1. Calculating Molar Mass from Freezing Point Depression
A simple Lab Experiment on Colligative Properties

2. Molecular Mass by Freezing Point Depression
Background
Vapor Pressure
The melting and freezing points for a substance are determined by the vapor pressure of the solid and liquid states.
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase.
Vapor pressure is determined by the ability of particles at the liquid / solid surface to escape into the vapor phase.

3. Molecular Mass by Freezing Point Depression
Boiling Point - Pure liquid
The boiling point of a pure liquid is the temperature at which the vapor pressure is equal to the pressure of its environment.
The “normal boiling point” is when the vapor pressure is exactly 1 atmosphere.
Melting Point – Pure Solid
The melting point of a substance is the temperature at which the liquid and solid have the same vapor pressure.
The normal melting point is the temperature at which the solid and liquid have the same vapor pressure and the total pressure is 1 atmosphere.

4. Molecular Mass by Freezing Point Depression
Solutions
When a solute (different substance than the solvent) is dissolved in a pure nonvolatile solvent, the vapor pressure of the resulting solution decreases.
In a pure liquid – the solvent - all particles at the liquid surface are the same.
In a solution (mixture of solvent & solute particles), the surface of the liquid is occupied by both solvent and solute particles.
Thus, there are fewer solvent particles in a solution to enter the vapor phase resulting in a lower vapor pressure (at all temperatures).

5. Molecular Mass by Freezing Point Depression
Boiling Point - Solutions
A solution with its lower vapor pressure than the pure solvent, must, therefore, be heated to a higher temperature before the vapor pressure equals atmospheric pressure.
The solution, therefore, has a higher boiling point than the pure solvent.
Thus, adding a solute to a solvent increases the boiling point of a solution.

6. Molecular Mass by Freezing Point Depression
Freezing Point - Solutions
The freezing point of a solution is the temperature at which the liquid solvent and the pure solid solute can coexist at equilibrium simultaneously, that is, they have the same vapor pressure.
A solute dissolved in water at 0oC has a higher vapor pressure than water, therefore, the solute will not dissolve.
The vapor pressure of a solid solute decreases faster than the solvent as the temperature decreases.
At some temperature lower than the melting point of solvent, the vapor pressures of the solute and solvent will become equal and the solute dissolves.
Thus, adding a solute to a solvent decreases the freezing point.

7. Phase diagram showing the effect of nonvolatile solute on freezing point and boiling point.

8. Molecular Mass by Freezing Point Depression
Colligative Properties
Colligative properties of solutions are properties that depend upon the concentration of solute molecules or ions, but not upon the identity of the solute.
Colligative properties include freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure.

9. Molecular Mass by Freezing Point Depression
Molality (not to be confused with Molarity, moles/liter)
The investigation of Colligative properties of solutions requires the use of a particular concentration unit.
Molality is defined as the number of moles of solute dissolved in a Kilogram (not liter) of solvent.
For each unique solvent, one mole of solute in a kilogram of solvent lowers the freezing point a specified amount.
The freezing point change is directly proportional to the amount of solute, not the identity of the solute.
For the solvent water, one mole of solute will lower the freezing point of a kilogram of water from 0oC to -1.86oC

10. Molecular Mass by Freezing Point Depression
The constant of proportionality between the molality of a solution and the change in freezing point is called the “Molal Freezing Point Depression Constant, Kf.
The molal freezing point depression constant is unique for each different solvent.
Water oC m-1 (units of “oC / molality”)
Camphor oC m-1
The relationship:

11. Freezing Point Depression
The freezing point depression, Tf is equal to the freezing point of the solvent minus the freezing point of the solution.
It is also proportional to the molal concentration of the solute Tf = Kf m
Where Kf, the freezing point depression constant, depends only on the solvent.

12. Molecular Mass by Freezing Point Depression
The Experiment
Determine the molecular (molar) mass of Isopropyl Alcohol, i.e., its molecular weight, by measuring its freezing point depression in an aqueous solution.
Isopropyl Alcohol (C3H7O)
For this experiment it is assumed that the vapor pressure of isopropyl alcohol (the solute) at 0oC is negligible.
This experiment is carried out in a water, ice, salt (NaCl) mixture, which itself is an example of freezing point lowering, where the freezing point of water is lowered to around -10oC.

