Topic 10: Solutions

1. To go with Chapter 13: Silberberg Principles of General Chemistry AP Chemistry, Mrs Peck Topic 10: Solutions

2. Perform calculations with different solution concentrations such as molarity, mass percent, molality, and mole fraction. Discuss the effects of temperature, pressure, and structure on solubility Perform calculations with Raoult’s law Understand colligative properties such as boiling point elevation, freezing point depression, and osmotic pressure Use colligative properties to determine the molar mass of a solute Objectives/Study Guide AP tip: Questions from this section may appear on both the F.R. and M.C. Sections of the AP exam. Question on the solubility or miscibility of substances In one another may appear in the essay section. Questions involving the Determination of molar mass by freezing point depression could also appear in The lab essay.

3. Types of solutions • A solution is a homogeneous mixture.. • We will focus on liquid solutions in this topic. • A solute, the substance being dissolved, is added to a solvent, which is present in the largest amount when referring to a liquid-liquid or gas-liquid solution. • Composition • Molarity: ‘M’ of a solution is the number of moles of solute per liter of solution • Mass percent: also called weight percent, is the percent by mass of solute in a solution. • Mole fraction: ‘x’ is the ratio of moles of a given component to the total number of moles of solution. For a two component solution, where nA and nB are the moles of the two components: • xA = nA/(nA + nB) • Molality: ‘m’ is the number of moles of solute per kilogram of solvent • m = moles solute/kg of solvent Solutions and their compositions

4. A solution is prepared by mixing 30.0mL of butane (C4H10, d= 0.600 g/ml) with 65.0 mL of octane (C8H18, d= 0.700 g/ml) Assuming that the volumes add in mixing, calculate the following for butane: A) molarity B) mass percent C) mole fraction D) molality Example 1 30.0ml x 0.600g x 1 mol = 0.310 mol C4H10 = 3.26 M 1 ml 58.1g 0.0950 L Mass solution = [30.0ml x 0.600g] + [65.0 ml x 0.700g] = 63.5g 1ml 1 ml Mass solute = 30.0 ml x 0.600g = 18.0g 1ml Mass % = 18 g x 100 = 28.3% butane 63.5g Moles octane = 65.0 ml x 0.700 g x 1 mol = 0.399 mol octane 1ml 114g Mole fraction butane = 0.310 mol butane = 0.437 (0.310 mol butane + 0.399 mol octane) 0.310 moles butane/0.0455 kg = 6.81m

5. The formation of a liquid solution begins by separating the solute into its individual components. Next, the solvent’s intermolecular forces must be overcome to make room for the solute. The solute and solvent then interact to form the solution. Factors affecting solubility

6. The phrase ‘like dissolves like’ means that solubility is favored if the solute and solvent have similar polarities, as determined by their structure. • The table summarizes the solubilities of different types of solutes in different types of solvents. Structural effects

7. Discuss the solubility of each of the following solutes in carbon tetrachloride: ammonium nitrate, 1-pentanol, and pentane. Explain why each solute will or will not dissolve. Example 2 Carbon tetrachloride is nonpolar and nonpolar solutes Will dissolve in it. Ammonium nitrate is ionic and will not Dissolve in a nonpolar solvent. It will dissolve in a polar solvent Like water. 1-propanol is polar and will not be soluble in carbon Tetrachloride. Pentane is nonpolar and will be miscible (will form A homogeneous mixture) with carbon tetrachloride.

8. Pressure has little effect on the solubilities of liquids and solids. Gases become more soluble in liquids when the pressure of the gas above the liquid decreases. Pressure effects

9. The solubility of most solids increases with temperature. • However, some solids decrease solubility with temperature. • The energy needed to separate the solute-solute particles is usually greater than the energy released from the solute-solvent attractions. • It is difficult to predict the temperature dependence of the solubilities of solids,. • Gases, on the other hand, always decrease in solubility with increasing temperature. Temperature effects

