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Substance that has no noticeable vapor pressure Colligative property Property of a solution that depends on the quantity of the solute, not the type of.

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Presentation on theme: "Substance that has no noticeable vapor pressure Colligative property Property of a solution that depends on the quantity of the solute, not the type of."— Presentation transcript:

1 Substance that has no noticeable vapor pressure Colligative property Property of a solution that depends on the quantity of the solute, not the type of solute. Not what it is, but how much it is. A.Lowering Vapor Pressure Nonvolatile substance Adding a nonvolatile substance to a solvent lowers the solvent’s vapor pressure. (Less evaporation when more impurities) Raoult’s Law Pressure of solution = pressure of pure solvent x mole fraction of solvent

2 P a = P tot x X a P a is the vapor pressure of the solution (Hey, isn’t this also Dalton’s law for gases?) Example – What is the vapor pressure of 18.02 g of water at 30.0 o C if we add 46.28 g of glycerine (C 3 H 8 O 3 )? 1.Find mole fraction of H 2 O 18.02. 18.02 g/mole = 1 mole 46.28. 92.11 g/mole = 0.5024 mole Total moles = 1.5024 Mole fraction = 1 1.5024 0.6656 2. Determine the percent VP of solvent P a = P tot x X a P a = 31.82 torr x 0.6656 21.18 torr

3 B. Boiling Point elevation As we add a nonvolitile solute, the boiling point of the solvent increases. Increase measured by molality (moles/kg) of solution. ΔT b = K b x m Change in boiling temp = bp elevation x molality K b for water is 0.512 K/molal C. Freezing point depression Adding a nonvolitile solute makes the freezing point of the solvent decrease Also measured per molal of solute ΔT f = K f x m m = molality of solution T f for water is 1.86 K/molal

4 D. Van’t Hoff Factor Some nonvolitile solutes create more than one impurity. Ionic substances and some covalent substances (electrolytes) break apart in water to create more impurities. NaCl  Na + + Cl - C 6 H 12 O 6  C 6 H 12 O 6(aq) Note – Actual van’t hoff factors for electrolytes are slightly less than the maximum. At any point, some of the ions reconnect, making the total number of impurities slightly less. Determine the maximum van’t hoff factors for the following. Ca(NO 3 ) 2 MgCl 2 LiBr i i = 2 i =1 3 3 2

5 Now we can rewrite all the equations including the van’t hoff factor ΔT b = iK b x m ΔT f = iK f x m We won’t worry about the slight decrease in i, or how Raout’s law is affected.


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