lect5_9Sept08

lect5_9Sept08 - CAPA 1 Out of 264 total students: 113 with...

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1 Teachers need reality checks on occasion CAPA 1 • Out of 264 total students: • 113 with 24/24 • 67 with GE 20/24 • 59 with 11 to 19 • 25 with LE 10/24, including 14 with 0 • CAPA 2 DUE Wednesday 10th • CAPA 3 – BEING handed out in LAB • Due Wednesday 17th NOW WE TURN TO A DISCUSSION OF SOLUTIONS AND The USE OF VOLUMETRIC TECHNIQUES SO FAR - DISCUSSION OF STOCHIOMETRY HAS EMPHASIZED WEIGHING AND GRAVIMETRIC TECHNIQUES OF QUANTITY SELECTION 1 Solutions Solution- homogenous mixture of two or more pure substances; composition can be varied. One or more solutes are dissolved in a solvent . Solute - substance dissolved in a solvent, present in smaller amount. Solvent - substance present in greater amount, contains dissolved solute. •W e w i l l usually use aqueous (aq) solution.- I.e., WATER will be the solvent Units of Concentrations of Solutions usually indicate the Amount of Solute present in a Given Quantity of Solution Concentration Units mass % = g solute g solution x 100 molarity (M) = moles solute L solution TWO QUITE COMMON UNITS ARE: Other units are also used
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2 Concentration of Solutions Calculate the mass of 8.00% (by weight) solution of NaOH that contains 32.0 g of NaOH. n sol' g . 400 NaOH g 8.00 solution g 100.0 NaOH g 32.0 = solution g ? = × Concentration of Solutions What mass of NaOH is required to prepare 250.0 g of solution that is 8.00% (by weight) NaOH? NaOH g 0 . 20 solution g 100.0 NaOH g 8.00 solution g 0 . 250 = Therefore 250.0-20.0 = 230.0 grams solvent Concentration of Solutions • Calculate the mass of NaOH in 250.0 mL of an 8.00% w/w NaOH solution. Density is 1.090 g/mL. Therefore The Molarity of This Solution is (21.8 g/ 40.0 g/mol) / 0.250 L = 2.18 M Mass Solution=250.0 mL x 1.090 g/mL=272.5 g Mass NaOH=0.0800 x 272.5 g =21.8 g Molarity Calculation • Calculate the molarity of these CuSO 4 solutions 4.57 g CuSO 4 in 100 mL 0.54 g CuSO 4 in 100 mL (25.0 gram / 250 grams/mole) = 0.100 moles Therefore Solution is 0.100 M CuSo 4 Molarity = (moles of solute )/ ( liter of solution ) 2.00 M HCl = 2.00 moles HCl 1.00 L HCl solution 6.00 M HCl = 6.00 moles HCl 1.00 L HCl solution molarity (M) = moles solute L solution
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3 Concentrations of Solutions
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This note was uploaded on 09/26/2008 for the course CHEM 107 taught by Professor Generalchemforeng during the Spring '07 term at Texas A&M.

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lect5_9Sept08 - CAPA 1 Out of 264 total students: 113 with...

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