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Prentice Hall Chemistry (c) 2005

Prentice Hall Chemistry (c) 2005. By Daniel R. Barnes Init: 12/17/2008 (?). Section Assessment Answers Chapter 16.

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Prentice Hall Chemistry (c) 2005

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  1. Prentice Hall Chemistry(c) 2005 By Daniel R. Barnes Init: 12/17/2008 (?) Section Assessment Answers Chapter 16 WARNING: some images and content in this presentation may have been taken without permission from the world wide web. It is intended for use only by Mr. Barnes and his students. It is not meant to be copied or distributed. Solutions

  2. S1 S2 S1 S2 = = P1 P2 P1 P2 P2 S1 (288 kPa)(0.16 g/L) S2 = P1 = 104 kPa 1. The solubility of a gas in water is 0.16 g/L at 104 kPa. What is the solubility when the pressure of the gas is increased to 288 kPa? Assume the temperature remains constant. S1 = 0.16 g/L P1 = 104 kPa T1 = T2 S2 = ? g/L P2 = 288 kPa P2 P2 16.1 Practice Problems = 0.443 g/L = 4.4 x 10-1 g/L

  3. S1 S2 S1 S2 = = P1 P2 P1 P2 S2 P1 (9.5 g/L)(1.0 atm) P2 = = S1 3.6 g/L 2. A gas has a solubility in water at 0oC of 3.6 g/L at a pressure of 1.0 atm. What pressure is needed to produce an aqueous solution containing 9.5 g/L of the same gas at 0oC? S1 = 3.6 g/L P1 = 1.0 atm T1 = T2 S2 = 9.5 g/L P2 = ? atm The unknown is on the bottom, so you’ve got to FLIP IT! 16.1 Practice Problems S2 S2 = 2.6 atm

  4. 3. What determines whether a substance will dissolve? What determines how fast a substance will dissolve? The chemical composition of the solute and the solvent determine whether or not the substance (solute) will dissolve. Such factors as agitation, temperature, and solute particle size affect the speed with which a solute dissolves. 16.1 Section Assessment

  5. 4. What units are usually used to express the solubility of a solute? Solubility is usually expressed in grams of solute per 100 g of solvent. 16.1 Section Assessment

  6. 5. What are two conditions that determine the mass of solute that will dissolve in a given mass of solvent? Temperature helps to determine the solubility of a solute. If the solute is a gas, pressure also affects solubility. 16.1 Section Assessment

  7. 6. What would you do to change . . . a. A saturated solid/liquid solution to an unsaturated solution? To make a saturated solid/liquid solution become unsaturated, I would add more solvent. b. A saturated gas/liquid solution to an unsaturated solution? 16.1 Section Assessment To make a saturated gas/liquid solution become unsaturated, I would increase the pressure.

  8. 7. The solubility of a gas is 0.58 g/L at a pressure of 104 kPa. What is the solubility if the pressure increases to 250 kPa at the same temperature? Henry’s Law says that solubility of a gas in a liquid is proportional to pressure. Henry’s Law, expressed as a mathematical equation, is . . . S2 = ? g/L S1 = 0.58 g/L S2 S1 16.1 Section Assessment = P1 = 104 kPa P2 = 250 kPa P2 P1 (0.58 g/L) (250 kPa) P2 S2 P2 S1 S2 = = 104 kPa P2 P1 S2 = 1.4 g/L

  9. 16. How do you calculate the molarity of a solution? To calculate molarity, divide moles of solute by liters of solution. (If grams of solute are given instead of moles, you’ll need the molar mass of the solute to help you convert grams of solute into moles of solute.) 16.2 Section Assessment

  10. 17. Compare the number of moles of solute before dilution with the number of moles of solute after dilution. Since dilution means adding more solvent, the number of moles of solute does not change. However, the concentration does go down. Therefore, even though the total number of moles of solute does not go down, the number of moles of solute per liter of solution does go down. 16.2 Section Assessment

  11. 18. What are two ways of expressing the concentration of a solution as a percent? Express the concentration as the ratio of volume of solute to the volume of solution (v/v) or as the ratio of the mass of the solute to the mass of the solution (m/m). 16.2 Section Assessment

  12. 19. Calculate the molarity of a solution containing 400 g CuSO4 in 4.00 L of solution. 6.27 x 10-1 M CuSO4. I apologize that I don’t have time to show how this answer is determined. (Or many of the others in the rest of this power point.) I have to submit 5 week progress report grades and my clock is ticking . . . If anyone is still interested in seeing HOW this answer is achieved, please e-mail me sometime after 3/14/2011. 16.2 Section Assessment

