Chem 170 1st Edition Lecture 14Outline of Last Lecture I. Last Lecture: EntropyII. What Affects SolubilityOutline of Current Lecture I. Last Lecture: Henry’s Law and Rault’s LawII. Ideal SolutionsIII. Colligative PropertiesCurrent LectureChapter 13 – SolutionsLast Lecture: Henry’s Law states that the solubility of a gas increases as pressure increases (Concentration=K Pgas , where K is a constant of a gas). French chemist F. M. Rault, discovered that a solute added to a solvent always reduced the vapor pressure of a solution and increased the boiling point.Let χB=mole fractionof solutePA0=vapor pressure of pure s olventPA=vapor pressure of A above the solutionΔ P=PA0−PA= χBPA0note: if there were 2 components then χA+ χB=1and χA=mol Amol A−mol BPA0−PA=(1− χA)PA0PA= χAPA0ex. There was a mixture of benzene (C6H6, P0=95.1 mmHg) and toluene (C7H8,P0=28.4 mmHg) and the mole ratio of each is 0.5.χtoluene=0.5χbenzene=0.5Benzene has a higher vapor pressure, because there is more benzene in the gas phase than toluene.Pbenzene= χAPA0=0.5× 95.1=47.6 mmHgPtoluene= χBPB0=0.5 ×28.4=14.2 mmHgtotal P=47.6 mmHg+14.2 mmHg=61.8 mmHgPbenezene= χbenezenePtotalThese 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.→χbenezene=PbenezenePtotal=47.6 mmHg61.8 mmHgAn “ideal solution” consists of a solvent and a solute. An idea solution should obey Rault’s law and all its intermolecular forces should have the same magnitude. A dilute solution mimics an idea solution. However the solute may not be ideal. But when solutes are considered non-volatile, the dissolution of a solute in a solvent lowers the vapor pressure, elevates the boiling point and depresses the freezing point. Also when a solute dissolves, the entropy increases in the solution and in the vapor. Non-volatile solutes are not likely to escape as a vapor (the solute vapor pressure is less than the solvent vapor pressure).Colligative properties depend on the collection of particles not their identity. This can be seen in the equationPA=(1− χB)PA0 which does not identify the particles. This equation solely asks for the numbers (pressure and moles).Lowering the vapor pressurePA=PA0− χAPA0PA0−PA=∆ PA= χAPA0ex. If there was a cup of 400g of water at 35°C and under 42.2 mmHg of pressure,then what would be the pressure after 50g of nicotine was added to the water. Nicotine’s molar mass is 162g/mol.400 g(1 mol18 g)÷ 50 g(1mol162 g)=22.2mol +0.31 mol=22.51molPwater=(22.2 mol22.51 mol)(42.2 mmHg)=41.6 mmHgOR∆ P= χnicotinePwater0=(0.31 mol22.51mol)(42.2 mmHg)=0.58 mmHgPsolution=42.2−0.58 =41.6 mmHgnote: when NaCl dissolves into Na+ and Cl-and if there are 0.31 moles of Na+ then there are 0.31 moles of Cl- and 0.61 moles
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