lect_12oct24f08

lect_12oct24f08 - Outline 12 24-OCT-2008 1. Review...

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1 Outline 12 24-OCT-2008 1. Review – Questions? % yield ( just mention ), # mol = V*M 2. Solution preparation 3. Solution Stoichiometry # mol = V*M , another way to get # mol
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2 Name______Quiz 11 TA Name______ (or disc. Sec.__) 1. (2) Calculate the number of mol of NaOH in 25.0 mL of 0.0400 M NaOH. 1. (3) Calculate mass of NaOH (40.0 g/mol) in 25.0 mL of 0.0400 M NaOH.
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3 REVIEW: Solutions, Dilution Molarity and volume are inversely proportional. So, adding water makes the solution less concentrated. M i *V i = M f *V f Adding water increases final volume, V f , and decreases final molarity, M f . initial moles solute = final moles solute
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4 REVIEW: Solutions, Dilution; Ex. 2 Ex: p. 3.97(b) , p. 129. What volume of 1.03 M calcium chloride soln must be diluted with water to make 750.0 mL of 2.66*10 -2 M chloride solution? Diagram. Given: 750.0 mL of 2.66*10 -2 M Cl Find: Volume 1.03 M CaCl 2 needed Soln: Use # mol Cl - before = # mol Cl - after or M i V i = M f V f or (V i M i = V f M f )
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5 REVIEW: Solutions, Dilution; Ex. 2 (cont. 1) Ex : p. 3.97(b) , p. 129. Given: 750.0 mL of 2.66*10 -2 M Cl - Find: Volume 1.03 M CaCl 2 needed Soln: Use # mol Cl - before = # mol Cl - after or V i *M i = V f * M f & (2 mol Cl - / 1 mol CaCl 2 ) V * (1.03 mol CaCl 2 /L)* (2 mol Cl - /mol CaCl 2 ) = 0.750 L * (2.66*10 -2 mol Cl - Rearrange V = 0.750* [2.66*10 -2 / (2* 1.03)]* L = 9.68 mL of 1.03 M CaCl 2 soln
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6 REVIEW: Solutions, Dilution; Ex. 2 (cont. 2) Ex (cont. 2): p. 3.97(b) , p. 129. Given: 750.0 mL of 2.66*10 -2 M Cl - Find: Volume 1.03 M CaCl 2 needed Soln: Use # mol Cl - before = # mol Cl - after or M i V i = M f V f So, measure out 9.68 mL (buret) of 1.03 M CaCl 2 soln and add ENOUGH water to make 750.0 mL soln.
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This note was uploaded on 02/18/2009 for the course CHEM 1A taught by Professor Okamura during the Fall '08 term at UC Riverside.

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lect_12oct24f08 - Outline 12 24-OCT-2008 1. Review...

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