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Solution. Solvent. Solute. A solution is a homogenous mixture of 2 or more substances. The solute is(are) the substance(s) present in the smaller amount(s). The solvent is the substance present in the larger amount. H 2 O. Soft drink ( l ). Sugar, CO 2. Air ( g ). N 2.

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4.1

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  1. Solution Solvent Solute A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount H2O Soft drink (l) Sugar, CO2 Air (g) N2 O2, Ar, CH4 Pb Sn Soft Solder (s) 4.1

  2. Solutions Solvent solute When the solvent is water the solution is said to be aqueous

  3. 9.1 General Properties of Aqueous Solutions A solution is a homogenous mixture of two or more substances. The substance present in the largest amount (moles) is referred to as the solvent. The other substances present are called the solutes. A substance that dissolves in a particular solvent is said to be soluble in that solvent.

  4. Concentration of Solutions 9.5 Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution. Other common rearrangements:

  5. Worked Example 9.8 For an aqueous solution of glucose (C6H12O6), determine (a) the molarity of 2.00 L of a solution that contains 50.0 g of glucose, (b) the volume of this solution that would contain 0.250 mole of glucose, and (c) the number of moles of glucose in 0.500 L of this solution. Strategy Convert the mass of glucose given to moles, and use the equations for interconversions of M, liters, and moles to calculate the answers. moles of glucose = Think About ItCheck to see that the magnitude of your answers are logical. For example, the mass given in the problem corresponds to 0.277 mole of solute. If you are asked, as in part (b), for the volume that contains a number of moles smaller than 0.277, make sure your answer is smaller than the original volume. 50.0 g 180.2 g/mol 0.277 mol C6H12O6 2.00 L solution 0.250 mol C6H12O6 0.139 M solution = 0.277 mol Solution molarity = = 0.139 M volume = = 1.80 L moles of C6H12O6 in 0.500 L = 0.500 L×0.139 M = 0.0695 mol

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  7. Moles of solute before dilution (i) Moles of solute after dilution (f) = Dilution Add Solvent = MfVf MiVi Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. 4.5

  8. How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? MfVf 0.200 x 0.06 Vi = = 4.00 Mi MiVi = MfVf Mi = 4.00 Vi = ? L Mf = 0.200 Vf = 0.06 L = 0.003 L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution

  9. Another Dilution Problem If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mL What would be the resulting concentration? M1*V1 = M2*V2 (6.5M) * (32 mL) = M2 * (500.0 mL) 6.5 M * 32 mL M2 = 500 mL M2 = 0.42 M

  10. Concentration of Solutions Dilutionis the process of preparing a less concentrated solution from a more concentrated one. moles of solute before dilution = moles of solute after dilution

  11. Concentration of Solutions In an experiment, a student needs 250.0 mL of a 0.100 M CuCl2 solution. A stock solution of 2.00 M CuCl2 is available. How much of the stock solution is needed? Solution: Use the relationship that moles of solute before dilution = moles of solute after dilution. (2.00 M CuCl2)(Lc) = (0.100 M CuCl2)(0.2500 L) Lc = 0.0125 L or 12.5 mL To make the solution: Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask. Carefully dilute to the calibration mark. Mc × Lc = Md × Ld

  12. Concentration of Solutions Because most volumes measured in the laboratory are in milliliters rather than liters, it is worth pointing out that the equation can be written as Mc × mLc = Md × mLd

  13. Worked Example 9.9 What volume of 12.0 M HCl, a common laboratory stock solution, must be used to prepare 150.0 mL of 0.125 M HCl? StrategyMc = 12.0 M, Md = 0.125 M, mLd = 250.0 mL Solution 12.0 M× mLc = 0.125 M × 250.0 mL mLc = 0.125 M × 250.0 mL 12.0 M = 2.60 mL Think About It Plug the answer into Equation 9.4, and make sure that the product of concentration and volume on both sides of the equation give the same result.

  14. Worked Example 9.10 Starting with a 2.0 M stock solution of hydrochloric acid, four standard solutions (1 to 4) are prepared by sequential diluting 10.00 mL of each solution to 250.00 mL. Determine (a) the concentrations of all four standard solutions and (b) the number of moles of HCl in each solution. Strategy (a) Md = ; (b) mol = M×L, 250.00 mL = 2.500×10-1 L 2.00 M × 10.00 mL 250.00 mL 8.00×10-2M × 10.00 mL 250.00 mL 3.20×10-3M × 10.00 mL 250.00 mL 1.28×10-4M × 10.00 mL 250.00 mL Mc× mLc mLd Solution (a) Md1 = Md2 = Md3 = Md4 = = 8.00×10-2M = 3.20×10-3M = 1.28×10-4M = 5.12×10-6M

  15. Worked Example 9.10 (cont.) Solution (b) mol1 = 8.00×10-2M × 2.500×10-1 L = 2.00×10-2 mol mol2 = 3.20×10-3M × 2.500×10-1 L = 8.00×10-4 mol mol3 = 1.28×10-4M × 2.500×10-1 L = 3.20×10-5 mol mol4 = 5.12×10-6M × 2.500×10-1 L = 1.28×10-6 mol Think About It Serial dilution is one of the fundamental practices of homeopathy. Some remedies undergo so many serial dilutions that very few (if any) molecules of the original substance still exist in the final preparation.

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