Name: JOANNA CELESTE M. QUINTANA Date performed: NOV. 12, 2012 Section: C-1L Date submitted: NOV. 21, 2012 Group Number: 3 Exercise No. 2 COLLIGATIVE PROPERTIES (Full Report) I. INTRODUCTION Colligative properties In liquid solutions, particles are close together and the solute molecules or ions disrupt intermolecular forces between the solvent molecules, causing changes in those properties of the solvent that depend in intermolecular attraction. For example, the freezing point of a solution is lower than that of the of the pure solvent and the boiling point is higher.
Colligative properties of solution are those that depend on the concentration of solute particles in the solution, regardless of what kinds of particles are present the greater the concentration of any solute, the lower the freezing point and the higher the boiling point of a solution. FREEZING POINT LOWERING A liquid begins to freeze when temperature is lowered to the substance’s freezing point and the first few molecules cluster together into a crystal lattice to form a tiny quantity of solid.
As long as both solid and liquid phases are present at the freezing point, the rate of crystallization equals the rate of melting and there is a dynamic equilibrium. When a solution freezes, a few molecules of solvent cluster together to form pure solid solvent and a dynamic equilibrium is set up between the solution and the solid solvent. In the case of a solution, the molecules in the liquid in contact with the solid solvent are not all solvent molecule.
The rate at which molecules move from solution to solid is therefore smaller that in the pure liquid to achieve dynamic equilibrium there must be a corresponding smaller rate of escape of molecules from solid crystal lattice. This slower rate occurs at a lower temperature and so the freezing point of the solution is lower than that of liquid solvent. The change in freezing point ? Tf is proportional to the concentration of the solute in the same way as the boiling point elevation. ?Tf = Kf ? msolute ? ?solute
Here also, the proportionality constant Kf depends on the solvent and not the kind of solute and isolute represents the number of particles per formula unti of solute. For water, the freezing point constant is -1. 86 oC-kg/mole. Freezing point or melting point is the temperature of transition between solid and liquid. Melting point can be measured more accurately than freezing points. This is becauses so in freezing point measurements, supercooling may occur which would yield a lower than sslkdjs freezing (melting point).
CHANGES IN VAPOUR PRESSURE: RAOULT’S LAW At the surface of an aqueous solution, there are molecules of water as well as ions or molecules from the solute. Water molecules can leave the liquid and enter the gas phase, exerting a vapour pressure. However, there are not as many water molecules at the surface as in pure water, because some of them have been displaced by dissolved ions or molecules/ therefore, not as many water molecules are available to leave the liquid surface, and the vapour pressure is lower than that of pure water at a given temperature.
From this analysis, it should make senses that the vapour pressure of the solovent above the solution, Psolvent, solution, that is , to their mole fraction. Thus, since Psolvent ? Xsolvent, we can write Psolvent = Xsolvent ? K (where K is a constant). This equation tells you that, if there are only half as many solvent molecules present at the surface of a solution as at the surface of the pure liquid, then the vapour pressure of the solvent above the solution will only be half as great as that of the pure solvent at the same temperature. If we are dealing only with pure solvent, the above equation becomes Posolvent = Xsolvent ?
K where Posolvent is the vapour pressure of the pure solvent and Xsolvent is equal to 1. This means that Posolvent = K; that is, the constant K is just the vapour pressure of the pure sovent. Substituting for K in the foremost equation, we arrive at an equation called Raoult’s law: Psolvent = Xsolvent ? Posolvent If the solution contains more than one volatile component, then Raoult’s law can be written for any one such component, A, as PA = XA ? PoA Like this ideal gas law, Raoult’s law is a description of a simplified model of a solution.
An ideal solution is one that obey Raoult’s law/ although most solution are not ideal, just as most gases are not ideal, we use Raoult’s law as good approximation to solution behaviour. In any solution, the mole fraction of the solvent will always be less than 1, so the vapour pressure of the solvent over an ideal solution (Psolvent) must be less than the vapour pressure of the pure solvent (Posolvent). This vapour pressure lowering, ? Psolvent, is given by ? Psolvent = Psolvent ? Posolventwhere Psolvent < Posolvent Boiling point elevation
Raoult’s law tells us that the vapour pressure of the solvent over a solution must be lower than that of the pure solvent. Assume for example that you have a solution of a non-volatile solute in the volatile solvent benzene ? ? ? ? II. MATERIALS A. Reagents 4. 00 g naphthalene 0. 20 g unknown solute A unknown solute B (assigned amount per group) distilled water B. Apparatus 250-mL beaker 400-mL beaker 100-mL graduated cylinder test tubes thermometer iron stand, iron ring, iron clamp hot plate C. Other Apparatus wire gauze tissue paper graphing paper timer III. PROCEDURE Freezing Point of Naphthalene
In a clean and dry test tube, 2. 0 g of naphthalene was weighed. To measure the temperature while heating, a thermometer was suspended by tissue paper at the mouth of the test tube. It was placed in a water bath with the water level above the sample in the test tube. To avoid the contact of the test tube to the bottom of the bath, it was supported by an iron clamp. The water bath was then heated until the entire sample has melted and until the temperature of the sample reached 90o C. The flame was put off and the temperature reading was recorded every 15 seconds until the temperature has fallen to 70oC.
