CHEM 170 1st Edition Lecture 16Outline of Last Lecture I. Last Lecture: Colligative PropertiesII. Boiling Point ElevationIII. Freezing Point DepressionIV. Osmotic PressureOutline of Current Lecture V. Last Lecture: Osmotic PressureVI. Van’t Hoff FactorVII. ElectrolytesCurrent LectureChapter 13 – SolutionsLast Lecture: Osmotic Pressure is the amount of pressure required to stop osmosis.Osmosis is the selective passage of solvents through a semipermeable membrane. The solute cannot pass through the semipermeable membrane.π =MRTwhere π is osmotic pressure; M is molarity; R is the constant 0.0821L ×atmmol × K ; K is the temperature in Kelvinex. What is the osmotic pressure of a 0.001 M sucrose solution at 25°C?π =(0.001 M)(0.0821L× atmmol × K)(298 K)=0.024 atmex. if there was 1.08g of serum albumin dissolved in 50cm3 of water when the osmotic pressure was 5.85mmHg at 298K, then what is the molecular weight (MW)?(MW)=mRTVπ=(1.08 g)(0.0821L × atmmol × K)(298 K)(50 g ×1 L1000 g)(5.85 mmHg ×1 atm760 mmHg)=68655gmolVan’t-Hoff factor= i=the actual number of particles in the solution after dissolution.ex. sucrose i = 1ex. MgCl2 i = 3These 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.Electrolytes are substances that disassociates into ions in water. Solutions with electrolytes are conductive. If you run a current through the solution with an electrolyte and place two wires and a lightbulb Weak electrolytes are molecules that do not disassociate 100%.ex. CH3COOH ↔ CH3COO- +H+Freezing point depression:∆ Tf=i KfMBoiling point elevation:∆ Tb= i KbMπ =MRTex. 0.01 M acetic acid = -0.0193°Cex. 0.01 M urea = -0.0180°CFun demonstrations:Heat up a copper rod and drop it into a recently opened 2 liter bottle of code. The increase in temperature will cause a decrease in pressure which will cause the gas (carbon-dioxide) to rush out of the solution. The fountain of soda easily reached the celling. This demonstrates Henry’s
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