Colligative Properties: Freezing Pointdepression and Molar Mass
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Submitted By Mizos1 Words 1635 Pages 7
Mlungisi Maphunta
147565
Bachelor of Science II
C202
Practical Report
Group 15
Experiment8
COLLIGATIVE PROPERTIES: FREEZING POINTDEPRESSION AND MOLAR MASS
Group Members
Mncedisi Mngomezulu
Sphiwe ndwandwe
Thandeka Mpila
EXPERIMENT 8: COLLIGATIVE PROPERTIES: FREEZING POINT DEPRESSION AND MOLAR MASS:
AIM
The aim of the experiment is to become familiar with the colligative properties and to use them to determine the molar mass of a substance.
THEORY
A solution consists primarily of solvent and therefore, most of the solution’s properties reflect the solute’s properties. The physical properties that the solution and solute do not share are known as colligative properties and they depend solely on the solute concentration. Some of these properties include vapor pressure lowering, boiling-point elevation, freezing point lowering, and osmotic pressure. The solvent boils when the vapor pressure, or tendency of solvent molecules to escape, is equivalent to the atmospheric pressure. At this moment, the gaseous and liquid states of the solvent are in dynamic equilibrium and the molecules change from the liquid to the gaseous states and from the gaseous to liquid states at equal rates. The dissolution of a solute with very low vapor pressure, or a nonvolatile solute, raises the boiling point and lowers the freezing point. Similarly, anti-freeze lowers the freezing point and lowers the boiling point. The colligative-property law describes these effects, stating that the "freezing point and boiling point of a solution differ from those of the pure solvent by amounts that are directly proportional to the molar concentration of the solute" (Brown, 203-204). The colligative-property law can be expressed using the equation: D T = Km, where D T is the change in freezing or boiling point, K is a solvent-specific constant, and m is the solution’s molality.
Adding a solute to a solvent lowers the freezing point and raises the boiling point, it also lowers the vapor pressure. The new freezing point of the solution can be determined by using the colligative property law: ΔTf = kf×m
The change in freezing point is equal to the molal freezing-point constant times the molality of the solution. The molal freezing-point constant used is the constant for the solvent, not the solute.
These effects are expressed quantitatively by the colligative-property law, which states that the freezing point and boiling point of a solution differ from those of the pure solvent by amounts that are directly proportional to the molal concentration of the solute. In its general form this equation is written:
ΔT = K • m
Where ΔT is the boiling point elevation or freezing point depression, K is a constant that is specific for each solvent, and m is the molality of the solution and is expressed as the number of moles of solute per kilogram of solvent. Some representative constants, boiling points, and freezing points are given in the table below. Consider first the cooling curve of pure solvent. As illustrated in the figure on the following page, the temperature of the pure liquid steadily decreases as heat leaves it, but at some point, perhaps after rising a bit, the temperature becomes constant. This temperature, the melting point, is the only temperature at which solid and liquid can coexist. At this temperature, removing heat does not cool the solvent, but rather, converts more liquid to solid. When the liquid has completely solidified, the heat removed again decreases the temperature of the solid solvent.
Cooling curves for pure solvent and solution
Cooling curves for pure solvent and solution
PROCEDURE:
An ice /salt / water bath was prepared using 50g solid NaCl in 400ml beaker adding 250ml of water to dissolve it and the adding it to a beaker 2/3 full with ice. In a nested test tube 10 ml of cyclohexane was measured using a burette in a fume cupboard and the mass of the cyclohexane was calculated, given the density of the cyclohexane to be 0.7726g/mL
In the cyclohexane a thermometer ranging between -10 and +110 ℃ was placed in the cyclohexane together with a stirrer around the thermometer, stirring at a rate of 1 stroke per second. The nested apparatus was lowered into the salt /water /ice bath and was gentle stirred until crystals formed at the bottom of the test tube. After the crystals had formed the test tube assembly it was raised out of the salt /water /ice bath temperature readings were made at 30 second interval and they were recorded until the temperature reached about 15℃ . The assembly was placed back into the ice bath and the same readings were taken again at the same time interval.
