Download presentation
Presentation is loading. Please wait.
Published byArnold Chambers Modified over 9 years ago
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
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.