Leather Flower Clematis crispa

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Leather Flower (Clematis crispa)
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Molarity and Calculations

• Dr. Smith
• 2007

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Volumes and Concentration

• Suppose you need to use 0.000549 g of a substance
  for a delicate reaction
• Could you use a balance?
• Better idea is to dissolve a weighable quantity,
  say 1.000 g, in a volume of water, say 1.000 L,
  and then pipet a known volume of liquid. How
  much liquid is needed? 0.549 mL from a 1.00 mL
  measuring pipet, or knowing the density you could
  weigh a portion of the solution, 0.551 g

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Solutions by Weight

• A solution of 1.000 g/L of sodium hydroxide and a
  solution of 1.000 g/L sulfuric acid can be
  prepared, but if the task is to determine how
  much sulfuric acid solution is needed to react
  completely with 50.00 mL of sodium hydroxide
  solution, it is tedious to do the calculation

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Preparing Solutions by Moles

• Weigh desired amount of material, place in flask,
  add some water to dissolve material, then dilute
  to the volume marking on the flask

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Molarity
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Other usable forms
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Example 1

• What is the molarity of a solution if 1.50 moles
  of substance is dissolved and diluted to 500 mL ?
• M mol/L 1.50/0.50 3.00 M

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Example 2

• How many moles are present in 1500 mL of 0.600 M
  solution?
• mol L x M 1.500 x 0.600 0.900 mol

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Example 3

• What is the final volume if 0.250 mol is used to
  make a 0.750 M solution?
• L mol/M 0.250/0.750 0.333 L

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Example 4

• How many g of sodium hydroxide are needed to make
  2.00 L of 0.100 M solution?
• First find mol neededmol L x M 2.00 x 0.100
  0.200 mol
• Second find gg Mwt x mol 40.01 x 0.200
  8.00 g

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Example 5

• 25.00 mL of 0.860 M barium chloride contains how
  many g of barium chloride?
• mol L x M 0.0250 x 0.860 0.0215 mol
• g Mwt x mol 208.23 x 0.0215 4.48 g

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Example 6

• What is Mwt of a material if 1.00 mL of 0.000413
  M solution contains 0.17984 g?
• mol L x M 0.00100 x 0.000413 4.13 x 10-7
  mol
• Mwt g/mol 0.17984/4.13 x 10-7 435,500
  daltons

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Problem 1

• 45.61 mL of 1.05 M solution contains how many
  mol?
• 0.0479 mol

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Problem 2

• What is the final volume when 0.482 mol is
  dissolved to make a 1.25 M solution?
• L 0.386

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Problem 3

• What is the molarity of a solution obtained from
  dilution of 0.0124 mol to 300 mL?
• M 0.0413

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Problem 4

• How many g of sodium sulfate are needed to make
  2.000 L of 0.100 M solution?
• 0.200 mol
• 28.41 g

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Problem 5

• What is the molarity of a solution if 4.16 g
  strontium nitrate is diluted to 500 mL?
• Mwt 211.64
• 0.0197 mol
• 0.0393 M