Chapter 11 Practice Worksheet Key. Solutions and Their Properties

Name: ____KEY___________________ Section: ___________ Chapter 11 Practice Worksheet Key: Solutions and Their Properties 1) Describe the 3 steps involved in the dissolution of a solid. Step 1: separation of solvent molecules (breaking intermolecular forces); Step 2: separation of solute particles (breaking ionic bonds); Step 3: combining solute and solvent particles. 2) Define the following terms of concentration: molarity, molality, mole fraction, and mass percent. Molarity = moles of solute / liters of solution (mol/L = M). Molality = moles of solute / kg of solvent (mol/kg = m). Mole fraction = moles of one component of a solution / total moles of solution (unitless). Mass percent = (mass of solute / (mass of solute + mass of solvent)) x 100 (unitless). 3) Define the terms unsaturated, saturated, and supersaturated. Unsaturated solution: more solute can be added to a solvent and it will still dissolve. Supersaturated solutions can be made by slowly heating and mixing solutions to add more solute than would be expected. Supersaturated solutions may look like the solvent but may immediately recrystallize if 1) removed from the heat source, 2) left to sit without stirring, or 3) adding even the smallest amount (1 crystal) of solute. 4) In general, how does temperature affect the solubility of solid solutes? How does temperature affect the solubility of gaseous solutes? Increasing temperature generally increases the solubility of solids. With more kinetic energy, particles are moving more quickly. Solvent-solvent attractions and solute-solute bonds are more likely to break, thus dissolving more solid. Conversely, at higher temperatures, gaseous solutes tend to escape from the liquid phase, thus decreasing their solubility. 5) How does pressure affect the solubility of a solid solute? How does pressure affect the solubility of a gaseous solute? A change in pressure does not noticeably affect the solubility of solid solutes. According to Henry’s Law, with increasing pressure, the solubility of gaseous solutes will increase. Increasing the pressing above a gas forces more atoms/molecules to go into solution, therefore increasing the pressure. 6) Define vapor pressure. Vapor pressure is the pressure exerted by atoms/molecules as they evaporate from the liquid phase into the gas phase (in a closed container). Eventually, an equilibrium is reached and the maximum number of particles that can go into the gas phase will. The distance these particles displace mercury in an evacuated tube can be measured (in mmHg) and the pressure exerted by the vapor phase determined. Name: ____KEY___________________ 7) How does addition of a solute to a solvent affect vapor pressure? Section: ___________ When a solute is added to a solvent, the vapor pressure will be lower than that of the pure solvent. If a solute is added to the solvent (pure substance), then there is less room for the solvent particles at the boundary between the liquid and gas phases. The solute particles taking up space at the boundary will reduce the number of particles allowed to escape to the gas phase. This lowers the number of particles contributing to the pressure and therefore lowers the vapor pressure. 8) How does the boiling point of a solution compare to the boiling point of a pure substance? How does the freezing point of a solution compare to the freezing point of a pure substance? Draw a phase diagram to defend your answers. (Hint: What happens to the vapor pressure of a solution compared to a pure solvent?) The boiling point of a solution is higher than the boiling point of a pure substance. As mentioned in question 5 above, adding a solute to a solvent lowers the vapor pressure. On a phase diagram, we can see that lowering the vapor pressure shifts the liquid-gas equilibria downward. This, in turn, will raise the boiling point of the solution (shift the temperature to the right). 9) Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees. (Kf of CHCl3 = 4.68 oC/m, normal freezing point = -63.5oC) ∆Tf = Kf · m m = (42.0 g * 1 mol / 154.1 g) / 0.600 kg = 0.454 m; Kf = 4.68 oC/m; ∆Tf = 2.12589oC The freezing point of a solution is lower than the freezing point of a pure substance, so we subtract 2.13oC from -63.5oC. -63.5oC – 2.13oC = 65.6oC 10) Calculate the boiling point of a solution containing 53.4 g of CCl4 and 1.23 kg of ethanol (CH3CH2OH, Kb = 1.22 oC/m, normal boiling point = 78.4 oC). ∆Tb = Kb · m m = (53.4.0 g * 1 mol / 153.8 g) / 1.23 kg = 0.282 m; Kb = 1.22 oC/m; ∆Tb = 0.3442oC The boiling point of a solution is higher than the boiling point of a pure substance, so we add 0.3442oC to 78.4oC. 78.4oC + 0.3442oC = 78.7oC 11) A layer of onion cells is placed on a microscope slide. A 1 M NaCl solution is added to the onion cells. Use the principle of osmosis to explain what happens to the onion cells. When NaCl is placed on the onion cells, the cells try to get the system back to concentration equilibrium. Water always flows from less concentrated (in solutes) to more concentrated solutions through a semipermeable membrane (the walls of the cells in this case). As onion cells lose water, they become hypotonic and begin to shrivel up (known as crenation).