B. DETERMINATION OF THE MOLAR MASS OF AN UNKNOWN
Using weighing paper 0.40g of the unknown were was weighed and was added to the cyclohexane test tube and was gentle stirred until all of the unknown had dissolved. It was then put in the ice bath again until some crystals formed at the bottom of the test tube. It was again removed from the ice bath and temperature readings were again taken and recorded twice doing as it was done to the pure cyclohexane.
CALCULATIONS
∆ T=kf ×m
Kf =20.4℃
Density of cyclohexane =0.7726 g/mL
Mass of the unknown =0.40g
Volume of cyclohexane =10mL
Before addition of the unknown
X1= 9.5oC
X2= 12.6 oC
(X1 +X2)/2= (9.5+12.6)/2= 11.05oC
After addition of the unknown
X1=10.8oC
X2=9.25oC
(X1+X2)/2= (9.25+10.8)/2=10.02oC
∆T= -1.025oC
∆T=KP X m m=∆T/Kf m=1.025/ (20.4oC/m) =0.050245m =0.050245moles/kg
MASS OF CYCLOHEXANE
0.7726g/ml x 10ml = 7.726g
Mr of cyclohexane: 84.16g/mol
Number of moles of C6H12 = 7.726g/(84.16g/mol) = 0.0918molesC6H12
MASS OF UNKNOWN 0.40g =0.40 x 10-3kg
0.40x 10-3 kg x 0.05024mol/kg= 0.0020096mol of unknown
Mr of the unknown
Mr = mass/ number of moles =0.40g/ 0.0020096mol =199.044g/mol
ANSWERS 1. The outer test tube of the nested test tubes was made of thick material, made the reaction to be very slow, since there was very little temperature change in these test tubes. Also we lacked a wire stirrer; the thermometer was used to stir the solution during the course of the experiment. The imprecision of the thermometer was chiefly responsible for the inaccuracy of the final result the transfer of the powdered particles of unknown from weighing paper to test tube may have been incomplete, as some particles may have been blown away or they may have gone unreacted due to the imperfections of the transfer method. 2. If the thermometer consistently read a 1.2 C lower than the correct temperature, there would be no change in the molar mass calculated. The calculation of the molar mass depends on the change in freezing point. As long as the thermometer was consistently wrong, the change in temperature would be the same regardless of the error. Thus, there would be no net impact on the molar mass. 3. Firstly the observed temperature change would be greater than it should have been this would be due to that the calculated molarity would be too high, thus the number of moles would be too high as well. An exaggerated number of moles would cause the calculated molar mass to be too low, as the same weight of substance will be distributed against more moles of the substance. Therefore the substance will seem lighter.
CONCLUSION
The experiment was fairly done and the change in temperature was obtained to be -1.025 oC and it was further used to calculate the number of moles of the unknown sample and it was found to be 0.0020096mol. This value was further used to calculate the molar mass of the unknown sample which was found to be 199.044g/mol which shows that this is a large molecule that makes up the unknown sample and that this could be an organic compound since it dissolves in cyclohexane and that is why it is of such a high molecular mass. Possible sources of error could be that the outer test tube of the nested test tubes was made of thick material, made the reaction to be very slow, since there was very little temperature change in these test tubes. There are multiple sources of error in this experiment. Human error was the largest error. By not measuring substances accurately, the calculations could be a bit off. As well, by not stirring enough to prevent super cooling, the freezing point could have been affected. Also we lacked a wire stirrer; the thermometer was used to stir the solution during the course of the experiment. The imprecision of the thermometer was chiefly responsible for the inaccuracy of the final result the transfer of the powdered particles of unknown from weighing paper to test tube may have been incomplete, as some particles may have been blown away or they may have gone unreacted due to the imperfections of the transfer method.
REFERENCES 1. Atkins P ,De Paula J . Atkin’s Physical Chemistry 8thEdition .(2006) United States of America New York W.H Freeman and company 2.
