1. 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.

2. 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.

3. 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).

4. 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. Molecular Mass by Freezing Point Depression

5. 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 0 o C 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. Molecular Mass by Freezing Point Depression

6. Colligative Properties  Boiling Point & Freezing Point are examples of “Colligative Properties” of solutions where the effect on the properties depends on the numbers of solute particles in a given mass of solvent. Molecular Mass by Freezing Point Depression

7. 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 0 o C to -1.86 o C Molecular Mass by Freezing Point Depression

8.  The constant of proportionality between the molality of a solution and the change in freezing point is called the “Molal Freezing Point Depression Constant, K f.  The molal freezing point depression constant is unique for each different solvent. Water - 1.86 o Cm -1 (units of “ o C / molality”) Camphor - 39.7 o Cm -1  The relationship: Molecular Mass by Freezing Point Depression

9. Calculations 1.The solvent in this experiment is water. Density = 1.0 g/mL 2.The solute in this experiment is Isopropyl Alcohol Density = 0.785 g/mL 3.The Molal Freezing Point Depression Constant, K f, for water is 1.86 o C/mole 4.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, o C, for this experiment. 5.From the Freezing Point Depression Expression calculate the molality of the water/alcohol solution.  t f = k f  m Molecular Mass by Freezing Point Depression

10. 5.Calculate the number of moles of solute (isopropyl alcohol) in the solution. molality = moles solute / kilogram solvent  moles solute = molality  kilograms 6.Molecular (Molar) Mass, aka, Molecular Wgt Solute Solute = Isopropyl Alcohol Mol Wgt = grams solute / moles solute grams solute = Vol solute  Density  Mol Wgt = (Vol solute  Density) / mole 7.Percent Error Accepted Value = 60.1 g/mole Molecular Mass by Freezing Point Depression