addition of glacial acetic acid + 0.05 mL + 0.10 mL + 0.15 mL + 0.20 mL Initial temp: 24.4oC 20.2 oC 23.1 oC 20.1 oC 21.5 oC 30 s interval: 12.9 oC 14.3 13.6 13.9 14.6 8.4 8.2 7.8 7.7 8.3 6.4 6.4 6.1 5.5 6.3 5.5 5.2 5.6 5.0 5.5 5.0 4.5 4.1 4.5 4.3 4.4 4.0 3.8 4.0 4.1 4.1 3.5 3.5 3.2 3.3 4.1 3.6 3.5 2.1 1.9 4.1 3.6 3.2 2.1 1.5
7. 8. Table 2: Calculated values for Van’t Hoff factor, apparent molecular weight, extent dissociation and polymerization after addition of concentrated glacial acetic acid. + 0.05 mL CH3COOH + 0.10 mL CH3COOH + 0.15 mL CH3COOH + 0.20 mL CH3COOH Δ Tf 0.5 0.9 2.0 2.6 Mass of acetic acid 0.05245 g 0.10490 g 0.15735 g 0.20980 g Molality 0.039893 m 0.079787 m 0.119681 m 0.15974 m Van’t Hoff factor ( i ) 2.448 2.203 3.264 3.182 Molecular weight 60.00189 g/mol 59.9964 g/mol 60.00189 g/mol 58.56455 g/mol α polymerization -2.896 -2.406 -4.528 -4.364 α dissociation 1.448 1.203 2.264 2.182
8. 9. Density of benzene: 0.8765 g/mL Volume of benzene: 25 mL Density of acetic acid: 1.049 g/mL Kf of benzene: 5.12 0C/m Mass of benzene: (vol of benzene x density of benzene) = = 21.9125 g Mass of acetic acid: (volume of acetic acid x density of acetic acid) For 0.05 mL= = 0.05245 g For 0.10 mL= = 0.10490 g
9. 10. For 0.15 mL= = 0.15735 g For 0.20 mL= = 0.20980 g Molality of solution= For 0.05 mL = = 0.039893 m For 0.10 mL = = 0.079787 m For 0.15 mL = = 0.119681 m For 0.20 mL = = 0.159574 m
10. 11. Van’t Hoff factor: For 0.05 mL = = 2.448 For 0.10 mL = = 2.203 For 0.15 mL = = 3.264 For 0.20 mL = = 3.182 Apparent molecular weight: For 0.05 mL = = 60.00189 g/mol
11. 12. For 0.10 mL = = 59.99644 g/mol For 0.15 mL = = 60.00189 g/mol For 0.20 mL = = 58.56455 g/mol Extent polymerization: α = For 0.05 mL = = -2.896 For 0.10 mL = = -2.406 F0r 0.15 mL = = -4.528 For 0.20 mL = = -4.364
12. 13. Extent dissociation: α = For 0.05 mL = = 1.448 For 0.10 mL = = 1.203 F0r 0.15 mL = = 2.264 For 0.20 mL = = 2.182
13. 14. The freezing point of a liquid is depressed when it contains a dissolved solid. The freezing point depression, or the difference between the freezing points of the pure solvent and solution, depends upon the number of particles in solution. The size of freezing point depression depends on two things: a) the size of Kf for a given solvent, which is well known, and b) the molal concentration of the solution which depends on the number of moles of solute and kg of solvent. The greater the concentration of the solution, the greater will be the freezing point depression. In the experiment, as time passes and more acetic acid is added in pure benzene, the freezing point decreases.
14. 15. 0 5 10 15 20 25 30 0 50 100 150 200 250 300 Temperature(C) TIME (s) Freezing point depression pure benzene 1st addition 2nd addition 3rd addition 4th addition
15. 16. Electrolytes have larger effects on boiling point elevation and freezing point depression. For a given concentration, a solute that dissociates will also bring about a greater freezing point depression. Acetic acid, a weak electrolyte, can dissociate or dimerize depending on the solvent. In aqueous solution, acetic acid dissociates into H+ and CH3COO-, thus, giving twice as many particles in solution. Van't Hoff factor ( i ) is used to introduce the effect of ion pairing or association of ions that prevents the effect of being exactly equal to the number of dissociated ions. It is the measure of the extent of ionization or dissociation of the electrolytes in the solution.
16. 17. It has an ideal value of 3 for 2:1 electrolyte like K2SO4 and CaCl2 and a value of 2 for 1:1 electrolytes like NaCl, KI, and also for acetic acid. In organic medium, the acetic acid tend to form dimers [(CH3COOH)2]. Dimerization occurs when two similar molecules join together by addition or condensation to form a larger molecule. The values of i calculated was then used to estimate the extent of dimerization of acetic acid in benzene.
17. 18. From the lowering of freezing point observed in the experiment, the computed values for the molecular weight of acetic acid are: 60.00189, 59.99644, 60.00189 and 58.56455 having an average of 59.64118. With this, the percent relative error from 60 g/mol molecular weight of acetic acid is -0.598%. The values for the degree of polymerization for 0.05 mL, 0.10 mL, 0.15 mL and 0.2 mL are: - 2.896, -2.406, -4.528, and -4.364 respectively. The computed values for the degree of dissociation are: 1.448, 1.203, 2.264 and 2.182.