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Chpt 11 Lec

# Chpt 11 Lec - Chemistry 8/e Steven S Zumdahl and Susan A...

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Chapter 11: Properties of solutions Chemistry 8/e Steven S. Zumdahl and Susan A. Zumdahl 1

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Chapter Contents Solution Composition Solubilities Colligative Properties Colloids and Emulsions 2
Solution Composition Molarity, M = moles solute / liter solution. Cannot be accurately predicted for mixtures because partial molar volumes vary. depends on the volume and changes with temperature Molality, m = moles solute / kg solvent Not useful in titration unless density known. Useful in colligative effects. depends only on the mass Mole fraction, X A = moles A / total moles 3

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4 Mass Percent Parts of solute in every 100 parts solution. If a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution. Or 10 kg solute in every 100 kg solution. Since masses are additive, the mass of the solution is the sum of the masses of solute and solvent. Solution of Mass Solvent of Mass Solute of Mass % 100 g Solution, of Mass g Solute, of Mass Percent Mass
5 Example Calculate the Mass Percent of a Solution Containing 27.5 g of Ethanol in 175 mL H 2 O.

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Conc. of 50% by wt. NaOH Density at 20 ° C is 1.5253 g/cm 3 each liter of solution weighs 1525.3 g ½ that mass is NaOH, or 762.6 5 g n NaOH = 762.65 g [ 1 mol/39.998 g ] = 19.067 [NaOH] = 19.067 M but also 19.067 mol / 0.76265 kg H 2 O = 25.001 m and n water = 762.65 g [ 1 mol/18.012 g ] = 42.332 X NaOH =19.067 /(19.067+42.332)= 0.31054 6
7 Various methods for describing solution composition If you mix 1 g C 2 H 5 OH with 100 g water to a final volume of 101 mL, calculate molarity, mass percent, mole fraction and molality of ethanol in this prepared solution. Mass percent = (part/whole)x100 = (1 g/ 101g) x100 = 0.990 % M = moles/volume = 2.17x10 -2 moles/0.101L = 0.215 Mole fraction = (moles A/moles A+B) = 0.00389 m = moles/kg solvent= 2.17x10 -2 moles/0.100kg = 0.217

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Various methods for describing solution composition Calculate the mass percent, molality and normality of a solution of 3.75M sulfuric acid with a density of 1.230g/ml. Mass percent = (part/whole)x100 = (368 g/ 1230g) x100 = 29.90 % m = moles/kg solvent = 3.75 moles/0.862kg = 4.35 Normality = equivalents/L solution Acid-base reaction, equivalent = mass of acid or base that can donate or accept 1 proton (H + ) HCl, NaOH, H 2 SO 4 , Ca(OH) 2 Ox-red reaction, equivalent = quantity of ox or red agent that can donate or accept 1 mole of electrons. MnO 4 - Mn 2+ 5 electrons Normality = 2 x 3.75moles= 7.5 1M = 2N
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