# ch11 - Chapter 11 Properties of Solutions 11.1 Solution...

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Unformatted text preview: Chapter 11- Properties of Solutions 11.1 Solution Composition molarity- # of moles L of solution mass percent- mass of solute x100 mass of solution mole fraction- moles of solute moles of solution molality- moles of solute kg of solvent * sample exercises 11.1 and 11.2 may be helpful 11.2 The Energies of Solution Formation ∆ H of solution- enthalpy of solution, which may be defined as + or & - ∆ H 1 + ∆ H 2 + ∆ H 3 ∆ H 1- expanding solute ∆ H 2- expanding solvent ∆ H 3- combining expanded solute and solvent to form solution ∆ hydration- enthalpy of hydration (interactions between solvent and solute) Ex. H 2 0 ( ∆ H 2 ) and ionic solids ( ∆ H 3 ) *sample exercise 11.3 Question: Decide whether liquid hexane (C 6 H 14 ) or liquid methanol (CH 3 OH) is the more appropriate solvent for the substances grease (C 20 H 42 ) and potassium iodide (KI). Answer: Hexane is a nonpolar solvent because it contains C-H bonds. Thus hexane will work best for the nonpolar solute grease. Methanol has an O-H group that makes it significantly polar. Thus it will serve as the better solvent for the ionic solid KI. 11.3 Factors Affecting Solubility A) structure B) pressure (Henry¡s Law) - P= KC P= partial pressure, k= constant characteristic of gas, C= concentration of dissolved Gas -Henry¡s law only works if there is no reaction between the solute and solvent *sample exercise 11.4 Question: A certain soft drink is bottled so that a bottle at 25 ° C contains CO 2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO 2 in the atmostphere is 4.0 x 10-4 atm, calculate the equilibrium concentrations of CO 2 in the soda both before and after the bottle is opened. Henry&s law constant for CO 2 in aqueous solution is 32 Lxatm/mol at 25 ° C. Answer: We can write Henry&s law for CO 2 as P CO2 = k CO2 C CO2 Where k CO2 = 32 L x atm/mol. In the unopened bottle, P CO2 = 5.0 atm and C CO2 = (P CO2 )/(k CO2 ) = 5.0atm/(32 L x atm/mol) = .16 mol/L In the opened bottle, the CO 2 in the soda eventually reaches equlibrium with the atmospheric CO 2 , so P CO2 = 4.0 x 10-4 atm and C CO2 = P CO2 /k CO2 = (4.0 x 10-4 )/(32 L x atm/mol) = 1.2 x 10-5 mol/L Note the large change in concentration of CO...
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## This note was uploaded on 09/27/2010 for the course CHEM 210 taught by Professor Mcomber during the Spring '10 term at Skyline College.

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ch11 - Chapter 11 Properties of Solutions 11.1 Solution...

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