Chem%20162-2011%20Lecture%202

Chem%20162-2011%20Lecture%202 - CHEMISTRY 162-2011 LECTURE...

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Chem 162-2011 Lecture 1 1 CHEMISTRY 162-2011 LECTURE 2 ANNOUNCEMENTS E-MAIL Students should have received my e-mail - Tindoy, Merschelle: merschy@pegasus.rutgers.edu, not working ATTENDANCE Sign in EXAMS QUIZZES Recitation quiz next week
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Chem 162-2011 Lecture 1 2 PLAN FOR TODAY : CONTINUATION OF LECTURE 1: Chapter 12.1 – 12.4 TYPES OF SOLUTIONS CONCENTRATIONS OF SOLUTIONS - %, M, m, X, P X, ppm, etc. ENERGETICS OF SOLUTIONS AND SOLUBILITIES EQUILIBRIUM IN SOLUTION FORMATION LECTURE 2 (CHAPTER 12.4-12.8) EQUILIBRIUM IN SOLUTION FORMATION (covered in lecture 1) THE SOLUBILITIES OF GASES - Henry’s Law COLLIGATIVE PROPERTIES - Vapor pressures of solutions (Raoult’s Law) - Freezing point depression - Boiling point elevation - Osmotic pressure
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Chem 162-2011 Lecture 1 3 ET note: Not everything goes through moles, e.g., volume to grams, grams to volume, atoms to molecules, molecules to atoms. CHAPTER 5 - STOICHIOMETRY moles A moles B atoms molecules (or atoms if particle is an a t o m ) M B or V B g B or MW B V B M A or V A V A g A or MW A MOLES ARE AT THE CENTER (FOR CONVERSIONS, WORK THROUGH MOLES) MW (D = g/V) mol A = M A L A moles = P A V A /RT A for gases (molecules x atoms/molecule = atoms) (moles A = g A /MW A ) (moles B = g B /MW B ) (D = g/V) (moles x Avog. No. = molecules (or atoms)) %A % B Empirical formula Molecular formula (MW/EW) x Emp form = molec form moles = g/MW PV = nRT moles = M x V P 1 V 1 /n 1 T 1 =P 2 V 2 /n 2 T 2 g/MW = M x V 6.022x10 23 molecule/mole D = g/L MW A or AW A = g A /mole A moles A = g A /MW A mol A = M A L A moles = P A V A /RT A for gases
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Chem 162-2011 Lecture 1 4 TYPES OF SOLUTIONS A solution is a homogeneous mixture of two or more substances. The solvent is the solution component that determines the state of matter of the solution; it is usually the component present in the greatest amount. The substance(s) dissolved in the solvent is the solute(s). g solute + g solvent = g solution g solute = g solution – g solvent g solvent = g solution – g solute ml solute + ml solvent ml solvent solute s u g a r solvent
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Chem 162-2011 Lecture 2 5 FORMULAS g solute + g solvent = g solution m L solute + mL solvent mL solution Mass percent = grams of solute/100 g solution Mole percent = moles of solute/100 moles solution Molarity = moles of solute/L of solution ET: Molarity is temperature dependent, but molality is not. Difficult but important interconversion Molality = moles of solute/kilogram of solvent PPM = grams of solute/1,000,000 grams solution Volume percent = volume of solute/100 mL of solution Proof = 2 x Vol. %; e.g., 2 x 40 mL/100 mL solution = 80 proof Mole fraction: X A = n A /(n A + n B ) X A + X B = 1 Particle fraction: P X A = in A /(in A + in B ) P X A + P X B = 1 Raoult’s law: Don’t use: P soln = X solvent P o solvent Use: P soln = P X solvent P o solvent For two volatile components: P soln = P X solventA P o solventA + P X solventB P o solventB P X
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This note was uploaded on 03/08/2011 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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Chem%20162-2011%20Lecture%202 - CHEMISTRY 162-2011 LECTURE...

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