1 / 8

Section 4.5

Section 4.5. Concentrations of Solutions. Concentration. Amount of solute dissolved in a given quantity of solvent or solution Amount of solute = Concentration. Concentration of a solution. moles of solute. Molarity ( M ) =. volume of solution in liters. Molarity (M).

afleckenstein
Download Presentation

Section 4.5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Section 4.5 Concentrations of Solutions

  2. Concentration • Amount of solute dissolved in a given quantity of solvent or solution • Amount of solute = Concentration

  3. Concentration of a solution moles of solute Molarity (M) = volume of solution in liters Molarity (M)

  4. SAMPLE EXERCISE 4.11 Calculating Molarity Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution. Solve: The number of moles of Na2SO4 is obtained by using its molar mass: Converting the volume of the solution to liters:

  5. Concentration of Electrolytes • For ionic compounds: • Look at chemical formula for relative concentration of ions • Ex 1: 1.0M NaCl is 1.0M Na+ and 1.0M Cl- • Ex 2: 1.0M Na2SO4 is 2.0 M Na+ and 1.0M SO42-

  6. Dilution Process of lowering the concentration of a solution by adding water

  7. Dilution Problem • We want: 250.0 mL of 0.100 M CuSO4 • We have: 1.00M CuSO4 Moles solute before dilution = Moles after Mol CuSO4 in dil soln = 0.250L soln x 0.100mol CuSO4 L soln = 0.0250 mol CuSO4 Liters of conc soln = (0.0250 mol CuSO4) x 1 L soln 1.00mol CuSO4 = 0.0250 L Pipet 0.0250L conc sol add dilute it to 250.0 mL

  8. Quick Calculation M conc x V conc = M dil x V dil Why can we ignore moles in dilution problems?

More Related