Download
chemistry 107 n.
Skip this Video
Loading SlideShow in 5 Seconds..
Chemistry 107 PowerPoint Presentation
Download Presentation
Chemistry 107

Chemistry 107

74 Views Download Presentation
Download Presentation

Chemistry 107

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chemistry 107 Unit #8 Instructor: Kristine A. Cooper Burlington County College Textbook chapters:48-53

  2. Solutions • Recall that a solution is a blend of two or more components. • Solvent = major fraction Dissolving agent, most cases H2O • Solute = minor fraction Being dissolved, usually a solid

  3. SOLUTION  Homogenous mixture: components blended together, uniform Heterogenous mixture: components remain separate, phases

  4. Solutions • Solutions do not have to be liquids! • They may be solids or gasses as well. Examples: Metal alloy – Brass (Zn & Cu) Air

  5. Solutions • Solutions are often identified by the percent solvent. • Ex: 5% NaCl solution • Sometimes the term concentrated is used to indicate a high percentage of solute is present. • Ex. Conc. NaCl, usually 40% • Dilute solutions contain low percentages of solute. • Ex: 3% H2O2

  6. Solutions • A saturated solution is one which contains the maximum dissolvable amount of solute • Beyond the point of saturation, added solute will not dissolve and will precipitate in solution.

  7. Solutions • An unsaturated solution is simply a solution which is not saturated and is capable of dissolving additional solvent. • Supersaturation : Larger amount of solute able to dissolve in solvent due to the heating of solution. ex. Rock Candy

  8. Percent Solutions • Two methods of measurement are used: Percent by mass Calculated on a gram basis Most common Percent by volume Calculated on a liter basis

  9. Percent Solutions • Percent by mass: gram to gram basis • Example: A solution is prepared by dissolving 65g of KOH in 100g water. What is the percent KOH in the solution? • %KOH= mass KOH / total mass • %KOH= 65g / 165g(65g + 100g) • %KOH= 39.39%

  10. More examples • A solution is prepared by dissolving 105g of NaCl in 425g water. What is the percent NaCl in the solution? • A solution is prepared by dissolving 85g of KCl in 260g of water. What is the percent KCl in the solution? • How many grams of sugar are present in 300g of a 15% solution?

  11. Percent Solutions • Percent by volume: mL to mL or liter to liter basis. • Example: A solution is prepared by dissolving 65mL of KOH in 100mL water. What is the percent KOH in the solution? • %KOH= volume KOH / total volume • %KOH= 65mL / 165mL • %KOH=39.39%

  12. More examples • A solution is prepared by dissolving 200mL of NaCl in 575mL water. What is the percent NaCl in the solution? • A solution is prepared by dissolving 135mL of KCl in 215mL of water. What is the percent KCl in the solution? • How many grams of sugar are present in 300mL of a 15% solution?

  13. Solvents • Recall that water is a very polar solvent, therefore is excellent for dissolving polar solutes. • If the solvent is not explicitly stated on the label, etc. it is assumed to be water. • Non-polar solutes require non-polar solvents, like alcohol.

  14. Molarity • Molarity (M)=moles per LITER M= moles present / liters of solution = (g/MW)/(liters) • Example: What is the molarity of a solution that is obtained by dissolving 35g of KOH in enough water to prepare 2L of solution? (Convert grams KOH to moles KOH)

  15. Molarity practice • How many moles of HNO3 are present in 500mL of a 5M solution? • If 60g of NaOH is dissolved in enough water to prepare a 0.75M solution, what volume will the solution occupy? • V=moles / M or L = mol / (mol / L)

  16. Molality • Molality (m)=moles per KILOGRAM M= moles present / kilograms of solution = (g/MW)/(kilograms) Example: If 8.6g of KBr is stirred into 500g of water, what is the molality of the solutions? g  mol KBr g  kg H2O = mol / kg

  17. Molality Examples • How many moles of H2SO4 are present in a 3.25 m solution that contains 700g of water? m = moles solute / kg solvent Moles = (m) (kg solvent); 3.25 m (moles/kg ) x 0.7kg = 2.28 moles

  18. Another molality example • How much water is needed to prepare a 0.275 m HNO3 solution from a 35g sample? • m = moles solute / kg solvent • (m) X (kg solvent) = moles of solute, now rearrange EQ ….. • kg solvent = moles solute / m • Convert 35 g HNO3 to moles HNO3 moles HNO3/mol / kg or moles HNO3X kg / mol = 2.02 kg water

  19. Mole Fraction • Mole fraction evaluates how many moles of each component make up a solution. Moles of compound X / total moles

  20. Mole Fraction Example A solution is prepared by dissolving 30.57g of NaCl in 140g of water. What is the mole fraction of NaCl and the mole fraction of water in the solution? Moles of NaCl=(g/MW)= 30.57g / 58.5g/mol =0.523 mol Moles of H2O=(g/MW)= 140g / 18g/mol = 7.78 mol Total mol = 8.30 mol ( .523 + 7.78 ) THEN… M.F. NaCl = mol NaCl / total mol =(0.523 / 8.30) = 0.0630 M.F. water = mol H2O / total mol =(7.78 / 8.30) = 0.937 Should be additive to ~1

  21. Colligative Properties • Properties observed when a solute is dissolved in a solvent and a change in solution characteristic occurs. • This change is concentration based, not identity based. # particles dependent • B.P. elevation • Freezing point depression • Osmotic pressure

  22. Colligative Properties • When an ionic compound is dissolved in water, the + and - ions interfere with the attractions between the water molecules. • This makes it more difficult for water molecules to attain a crystalline structure, particles get in the way of the bonds…… therefore lowers the freezing point. • This is why salt can be used to melt ice.

  23. Freezing Point • The freezing point is related to the molality of the dissolved particles, not the identity of the particles themselves. • F.P. depression: (subtract change in temp from original) Δ Tf=(Kf)(me)(Df) Δ Tf =change in F.P. Kf =F.P. constant of solvent me=molality of solution Df = Dissociation factor

  24. Ionic vs. Molecular • Recall the differences in dissociation between these compounds. • me C6H12O6 = 1.0 -molecular • me NaCl = 2.0 Na (1) and Cl (1) -ionic • me CaCl2 = 3.0 Ca (1) and Cl (2) -ionic

  25. Boiling Point • Just as solute can depress freezing point, it can elevate boiling point. • Water with ions interfering with intermolecular attractions cannot escape solution as easily, therefore boils at a higher temperature. • B.P. elevation: ΔTb=(Kb)(me)(Df) ΔTb = change in B.P. (add change in temp to original) Kb = B.P. constant of solvent me = molality of solution Df = Dissociation factor

  26. Osmotic Pressure • Particles (ions) move from areas of high concentration to areas of lower concentration. • Osmotic pressure is the force generated by these moving particles. • The greater the difference in molality, the greater the osmotic pressure. • These particles are termed electrolytes.