Molecular Mass by Freezing Point Depression

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. 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.

5. 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.

6. Molecular Mass by Freezing Point Depression 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.

7. 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

8. 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 - 1.86oCm-1 (units of “oC / molality”) Camphor - 39.7oCm-1 • The relationship:

9. 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 = 0.785 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

10. 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, aka, Molecular Wgt Solute Solute = Isopropyl Alcohol Mol Wgt = grams solute / moles solute grams solute = Vol solute  Density  Mol Wgt = (Vol solute  Density) / mole • Percent Error Accepted Value = 60.1 g/mole