# Solutions

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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
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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 views: 42 posted: 8/11/2012 language: pages: 17