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					     Solutions
Concentrations of Solutions
         Solute versus Solvent
   Solute – the substance (usually a solid,
    sometime a liquid) being dissolved or
    broken down by the solvent

   Solvent - the substance (usually a liquid,
    sometimes can be a gas) that dissolves or
    breaks down the solute

   Water is the universal solvent. Why???
                 Molarity
   From your own experience, what does
    “concentration” mean?
                       Molarity
   Concentration:
       Used to quantitatively tell us how much solute
        per solvent we have in a given solution
   Dilute solution:
       Only has a little solute
   Concentrated solution:
       Has a lot of solute
                         Molarity
   Molarity:
       A form of expressing concentration in which 1
        molar = 1 mole of solute per 1 liter of solution
   Calculating molarity:
                    moles of solute
        Molarity 
                   liters of solution
            M    n
                   V
                         Molarity
    Example 1: a chemist makes a 400 mL solution
     by adding 67.4 g of AgNO3 to some water.
     What is the molarity of this solution?
        Step 1 – convert from grams to moles

                 mol AgNO 3
67.4 g AgNO 3                0.3965 mol AgNO 3
                170 g AgNO 3
                         Molarity
   Example 1: a chemist makes a 400 mL solution
    by adding 67.4 g of AgNO3 to some water.
    What is the molarity of this solution?
       Step 2 – calculate molarity

0.3965 mol AgNO3
                   1 M AgNO3
   0.4 L solution
                    Molarity
   Example 2: how many moles of a solute are
    present in 34 mL of 3.64 M Ba(OH)2?


M   n
      V
  n  M V
      3.64 M  0.034 L  0.090 mol Ba OH 2
               Making Dilutions
   Would adding water to a concentrated
    solution dilute it?
       Yes
   Would diluting a solution change the
    number of moles of solute present in the
    sample?
       No, it would only make the volume bigger.
           Making Dilutions

M   , by solving for n weget
       n
      V
n  M  V
n1  M1  V1 what you start with
n 2  M 2  V2 after you add more water
n1  n 2 so M1  V1  M 2  V2
   2.3 moles of sodium chloride in 0.45 liters
    of solution.

   1.2 moles of calcium carbonate in 1.22
    liters of solution.

   0.095 moles of sodium sulfate in 12.0 mL
    of solution.
    120 grams of calcium nitrite in 240 ml of
    solution.

   2L of 6M HCI           1.5L of 2M NaOH

   45 ml of 0.12 M sodium carbonate
             Making Dilutions
   Example: Suzie needs 3L of a 0.05M HCl. She
    only has 12M HCl available. How much does she
    need to dilute to make her solution?

         M1  V1  M 2  V2
     0.05M 3L  12M  V2
                  V2  0.01L  10 mL
               Percent Solutions
   Two ways of expressing concentrations by
    percent:
       Volume/volume (%(v/v)):
         volume solute
                        100
        volume solution
       Mass/volume (%(m/v)):
            mass solute in grams
                                        100
        volume solution in milliliter s
           Percent Solutions
   Example 1: a bottle of juice claims to be
    12%(v/v) pure fruit juice. You run
    expensive tests and determine that, in
    500 mL of juice, there is 58 mL of pure
    juice. Does the company have a law
    suit coming up?
                  Percent Solutions
   Example 1: a bottle of juice claims to be 12%(v/v) pure fruit juice. You
    run expensive tests and determine that, in 500 mL of juice, there is 58 mL
    of pure juice. Does the company have a law suit coming up?


              volume solute
    %(v/v)                    100
             volume solution
             58mL
                   100  11.6%
             500ml
            yes, they can be sued!
            Percent Solutions
   Example 2: A student needs to make a 25%
    (m/v) NaCl solutions. To make 367 mL, how
    much NaCl is needed?
                 mass solute in grams
    %(m/v)                                  100
             volume solution in milliliter s
             %(m/v)
        m           V
               100
            0.25 367 mL  92 g NaCl

				
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