CHEM 1332 1st Edition Lecture 4 Outline of Last Lecture - Discrete molecules- Affects of size- Solutions- ConcentrationOutline of Current Lecture - Vapor pressure- Boiling point- Freezing point- Determination of massCurrent Lecture- Physical properties of a solution are color, melting point, boiling point, shape, size- Colligative properties depend upon the amount of solute- We can take a look at the first property vapor pressure but in this instant of a non-volatile, non-electrolyte solute- According to Raoult’s Law Vapor pressure of a solution is equal to mole fraction of solvent times vapor pressure of solute- E.g. mole fraction of water is .98 at 25 degrees Celsius and vapor pressure of solvent is 24mmHg so the vapor pressure of the solution is 23.52 mmHg- The vapor pressure of the solution is than the vapor pressure of the pure solvent because the solute particles get in the way - In a volatile, non-electrolyte soluteThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- In this case the vapor pressure of a solution is equal to the mole fraction of solvent timesthe vapor pressure of the solvent plus mole fraction of solute times vapor pressure of the solute- In this case there are solute particles that are trying to get out that block the solvent particles but the solvent particles block the solute particles as well- Boiling pressure of a solution is higher than a solute that is non volatile- bp=kbpxm- kbp is a boiling point constant that depends upon the solute - Example equation molality= 1.1m constant= .512  bp=.512x1.1=.1563- You need to add it to the boiling point for final answer so bp+ bp=100+.1563=100.1563 Celsius - Freezing point of solution is less than freezing point of solvent- fp=kfpxm where kfp is the freezing point constant- If we take an example where the kfp=1.86 fp=1.86x1.1= 2.046- In this case you have to subtract it from the freezing point so fp-fp=0-2.046 is about 2.5Celsius- Molar mass determination allows us to use for example the freezing point to determine the molar mass of an unknown compound- For example you are given the mass in grams of an unknown compound- You dissolve the compound in a solvent that you know the mass of- Then you measure the freezing point but it could be boiling point as well- So now you take what you know for example you know the fp so you try to find the m or molality deriving it from the equation fp=kfpxm- Once you find the molality you try to find the mole of the solute deriving it from the equation molality=mole solute/ kg solvent- Now you can use the equation Molar mass= g of unknown solute/ mole of soluteKey TermsColligative properties- collective physical properties that are affected by the soluteVolatile- easily