Review 5'Moles in solution

1
Review 5.Moles in solution

• In this lesson you will review molarity and
  litermole calculations

2
Definition of molarity

• A solution is a mixture of solute dissolved in
  solvent. The proportions of the components are
  the solution concentration
• The concentration of a solution is usually
  expressed in molarity
• Molarity M moles solute liters solution

3
Find the molarity

• A solution contains 15.6 g MgCl2 in 300. mL (3.00
  x 102 mL) of solution. What is the concentration
  of the solution?
• 15.6 g 1 mol 0.164 mol MgCl2
  95.21 g
• 3.00 x 102 mL 1 L 0.300 L
  solution 1000 mL
• M 0.164 mol MgCl2 0.546 M (3 sigfigs)
  0.300 L solution

4
Litermole conversions

• Molarity is a conversion factor
• moles 1L L L mol moles
  mol 1 L

5
Litermole conversions

• How many moles of CaCl2 are in 115 mL of 0.840 M
  CaCl2 solution?
• 115 mL 1 L 0.840 mol 0.0966 mol
  1000 mL 1 L
• How many mL of 3.00 M NH3 solution will supply
  0.135 mol NH3?
• 0.135 mol 1L 1000 mL 45.0 mL
  3.00 mol 1 L

6
More complex problems

• How many grams of KNO3 are needed to make 750 mL
  of 0.20 M KNO3 solution?
• 750 mL 1 L 0.20 mol 101.1 g
  1000 mL 1 L 1 mol 15.165 g 15
  grams (2 sigfigs)

7
Even more complex problems

• 10.0 mL of glycerol (C3H8O3, d 1.26 g/mL) are
  added to enough water to make 250 mL of solution.
  What is the molarity of the solution?
• 10.0 mL 1.26 g 1 mol 0.1368
  mol mL 92.0944 g glycerol
• 250 mL 1L 0.25 L solution 1000
  mL
• M 0.1368 mol glycerol 0.5473 M 0.55 M
  0.25 L solution

8
Way more complex problems

• Concentrated hydrochloric acid solution is 37
  HCl by mass and has a density of 1.184 g/mL.
  What is the molarity of concentrated HCl?
• HINT first find the g HCl in 1 L of the
  solution.
• 1000 mL soln 1.184 g soln 37 g HCl
  mL soln 100 g soln 1 mol
  HCl 12.0 mol HCl in 1 L soln 36.46 g
  HCl 12 M HCl
• Note the importance of labels in the setup!

9
Dilution problems

• Dilution making a new solution by adding water
  to the original solution.
• The number of moles of solute stays the same in
  both solutions.
• moles solute M V (molarity x volume)
• M1V1 M2V2 (molarity x volume)original
  soln (molarity x volume)new soln

10
Dilution example

• If 300. mL of water are added to 600. mL of 0.960
  M H2SO4, what is the molarity of the diluted
  acid?
• The volume of the new solution is 900. mL
• M1V1 M2V2 (0.960 M)(600 mL) (? M)(900
  mL)
• The diluted acid is 0.640 M

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More dilution problems

• How many mL of 3.00 M HCl should be diluted to
  make 500. mL of 0.100 M HCl?
• This time V1 is unknown.
• M1V1 M2V2 (3.00 M)(? mL) (0.100
  M)(500 mL)
• To make the desired solution, dilute 16.7 mL of
  3.00 M HCl to 500. mL

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The bottom line

• You should be able to define molarity calculate
  it given the amounts of solute and solution
• You should be able to use molarity to convert mL
  solution to moles and moles to mL solution
• You should be able to integrate this with molar
  mass, density, and composition information
• You should be able to calculate molarity and
  volume for dilution problems

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Where to get more information

• For more about molarity and molarity conversions,
  see Chapter 4, section 4 in your text
• Problems involving composition and density are
  not directly discussed in your text. Just
  remember to label carefully!
• For more about dilution problems, see Chapter 4,
  section 4 in your text