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Chapter 12 Solutions

Chapter 12 Solutions. A solution is composed of two parts: the solute and the solvent. Solute The gas (or solid) in a solution of gases (or solids), or the component present in the smaller amount. Solvent

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Chapter 12 Solutions

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  1. Chapter 12 Solutions

  2. A solution is composed of two parts: the solute and the solvent. • Solute • The gas (or solid) in a solution of gases (or solids), or the component present in the smaller amount. • Solvent • The liquid in the case of a solution of gases or solids, or the component present in the larger amount.

  3. Fluids that mix with or dissolve in each other in all proportions are said to be miscible (left). Fluids that do not dissolve in each other are said to be immiscible (right).

  4. The 5 g of Mo is the solute; the 80 g of Cr is the solvent. • MgCl2 is the solute; H2O is the solvent. • O2 and N2 are the solutes; Ar is the solvent.

  5. A saturated solution is in equilibrium with respect to the amount of dissolved solute. • The rate at which the solute leaves the solid state equals the rate at which the solute returns to the solid state.

  6. The solubility of a solute is the amount that dissolves in a given quantity of solvent at a given temperature. • An unsaturated solution is a solution not in equilibrium with respect to a given dissolved substance and in which more of the substance can be dissolved.

  7. A supersaturated solution is a solution that contains more dissolved substance than a saturated solution does. This occurs when a solution is prepared at a higher temperature and is then slowly cooled. This is a very unstable situation, so any disturbance causes precipitation.

  8. Solubility can be understood in terms of two factors: • The natural tendency toward disorder favors dissolving. • The relative forces between and within species must be considered. • Stronger forces within solute species oppose dissolving. • Stronger forces between species favor dissolving.

  9. For molecular solutions, this can be summarized as “Like dissolves like.” In other words, solutes dissolve in solvents that have the same type of intermolecular forces. • An immiscible solute and solvent are illustrated at right.

  10. When considering ionic solutes in water, we need to examine the hydration energy and the lattice energy.

  11. The stronger ion-dipole force between the ion and the solvent—that is, hydration energy—favors dissolving. • A stronger force between ions—that is, lattice energy—opposes dissolving.

  12. The force of attraction between water and both a cation and an anion is illustrated to the left with lithium fluoride, LiF.

  13. The process of dissolving occurs at the surfaces of the solid. Here we see water hydrating (dissolving) ions.

  14. The hydration energy for AB2 must be greater than the hydration energy for CB2.

  15. In general, solubility depends on temperature.

  16. In most cases, solubility increases with increasing temperature. However, for a number of compounds, solubility decreases with increasing temperature. • The difference is explained by differences in the heat of solution.

  17. When dissolving absorbs heat (is endothermic), the temperature of the solution decreases as the solute dissolves. The solubility will increase as temperature increases. • When dissolving releases heat (is exothermic), the temperature of the solution increases as the solute dissolves. The solubility will decrease as temperature increases.

  18. Cold packs use an endothermic solution process. • Hot packs use an exothermic solution process.

  19. Henry’s law describes the effect of pressure on gas solubility: The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution. • This is expressed mathematically in the equation • S = kHP • where • S = gas solubility • kH = Henry’s law constant for the gas • P = partial pressure of the gas over the solution

  20. In general, pressure has little or no effect on the solubility of solids or liquids in water. • The solubility of a gas increases as pressure increases, as illustrated at right.

  21. Helium–oxygen mixtures are sometimes used as the breathing gas in deep-sea diving. At sea level (where the pressure is 1.0 atm), the solubility of pure helium in blood is 0.94 g/mL. What is the solubility of pure helium at a depth of 1500 feet? • Pressure increases by 1.0 atm for every 33 feet of depth, so at 1500 feet the pressure is 46 atm. (For a helium–oxygen mixture, the solubility of helium will depend on its initial partial pressure, which will be less than 1.0 atm.)

  22. P1 = 1.0 atm P2 = 46 atm • S1 = 0.94 g/mL S2 = ?

  23. At high elevations, the partial pressure of oxygen decreases, decreasing the solubility of oxygen in water. Fish require a certain minimum level of dissolved oxygen to survive.

  24. The concentration of a solute can be quantitatively expressed in several ways: • Molarity • Mass percentage of solute • Molality • Mole fraction

  25. Molarity is the moles of solute per liter of solution. It is abbreviated as M.

  26. Mass percentage of solute is the percentage by mass of solute in a solution.

  27. An experiment calls for 36.0 g of a 5.00% aqueous solution of potassium bromide. Describe how you would make up such a solution.

