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Concentration of Solutions

Concentration of Solutions. Concentration of Solutions. Measure of the amount of solute per amount of solvent Dilute – solution with small amount of solute Concentrated – solution with large amount of solute. Percent by Mass. _mass of solute_    X   100  =  % mass of solution

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Concentration of Solutions

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  1. Concentration of Solutions

  2. Concentration of Solutions • Measure of the amount of solute per amount of solvent • Dilute – solution with small amount of solute • Concentrated – solution with large amount of solute

  3. Percent by Mass _mass of solute_   X   100  =  % mass of solution                         _part   X  100  =  %                         whole                          remember:         part  +  part  =  whole                                                 solute + solvent  =  solution Example: A solution contains 54.5 g NaCl in 145.5 g water.  What is the percent NaCl by mass?             54.5 g NaCl               =          54.5 g NaCl  +  145.5 g H2O    54.5 g NaCl    X   100    =  200.0 g solution 27.3%

  4. Molarity molarity (designated by M) = molar concentration = moles of solute liter of solution Examples: What is the molarity of a solution that contains 90.0 g NaCl in 3.50 liters of solution? 90.0 g NaCl   X   1 mol NaCl     = 58.5 g NaCl             1.54 mol NaCl           = 3.50 liters solution 0.440M You have 0.8 L of a 0.5 M HCl solution.  How many moles of HCl does this solution contain? M     =     mols   rearrange the equation and                        liter mols = Molarity X liters                 mols = 0.5 M  X  0.8 L  = 0.4 mols HCl

  5. Molality molality (designated by m) = molal concentration = moles solute kg of solvent for water     1 mL = 1 g  because density of H2O = 1 g/mL Examples: What is the molality of a solution that contains 165.0 g NaCl in 2 liters of water? 2 liters of water = 2000 mL of water = 2000 g water = 2 kg 165.0 g NaCl   X   1 mol NaCl  = 58.5 g NaCl                                2.821 mol NaCl     = 2 kg  1.411 m How much iodine must be added to prepare a 0.48 m solution of iodine in CCl4 if 100.0 g of CCl4 is used? molality    =    mol solute   rearrange and kg solvent                                                   mols iodine = molality  X  kg solvent =   0.48 m  X  .1000 kg CCl4 = 0.048 mol iodine change mols iodine to grams    0.048 mol I2   X    253.8 g I2   =   1 mol I2 12.2 g I2

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