13. Molecular Mass by Freezing Point Depression
The Apparatus
Test Tube fitted with a rubber stopper into which is inserted a thermometer.
Thermometer is held in place by the rubber stopper and the wire is free to move up and down for stirring.
The cold bath, into which the test tube is inserted, is a large beaker containing the mixture of water, ice, salt.
The water, ice, salt mixture needs to be at a temperature of at least -10.0oC.
*** Do not use the thermometer to stir the solution.

14. Molecular Mass by Freezing Point Depression
The Procedure – Part I
In a large beaker place equal amounts of ice and salt, and a small amount of water.
Stir mixture.
Adjust salt content so that the temperature of the solution is at least -10oC.
Rinse the freezing point apparatus – test tube – with distilled water.
Place approximately 25.0 mL of distilled water, measured to the nearest 0.1 mL, in the test tube.
Stopper the test tube with rubber stopper holding the thermometer and wire stirrer.
Clamp test tube in position so that the contents are below the water level in the beaker.

15. Molecular Mass by Freezing Point Depression
The Procedure (Con’t)
Move the stirrer up and down constantly to agitate the water.
Record at least 5 temperature readings at 30 second intervals or until stirring the freezing mixture become difficult.
Record the freezing the point of distilled water.
Extract the test tube from the water/ice bath and allow the freezing mixture to thaw (use warmth of your hand to facilitate the thawing).
Repeat the freezing procedure for a second freezing point value.
Calculate the mean freezing point of distilled water.

16. Molecular Mass by Freezing Point Depression
The Procedure – Part II
Use the buret to dispense 6-7 milliliters of the isopropyl alcohol to a small test tube, recording the exact initial and final volumes of the buret. Note: Burets can and must be read to the nearest 0.01 mL.
Transfer the contents of the small test tube to the freezing point apparatus containing the thawed distilled water used in the first part of the experiment.
Continuously agitate the mixture with the stirrer.
Record the temperature every 30 seconds

17. Molecular Mass by Freezing Point Depression
Continue the agitation until the temperature readings have been constant for five (5) consecutive readings or until stirring the frozen mixture becomes difficult.
Record the freezing point of the solution.
Thaw the mixture and repeat the freezing process.
Calculate the mean freezing point.

18. Molecular Mass by Freezing Point Depression
Supercooled Solutions
If the water/alcohol mixture does not appear to freeze at a constant temperature, the mixture may be Supercooled.
A Supercooled liquid freezes slightly below its theoretical freezing point.
To obtain the freezing point, construct a graph of temperature (y-axis) vs. time (x-axis).
The resulting plot should show a curve with two relatively straight segments.
The intersection of these two straight lines is the freezing point of the solution.

19. To find the Freezing point temperatures when supercooling occurs, Ignore the dip in the line graph and find the intersection point between the two lines and then extrapolate to the temperature axis.

20. Molecular Mass by Freezing Point Depression
Calculations
The solvent in this experiment is water.
Density = 1.0 g/mL
The solute in this experiment is Isopropyl Alcohol
Density = g/mL
The Molal Freezing Point Depression Constant, Kf, for water is 1.86 oC/mole
From the average experimental melting point of ice (distilled water) and the average freezing point for the alcohol/water mixture calculate the freezing point depression, oC, for this experiment.
From the Freezing Point Depression Expression calculate the molality of the water/alcohol solution.
tf = kf  m

21. Molecular Mass by Freezing Point Depression
Calculate the number of moles of solute (isopropyl alcohol) in the solution.
molality = moles solute / kilogram solvent
 moles solute = molality  kilograms
Molecular (Molar) Mass, or Molecular Weight of Solute
Solute = Isopropyl Alcohol
MWt = grams solute / moles solute
grams solute = Vol solute  Density
 MWt = (Vol solute  Density) / mole
Percent Error
Accepted Value = g/mole