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Solutions – Chapter 16

Solutions – Chapter 16 . Mr.Yeung. Lesson 4 - Objectives. Take up questions Must concentrate….in liquid form! And gas (Concentrations) Molarity. Molarity. (Molarity) = moles of solute / litres of solution Example: putting 2 marbles in a can of Coke. The concentration would be:

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Solutions – Chapter 16

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  1. Solutions – Chapter 16 Mr.Yeung

  2. Lesson 4 - Objectives • Take up questions • Must concentrate….in liquid form! And gas (Concentrations) • Molarity

  3. Molarity • (Molarity) = moles of solute / litres of solutionExample: putting 2 marbles in a can of Coke. The concentration would be: • 2 marbles / a can of Coke • In mole terms… if you have 0.5moles and you want to find the molarity. It would be • 0.5moles / 1L = .25M • Molarity can sometimes be denoted as C for concentration or M for molarity • Units for molarity = moles/litre

  4. Questions • Example 1.  What is the molarity of a 5.00 liter solution that was made with 10.0 moles of  KBr ? •                     # of moles of soluteMolarity = ----------------------                     Liters of solution • Given:  # of moles of solute = 10.0 moles           Liters of solution = 5.00 liters •                    10.0 moles of KBrMolarity = --------------------------  = 2.00 M                     5.00 Liters of solution • Answer = 2.00 M

  5. Questions • Example 2.  A 250 ml solution is made with 0.50 moles of NaCl.  What is the Molarity of the solution? • We must change the ml to Liters as shown below: • 250 ml      1 liter             x     --------    = 0.25 liters                   1000 ml •                    # of moles of soluteMolarity = ----------------------                     Liters of solution • Given:  Number of moles of solute = 0.50 moles of NaCl            Liters of solution = 0.25 L of solution •                   0.50 moles of NaClMolarity = --------------------- = 2.0 M solution                   0.25 L • Answer = 2.0 M solution of NaCl

  6. Questions • Example 1.  What would be the volume of a 2.00 M (moles/L) solution made with 6.00 moles of LiF? • Solution: •                                 # of moles of soluteLiters of solution =   --------------------                                        Molarity • Given:  # of moles of solute = 6.00 moles           Molarity = 2.00 M (moles/L) • Liters of solution = 6.00 moles                               -----------                               2.00 moles/L • Answer = 3.00 L of solution

  7. Questions • Example 1.  How many moles of CaCl2 would be used in the making 0.500 L of a 5.0M solution? How many grams of CaCl2 were used? • Solution: • # of moles of solute = Molarity x Liters of solution. • Given: Molarity = 5.0 M (moles/L)           Volume = 0.500 L • # of moles of CaCl2 = 5.0 moles/L x 0.500 moles • Answer = 2.5 moles of CaCl2

  8. Example 2.  What is the volume of 3.0 M solution of NaCl made with 526g of solute? • Solution: • First find the molar mass of NaCl. • Na = 23.0 g x 1 ion per formula unit = 23.0 gCl = 35.5 g x 1 ion per formula unit  = 35.5 g                                                       ----------58.5 g • Now find out how many moles of NaCl you have: •                      mass of sample# of moles = -----------------                      Molar mass • Given:  mass of sample = 526 g                 Molar mass = 58.5 g •                                    526 g# of moles of NaCl = ------------                                   58.5 g • Answer:  # of moles of NaCl = 8.99 moles • Finally, go back to your molarity formula to solve the problem: •                               # of moles of soluteLiters of solution =   --------------------                                        Molarity • Given:  # of moles of solute = 8.99 moles           Molarity of the solution = 3.0 M (moles/L) •                                       8.99 moles# of Liters of solution = -------------                                     3.0 moles/L • Final Answer = 3.0 L

  9. Various representations of concentration • For liquids- • Grams per litre (g/L) • %w/v = (mass of solute/100ml of solution) * 100 • %v/v = (volume of solute (ml)/100ml of solution) * 100 • For solids • %w/w = (mass of solute/100g of solution) * 100

  10. Representations of concentrations • For gas • Ppm = parts per million • Ex: 100 ppm sodium ions in water = 10 sodium ions in 1 million particles of water • Ppb = parts per billion • Ex: 10 ppb iron in water = 10 particles of iron in 1 billion particles of water

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