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Solutions

Solutions. Chapter 15. Mixtures. Heterogeneous mixture- unevenly mixed substance (separation can be seen) Homogeneous mixture- evenly mixed substance (no separation can be seen). Suspensions. ~Small but visible particles suspended or floating in a gas or liquid (heterogeneous mixture)

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Solutions

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  1. Solutions Chapter 15

  2. Mixtures • Heterogeneous mixture- unevenly mixed substance (separation can be seen) • Homogeneous mixture- evenly mixed substance (no separation can be seen)

  3. Suspensions • ~Small but visible particles suspended or floating in a gas or liquid (heterogeneous mixture) • Like a snow globe or dust or “shake before using” • the particles are too big to float forever without being stirred • If a suspension sits, the particles will settle • Can be filtered out

  4. Colloids or Colloidal Suspension • ~mixture that appears uniform unless under a high powered microscope. • Particles are a little larger than the wavelength of light • Extremely light particles float almost indefinitely. • Milk, blood, smoke • These can be separated in a centrifuge

  5. Tyndall Effect • ~Scattering of light by a colloid or suspension • Both a colloid and a suspension have particles larger than the wavelength of light, so when light shines through it should be deflected every which way. • This will make the beam of light visible.

  6. Solutions • Particles are smaller than the wavelength of light. Therefore, it will not scatter light. • With solutions, no separation can be seen even under a high powered microscope. • Cannot be separated by any filter or by a centrifuge. • Can be separated by boiling/ melting points. • salt water, metal alloys, air

  7. Tyndall Effect Colloid/suspension solution

  8. Parts of a solution • Solvent- what the substance is dissolved in • Solute- what is being dissolved • Water is called the “universal solvent” • because it dissolves a lot of substances and is very common. • Water solutions are called aqueous.

  9. Solution misconceptions • Solutions don’t have to be a solid in a liquid. • carbonated water is CO2 dissolved in water, streams have dissolved O2 in them. • The solvent doesn’t have to be water or even a liquid. • Alloys (two or more metals) are a solution as is air. Several things dissolve in oils.

  10. Concentration • ~How much solute is present in a solution compared to the solvent. • Molarity (M)- moles of solute per liter of solution. M = mol/L • 2.1 M AgNO3 means 2.1 mol of AgNO3 for every one liter of solution

  11. Other measures of concentration

  12. Molarity Problems Molarity = mol/L Molarity = moles of solute / Liters of solution

  13. How many moles of HCl are in 125 mL of 2.5 M HCl? Molarity problems .125 L of soln = .31 mol HCl

  14. Here we go • What concentration solution would be prepared if 39 g of Ba(OH)2 were mixed in a 450 mL solution? =.2276 mol Ba(OH)2 M = mol/L .2276 mol Ba(OH)2 .45 L of solution =.51 M Ba(OH)2

  15. More • For a lab in this chapter, I need to make .60 L of 3.0 M NaOH, what mass of NaOH did I need? • .6 L x 3.0 M NaOH = 1.8 mol NaOH • 1.8 mol NaOH x 39.998 g/mol • = 72 g NaOH

  16. Molarity Problems • A 0.24 M solution of Na2SO4 contains 0.36 moles of Na2SO4. How many liters were required to make this solution? 0.36 mol Na2SO4 1 L soln 0.24 mol = 1.5 L Na2SO4

  17. Getting tougher 2 2 • AgNO3 + BaCl2 AgCl + Ba(NO3)2 • Balance the equation. If 1.2 L of .50 M AgNO3 is reacted completely, what molarity solution of Ba(NO3)2 will be created if the volume increased to 1.5 L? 1.2 L x .5 M AgNO3 = .6 mol AgNO3 = .3 mol Ba(NO3)2 1.5 L = .20 M Ba(NO3)2

  18. 2 • HNO3 + Zn  H2 + Zn(NO3)2 • If you have .65 L of 1.2 M HNO3 and you react it completely what volume of H2 gas will you produce at STP? 1.2 M HNO3 x .65 L = .78 mol HNO3 =.39 mol H2 = 8.7 L at STP

  19. 2 • HNO3 + Zn  H2 + Zn(NO3)2 • If you have .65 L of 1.2 M HNO3 and you react it completely, what conc. of Zn(NO3)2 will be left if the volume increases to .75 L? 1.2 M HNO3 x .65 L = .78 mol HNO3 = .39 mol Zn(NO3)2 .75 L = .52 M Zn(NO3)2

  20. 2 3 3 • Fe + H2SO4 Fe2(SO4)3 + H2 • If 350 mL of 2.3 M H2SO4 is completely reacted, what is the volume of hydrogen gas produced at 24o C and 114 kPa? .35 L x 2.3 M = .805 mol H2SO4 =.805 mol H2 PV = nRT =17 L H2 114 kPa V = .805 mol (8.31) 297 K

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