  13. 20. How many moles of solute are present in 50.0 mL of 0.20 M KNO3? 1.00 x 10-2 M KNO3. 16.2 Section Assessment

  14. 21. How many milliliters of a stock solution of 2.00 M KNO3 would you need to prepare 100.0 mL of 0.150 M KNO3? 7.50 mL 16.2 Section Assessment

  15. 22. What is the concentration, in percent (v/v), of a solution containing 50 mL of diethyl ether (C4H10O) in 2.5 L of solution? 2.0% (v/v) diethyl ether 16.2 Section Assessment

  16. 23. How many grams of K2SO4 would you need to prepare 1500 g of 5.0% K2SO4 (m/m) solution? 7.5 x 101 g K2SO4 = 75 g 16.2 Section Assessment

  17. 24. What are three colligative properties of solutions? Vapor-pressure lowering, boiling-point-elevation, and freezing-point depression. (The book doesn’t mention it, but I think osmotic pressure is also one.) 16.3 Section Assessment

  18. 25. What factor determines how much the vapor pressure, freezing point, and boiling point of a solution differ from those properties of the pure solvent? The number of solute particles dissolved in the solvent 16.3 Section Assessment

  19. 26. Would a dilute or a conentrated sodium fluoride solution have a higher boiling point? Explain. Concentrated sodium fluoride, because the magnitude of the boiling point elevation is proportional to the number of solute particles dissolved in the solvent. 16.3 Section Assessment

  20. An equal number of moles of KI and MgI2 are dissolved in equal volumes of water. Which solution has the higher • a. Boiling point? • b. Vapor pressure? • c. Freezing point? a. The MgI2 solution has a higher boiling point because one mole of MgI2 breaks into three moles of ions (one mole of Mg2+ and two moles of I-). One mole of KI only breaks into two moles of ions (one mole of K+ and one mole of I-). 16.3 Section Assessment b. More dissolved ions means lower vapor pressure, so the KI solution, which has fewer ions per mole, has its vapor pressure depressed less than the MgI2 solution does. Therefore, the KI solution’s vapor pressure is higher (closer to that of pure water). c. The KI solution’s freezing point isn’t lowered as much as that of the MgI2 solution, so it remains higher (closer to the freezing point of water) than that of the MgI2 solution.

  21. 28. Explain why the vapor pressure, boiling point, and freezing point of an aqueous solution of a nonvolatile solute are not the same as those of a pure solvent. “Formation of solvation shells around solute particles reduces the number of water molecules with sufficient kinetic energy to escape the solution. Therefore, the vapor pressure is lower relative to pure solvent. More energy must be supplied to reach the boiling point; therefore, the boiling point is elevated relative to pure solvent. Solvation shells interfere with the formation of hydrogen-bonded ice structure. This results in depression of the freezing point relative to pure solvent.” 16.3 Section Assessment Well, that’s what the Teacher’s Edition says. I just want to go on record as saying that I’m not 100% satisfied with that explanation. I think there’s also a phase interface solvent particle-to-area ratio difference that affects the relative rate of interphasic particle transmigration. Yeah. I just said that. Run and tell that. Don’t worry. I barely understand what I just said myself.

  22. 37. What are two ways of expressing the ratio of solute particles to solvent particles? Molality and mole fractions 16.4 Section Assessment

  23. 38. How are freezing point depression and boiling point elevation related to molality? “The magnitudes of freezing point depression (DTf) and boiling point elevation (DTb) of a solution are directly proportional to the molal concentration (m), when the solute in molecular, not ionic.” 16.4 Section Assessment

  24. 39. How many grams of sodium bromide must be dissolved in 400.0 g of water to produce a 0.500 molal solution? 20.6 g NaBr 16.4 Section Assessment

  25. 40. Calculate the mole fraction of each component in a solution of 2.50 mol ethanoic acid (CH3COOH) in 10.00 mol of water. XCH3COOH = 0.200; XH2O = 0.800 Notice that the two mole fractions add up to 1.00. They always do. 16.4 Section Assessment

  26. 41. What is the freezing point of a solution of 12.0 g of CCl4 dissolved in 750.0 g of benzene? The freezing point of benzene is 5.48oC; Kf is 5.12oC/m. 4.95oC 16.4 Section Assessment

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