CHEM 201 - 9.5.07

CHEM 201 - 9.5.07 - V 1 = 0.03325 o Example 1 L of 0.1 M...

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CHEM 201 – September 5, 2007 Molarity (M) = moles/liters Solution Dilution – number of moles of solute is always constant where o moles solute = Molarity x Liters solvent o moles = M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 / V 2 o Example: have 150.0 mL of 3.008 M of a substance diluted to 1.000L M 2 = (3.008M x 0.1500)/1.000L M 2 = 0.4512 o Example: 150mL of 3.008 NaOH. Want 1.000Lof 0.1000M NaOH M 1 V 1 = M 2 V 2 3.008(V 1 ) = (0.1000)(1.000L)
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Unformatted text preview: V 1 = 0.03325 o Example: 1 L of 0.1 M NaOH M 1 V 1 = M 2 V 2 0.1(V 1 ) = (0.0001)(.01L) V 1 = .000010 L = 10μL Titration – procedure to find an unknown solution concentration by reacting it with a known Hygroscopic – absorbs water vapor from the air Indicator – changes the color of the solution based on the progress of the reaction...
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This note was uploaded on 04/07/2008 for the course CHEM 201 taught by Professor Harris during the Fall '07 term at Bucknell.

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