Week 6 notes
Week 6 notes CHEM 120
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This 2 page Class Notes was uploaded by Leslie Pike on Monday September 28, 2015. The Class Notes belongs to CHEM 120 at Western Kentucky University taught by Dr. Darwin Dahl in Summer 2015. Since its upload, it has received 47 views. For similar materials see College Chemistry I in Chemistry at Western Kentucky University.
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Date Created: 09/28/15
Chem 120 notes Week 6 Important No class Wednesday or Friday Class today was spent working molarity problems Problem 1 You have 02 grams leadll chloride in 250 mL of solution What is the molarity of this solution C12gtIlt mol PbCl2 gtllt 2779g PbCl2 02 g Pb 00029 M Pb Cl2 0250L To calculate the molarity of lead in the solution compare moles of lead to moles of leadll chloride There is one mole of lead per one mole of leadll chloride therefore lead has a concentration of 00029M To calculate the molarity of chlorine in the solution compare moles of chlorine to moles of leadll chloride There are two moles of chlorine per one mole of leadll chloride Therefore the molarity of chorine is double the molarity of leadll chloride and is 00058M mm means mass solutemass solution DOES NOT MEAN MASS OF SOLVENT The bottom m is for the mass of the ENTIRE solution Speci c Gravity Speci c gravity is measured in units of grams per cubic centimeter or grams per mL By de nition water has a speci c gravity of 1 Everything else is based off of that mass of solution g S 39 G 39 pea c rawzy volumeof solutionmL Calculating molarity from speci c gravity is yet another factorlabel problem like most other chemistry problems Dr Dahl has gone over so far Sample problem what is the molarity of a 35 hydrochloric acid solution with a speci c gravity of 12 Hint in a 100 g sample that is 35 HCI by weight 35 g will be HCI and 100 g will be the weight of the entire solution 12Lg sollLZrzon 35g HCI m so u 101 molHCl 100 g solution 3645 g HCI L 1000 mL 115MHCl Parts per Million ppm The concentration of very dilute solutions is measured in parts per million ppm De nition as follows mass of solute mg m pp mass of solution kg Note the solute weighs so little that quotmass of solventquot and quotmass of solutionquot can be used interchangeably here If the solvent is water we can take a shortcut because by de nition water weighs one kilogram per liter mass of solute mg m pp volume of solution L Sample problem You need to prepare a 300 mL solution of 50ppm Cl All you have available is barium chloride and water How much barium chloride in grams do you need 50m3 Cl 03 L Cl 1 8 1000 mg Cl 3454 g Cl 2mol Cl mol Ba Cl2 mol Cl mol Ba Cl2 2079 g Ba Cl2 20044 g Ba Cl2 This gives you the weight of barium chloride Dilutions The relationship of dilutions is de ned as follows MiViszVf This is a plugandchug type of problem Plug in the knowns and use algebra to solve for the unknown
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