The set up was put aside for the next part of the experiment. Data gathered were tabulated and plotted for analysis and evaluation. Freezing Point Depression of Naphthalene Pre weighed 0. 20 g of unknown solute A was added to the previous set up of naphthalene. The same procedure was done with it. The thermometer was again suspended at the mouth of the test tube by tissue paper. With the help of iron clamp, it was again placed in a water bath, with the water level above the sample in the test tube, to avoid contact to the bottom of the bath.
The water bath was then heated until the entire sample of unknown solute A and naphthalene has melted. When the temperature reached 90oC, the flame was put off. The temperature reading was recorded every 15 seconds until the temperature has fallen to 70oC. Data was also tabulated and plotted together with the data from freezing point of naphthalene. Boiling Point of Water In a 250-mL beaker, 100-mL of distilled water was boiled until it completely boiled. The temperature reading was recorded. In a separate 250-mL beaker, 0. 20 g of unknown solute B was dissolved in 100-mL distilled water.
This was also heated until it finally boiled. The boiling point was also recorded. It was tabulated together with the boiling points of solutions with varying amounts of solute from other groups. Comparison was made for evaluation of the results. IV. DATA/OBSERVATIONS Table 1. 1. Observations on cooling of naphthalene at 15-second intervals. Time (sec. )Temperature (oC)Appearance 1590clear liquid 3090clear liquid 4587clear liquid 6086clear liquid 7585clear liquid 9085clear liquid 10584clear liquid 12084clear liquid 13583clear liquid 15083clear liquid 16582clear liquid 18081clear liquid 9581clear liquid 21080clear liquid 22580clear liquid 24079clear liquid 25579clear liquid 27078clear liquid 28577clear liquid 30077clear liquid 31576solidification 33075solidification 34575 36075 37575 39075 40575 42075 43575 45075 46575 48075 49575 51075 52575 54075 55575 57075 58575 60075 61575 63075 64574 66074 67574 69074 70574 72073 73573 75073 76572 78072 79571 81070 Mass of naphthalene used (g): 2. 00 g Table 1. 2. Observations on cooling of solution of naphthalene and unknown solute at 15-second interval. Time (sec. )Temperature (oC)Appearance 1590clear liquid 3090clear liquid 587clear liquid 6086clear liquid 7585clear liquid 9085clear liquid 10584clear liquid 12084clear liquid 13583clear liquid 15083clear liquid 16582clear liquid 18081clear liquid 19581clear liquid 21080clear liquid 22580clear liquid 24079clear liquid 25579clear liquid 27078clear liquid 28577clear liquid 30077clear liquid 31576clear liquid 33075 34575 36075 37575 39075 40575 42075 43575 45075 46575 48075 49575 51075 52575 Mass of naphthalene used (g): 2. 00 g mass of unknown solute B (g): 0. 20 g Table 1. 3. Data on freezing point depression of naphthalene. Mass of naphthalene used (g)2. 0 g Mass of unknown solute A used (g)0. 20 g Mass of solution (g)2. 20 g Freezing point of pure naphthalene (oC)75 oC Freezing point of solution (oC)73 oC Freezing point difference of pure naphthalene and of solution (oC) Molality of solution (mol/kg) Moles of solute used (mole) Molecular mass of solute (g/mole) Table 1. 4. Summary of data on boiling points of solutions with varying amounts of solute. Group No. Amount of solute B used (g)Boiling point (oC) --100 10. 5099. 0 21. 0090. 0 31. 5099. 5 42. 0099. 5 52. 50100 V. DISCUSSION ? ? ? ? VI. CONCLUSION ? VII. LITERATURE CITED/BIBLIOGRAPHY
