1 / 0

Properties of Solutions Chp 10 and 11

Properties of Solutions Chp 10 and 11. Homogeneous Looks same throughout Elements Compounds Some mixtures ex: koolaid. Heterogeneous Parts can be seen Some mixtures Ex: rocky road ice cream. Classification of Matter.

sasson
Télécharger la présentation

Properties of Solutions Chp 10 and 11

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. Properties of SolutionsChp 10 and 11
  2. Homogeneous Looks same throughout Elements Compounds Some mixtures ex: koolaid Heterogeneous Parts can be seen Some mixtures Ex: rocky road ice cream Classification of Matter
  3. Mixtures2+ elements, compounds or bothfairly easily dividedproportions can varyretain properties Elements Single substance Can’t be divided Unique properties Representative particle is atom Compounds 2 + elements Not easily divided Chemically bonded Same proportions New properties Representative particle is molecule
  4. Mixtures Solutions Colloids Suspensions < 1 nm > 1nm < 1000 nm > 1000 nm Do not settle out do not settle out settle out No Tyndall effect Tyndall effect Tyndall effect
  5. Solute substance dissolved least amount can be: solid liquid gas Solvent dissolving medium largest amount can be: solid liquid gas Universal solvent is water Parts of solution solute + solvent = solution
  6. electrolyte conducts electricity ions ionic bonds Ex: salt in water nonelectrolyte does not conduct electricity no source of ions covalent bonds Ex: sugar in water Nature of solutions Na+ C12H22O11 C12H22O11 Na+ Cl- Cl- Na+ Cl- C12H22O11
  7. Three steps of solution formation Step 1 DH1 Separate the solute (expand the solute) endothermic Step 2 DH2 Overcome IMF in solvent to make room for the solute (expand the solvent) endothermic Step 3 DH3 Allow solute and solvent to interact to form the solution. exothermic

  8. Solution Formation

    DH3 DHsoln = enthalpy (heat) of solution DHsoln= DH1 + DH2 + DH3 DH1 + DH2 DH1 + DH2 Energy of solution DH3 DHsoln Energy of separate solute and solvent Energy (H) Energy (H) Energy of solution Positive DHson (energy absorbed) Negative DHsoln (energy released)
  9. Energy terms for various types of solutes and solvents
  10. Factors affecting solubility Polarity Pressure solids and liquids-little effect gases-significant effect Temperature solids and liquids-temperature↑, solubility↑ gases –temperature ↑, solubility ↓
  11. ‘Like dissolves like” Polar solute dissolves in polar solvent Nonpolar solute dissolves in nonpolar solvent Which solvent would be the best for dissolving grease C20H42? Water or hexane? Which solvent would be best for dissolving KI? Water or hexane?
  12. Henry’s Law The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. P = kC P = pressure k = Henry’s law constant C = concentration of dissolved gas

  13. Solution Concentration

    molarity mols solute/liter of solution A solution is prepared by mixing 1.00 g ethanol with 100.0 g water to give a final volume of 101 mL. Calculate the molarity. 1.00 g C2H5OH X 1 mol C2H5OH = 2.17 X 10-2 mol C2H5OH 46.07 g C2H5OH Volume = 101 mL = 0.101L Molarity of C2H5OH = moles of C2H5OH = 2.17 X 10-2 mol liters of solution 0.101 L = 0.215 M
  14. molality mols solute/kg of solvent Molality of C2H5OH = moles of C2H5OH = 2.17 X 10-2mol kilogram of H2O 100.0 g X 1 kg 1000 g = 2.17 X 10-2mol 0.1000 kg = 0.217 m
  15. mass percent mass of solute X 100 % mass of solution Mass percent of C2H5OH = 1.00 g C2H5OH X 100% 100.0 g H2O + 1.00 g C2H5OH = 0.990% C2H5OH
  16. mole fraction mole fraction of component A = XA = nA nA+ nB mole fraction of C2H5OH = nC2H5OH nC2H5OH + nH2O nH2O = 100.0 g H2O X 1 mol H2O = 5.56 mol 18.0 g H2O XC2H5OH = 2.17 X 10-2 mol 2.17 X 10-2mol + 5.56 mol = 0.00389
More Related