Chapter 13. This ratio is the concentration of the solution.
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1 Concentration Calculation Concentration In a solution, the solute is distributed evenly throughout the solvent. This means that any part of a solution has the same ratio of solute to solvent as any other part of the solution. This ratio is the concentration of the solution. The concentration is the amount of a particular substance in a given quantity of a solution
2 Concentration, continued Calculating Concentration, continued Concentrations can be expressed in many forms.
3 Calculating Parts per Million Sample Problem A A chemical analysis shows that there are 2.2 mg of lead in exactly 500 g of water. Convert this measurement to parts per million.
4 Calculating Parts per Million, continued Sample Problem A Solution mass of solute: 2.2 mg mass of solvent: 500 g parts per million =? 1 g 1000 mg mg g g Pb parts 500 g H O 1 million ppm (parts Pb per million parts H2O)
5 Since the mole is the unit chemists use to measure the number of particles, they often specify concentrations using molarity. (M) is a concentration unit of a solution expressed as moles of solute dissolved per liter of solution.
6 , continued Preparing a Solution of Specified Note that molarity describes concentration in terms of volume of solution, not volume of solvent. If you simply added mol solute to L solvent, the solution would not be M. The added solute will change the volume, so the solution would not have a concentration of M. The solution must be made to have exactly the specified volume of solution.
7 Preparing L of a M Solution
8 , continued Calculating In working with solutions in chemistry, you will find that numerical calculations often involve molarity. The key to all such calculations is the definition of molarity, which is stated as an equation below. moles of solute molarity = liters of solution
9 Calculating Given Mass of Solute and Volume of Solution
10 Calculating Mass of Solute Given and Volume of Solution
11 Calculating Sample Problem B What is the molarity of a potassium chloride solution that has a volume of ml and contains 85.0 g KCl?
12 Calculating Sample Problem B Solution volume of solution = ml mass of solute = 85.0 g KCl molarity of KCl solution =? 1 mol 85.0 g KCl 1.14 mol KCl g KCl 1 L ml L 1000 ml 1.14 mol KCl L 2.85 mol/l M KCl
13 , continued Using in Stoichiometric Calculations There are many instances in which solutions of known molarity are used in chemical reactions in the laboratory. Instead of starting with a known mass of reactants or with a desired mass of product, the process involves a solution of known molarity. The substances are measured out by volume, instead of being weighed on a balance.
14 Solution Stoichiometry Sample Problem C What volume (in milliliters) of a M solution of copper(ii) sulfate, CuSO 4, is needed to react with an excess of aluminum to provide 11.0 g of copper?
15 Solution Stoichiometry, continued Sample Problem B Solution [CuSO 4 ] = M mass of product = 11.0 g Cu solution volume =? L molar mass of Cu = g /mol 3CuSO 4 (aq) + 2Al(s) 3Cu(s) + Al 2 (SO 4 ) 3 (aq)
16 Solution Stoichiometry, continued Sample Problem B Solution, continued 3CuSO 4 (aq) + 2Al(s) 3Cu(s) + Al 2 (SO 4 ) 3 (aq) 1 mol Cu 3 mol CuSO 11.0 g Cu g Cu 3 mol Cu 4 1 L solution 1000 ml solution mol CuSO 1 L solution ml CuSO 4 solution
17
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