 # Title: Lesson 12 Solutions Learning Objectives: – Understand the relationship between concentration, volume and moles – Pose and solve problems involving.

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Title: Lesson 12 Solutions Learning Objectives: – Understand the relationship between concentration, volume and moles – Pose and solve problems involving solutions (of the chemical kind not the answers kind)

Main Menu Lesson 12: Solutions  Objectives:  Understand the relationship between concentration, volume and moles  Prepare a standard solution of silver nitrate.  Pose and solve problems involving solutions (of the chemical kind not the answers kind)

Main Menu Solutions Basics  Aqueous copper sulfate solution: + SOLUTE SOLVENT SOLUTION

Main Menu Concentration  This is the strength of a solution. Most ConcentratedLeast Concentrated

Main Menu Concentration  The amount of solute dissolved in a unit of solution.  The volume that is usually taken is 1dm 3. The amount of solute may be expressed in g or mol therefore the units of concentration are g dm -3 or mol dm -3.  Unlike gases, the volume of a liquid is not directly related to it’s amount. For solutions, we express the amount through it’s CONCENTRATION.

Main Menu Molarity  The number of moles of a substance dissolved in one litre (dm 3 ) of a solution.  Units: mol dm -3  Pronounced: moles per decimetre cubed  Units often abbreviated to ‘M’ (do not do this in an exam!)  Volume must be in litres (dm 3 ) not ml or cm 3  This is the most useful measure of concentration but there are others such as % by weight, % by volume and molality.

Main Menu Example 1:  25.0 cm 3 of a solution of hydrochloric acid contains 0.100 mol HCl. What is it’s concentration?  Answer:  Concentration = moles / volume = 0.100 / 0.0250 = 4.00 mol dm -3  Note: the volume was first divided by 1000 to convert to dm 3

Concentration can also be expressed in mass (g dm -3 )

Main Menu Example 2:  Water is added to 4.00 g NaOH to produce a 2.00 mol dm -3 solution. What volume should the solution be in cm 3 ?  Calculate quantity of NaOH:  n(NaoH)= mass / molar mass = 4.00/40.0 = 0.100  Calculate volume of solution:  Volume = moles / concentration = 0.100 / 2.00 = 0.0500 dm 3 = 50.0 cm 3

Main Menu More questions…  Complete the test yourself questions on page 45.  Questions 40 and 41.  Check your answers on page 559.

Parts per million (ppm) Another unit of concentration Denotes one part per 10 6 by mass. Useful for very low concentrations such as found in air and water pollution A concentration of 1 ppm for a substance means that each kilogram of solution contains 1 milligram of solute. Assuming a density of 1 g dm -3, 1ppm also means each dm 3 of solution contains 1 mg of solute.

You can make a dilution from a more concentrated starting solution, called a stock solution, by adding a solvent. As a solution is diluted, the number of moles of solute remains the same, but now they are spread over a larger volume. Hence, concentration is decreased. Number of moles is constant. n=cV Concentration x volume must be constant through dilution

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