  28. A 5.00% aqueous solution of KBr has 5.00 g KBr per 100. g solution. The remainder of the 100. g is water: 95 g. • We can use this ratio to determine the mass of KBr in 36.0 g solution: • Since 1.8 g KBr is required for 36.0 g of solution, the remainder consists of 34.2 g water. • We make the solution by mixing 1.8 g KBr in 34.2 g water.

  29. Molality is the moles of solute per kilogram of solvent. It is abbreviated as m.

  30. Iodine dissolves in a variety of organic solvents. For example, in methylene chloride, it forms an orange solution. What is the molality of a solution of 5.00 g iodine, I2,in 30.0 g of methylene chloride, CH2Cl2?

  31. Mass of solute = 5.00 g I2 • Mass of solvent = 30.0 g CH2Cl2

  32. Mole fraction is the moles of component over the total moles of solution. It is abbreviated C.

  33. A solution of iodine, I2, in methylene chloride, CH2Cl2, contains 5.00 g I2 and 56.0 g CH2Cl2. What is the mole fraction of each component in this solution?

  34. Mass of solute = 5.00 g I2 • Mass of solvent = 56.0 g CH2Cl2

  35. A bottle of bourbon is labeled 94 proof, meaning that it is 47% by volume of alcohol in water. What is the mole fraction of ethyl alcohol, C2H5OH, in the bourbon? The density of ethyl alcohol is 0.80 g/mL. One liter of bourbon contains 470 mL of alcohol and 530 mL of water. To solve this problem, we will convert the volume of ethyl alcohol to mass using density, and then convert to moles using molar mass.

  36. A 3.6 m solution of calcium chloride, CaCl2, is used in tractor tires to give them weight. The addition of CaCl2 also prevents water in the tires from freezing at temperatures above –20°C. What are the mole fractions of CaCl2 and water in such a solution? The 3.6 m solution contains 3.6 mol CaCl2 in 1.0 kg of water. To solve this problem, we will convert the mass of water to moles, and then compute the mole fractions.

  37. Moles solute = 3.6 mol CaCl2

  38. Converting between molality and mole fraction is relatively simple, because you know the masses or moles of both the solute and the solvent.

  39. A solution contains 8.89 × 10-3 mole fraction of I2 dissolved in 0.9911 mole fraction of CH2Cl2 (methylene chloride). What is the molality of I2 in this solution? We will assume we have 1 mole of the solution, so we begin with 8.89 × 10-3 mol I2 and 0.9911 mol CH2Cl2. Next, we will convert the moles of solvent into grams, and then into kilograms. Finally, we will compute the molality of the solution.

  40. Moles solute = 0.00889 mol I2

  41. Converting between molality and molarity requires knowing the density of the solution. This enables you to calculate the mass or volume of the solution. You can then distinguish the amount of solute from the amount of solvent, or combine them to find the volume of solution. • These types of conversions are illustrated in the following problems.

  42. Citric acid, HC6H7O7, is often used in fruit beverages to add tartness. An aqueous solution of citric acid is 2.331 m HC6H7O7. What is the molarity of the solution? The density of the solution is 1.1346 g/mL. A 2.331 m solution contains 2.331 mol solute in 1.000 kg solvent. We will use this relationship first to convert the moles of citric acid to grams and then to find the mass of solution. Using density, we can then find the volume of solution. Finally, we will compute the molarity of the solution.

  43. Moles solute = 2.331 mol HC6H7O7

  44. An aqueous solution of ethanol is 14.1 M C2H5OH. The density of the solution is 0.853 g/cm3. What is the molality of ethanol in the solution? We will work with 1.00 L of solution. First, we will convert volume to mass using the density. Then, we will find the masses of the solute and the solvent. Finally, we will compute the molality.

  45. Colligative properties of solutions are properties that depend on the concentration of the solute molecules or ions in solution but not on the chemical identity of the solute. • Vapor-pressure lowering • Boiling-point elevation • Freezing-point lowering • Osmotic pressure

  46. The vapor pressure of a solution, P, is less than the vapor pressure of the pure solvent, P°. • When the solute is nonvolatile, the vapor pressure of a solution is the mole fraction of the solvent times the vapor pressure of pure solvent.

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