10. A nonvolatile solute lowers the vapor pressure of the solvent. • The nonvolatile solute decreases the escaping tendency of the solvent molecules. • Raoult’s law: • State that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent present. • Psoln = xsolventP0solvent • Psoln is the observed vapor pressure of the solution • Xsolvent is the mole fraction of the solvent • P0solvent is the vapor pressure of the pure solvent. • For an ideal solution, a plot of Psolnvs xsolvent at constant temperature gives a straight line with a slope equal to P0solvent. The vapor pressures of solutions

11. The vapor pressure of a solution containing 53.6g of glycerin, C3H8O3, in 133.7 g of ethanol, C2H5OH, is 113 torr at 40*C. Calculate the vapor pressure of the pure ethanol at 40*C assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol. Example 3 Xethanol = mol ethanol/ total mol in solution Pethanol = xethanol P0ethanol 133.7 g C2H5OH x 1 mol = 2.90 mol C2H5OH 46.07 g 53.6 g C3H8O3 x 1 mol = 0.582 mol C3H8O3 92.09 g Total mol = 0.582 mol + 2.90 mol = 3.48 mol P0ethanol = 113 torr = 136 torr (2.90 mol/3.48 mol)

12. Ideal solutions obey Raoult’s law. • Real solutions best approximate the behavior of ideal solutions when the solute concentration is low and the solute and solvent have similar types of intermolecular forces and molecular sizes. • Deviations from Raoult’s law occur when the interactions between the solute and solvent are extremely strong, as in hydrogen bonding, and the vapor pressure of the solution is lower than predicted. • Also, the vapor pressure of a real solution is greater than predicted when the intermolecular forces between the solute and solvent are weaker than the intermolecular forces between the solute-solute or solvent-solvent. Real vs ideal solutions

13. Colligative properties depend on the number, not the identity, of the solute particles in an ideal solution. • Boiling point elevation • A nonvolatile solute elevated the boiling point of a solvent. • Because a nonvolatile solute lowers the vapor pressure of a solution, it must be heated to a temperature higher than the boiling point of the pure solvent. • The normal boiling point is the point what the vapor pressure of liquid equals 1 atm. • The change in boiling point due to the presence of the nonvolatile solute is represented by • ΔTb = Kbm • ΔT is the difference between the boiling point of the solution and that of the pure solvent • Kb is the molal boiling point elevation constant of the solvent • m is the molality of the solute in solution. Colligative properties

14. 2.00 g of a large biomolecule was dissolved in 15.0g of carbon tetrachloride. The boiling point of this solution was 77.85*C. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling point constant is 5.03*C kg/mol, and the boiling point of pure carbon tetrachloride is 76.5*C Example 4 ΔTb = 77.85*C – 76.5*C = 1.35*C m = 1.35*C = 0.268 mol/kg (5.03*C Kg/mol) Mol of biomolecule = 0.0150 kg solvent x 0.268 mol hydrocarbon/kg solvent = 4.02 x 10-3 mol Molar mass = 2.00 g = 498 g/mol 4.02 x 10-3 mol

15. A nonvolatile solute decreases the freezing point of a solvent. ΔTf = Kfm ΔT is the difference between the boiling point of the solution and that of the pure solvent. Kf is the molal freezing point depression constant of the solvent. Freezing point depression

16. Which of the following solutions would have the largest freezing point depression? Explain your answer. • A) 0.010 m NaCl • B) 0.010 m Na3PO4 • C) 0.010 m Al2(SO4)3 • D) 0.010 m C6H12O6 Example 5 The solution with the lowest freezing point depression is 0.010 m Al2(SO4)3, the solution which yields the highest molality (or the Most) particles when dissolved. 0.010 m Al2(SO4)3 yields five particles (ions) when dissolved or the total concentration of solute particles is 0.050 m. Two particles (or 0.020 m) are present in a solution of 0.010 m NaCl. Four particles (or 0.040 m) are present in a solution of 0.010 m Na3PO4. One particle (or 0.010 m) is present in a solution of 0.010 m C6H12O6

17. Osmosis is the flow of a pure solvent into a solution through a semipermeable membrane. • Osmotic pressure is the pressure that must be applied to a solution to stop osmosis • π = MRT • Π is the osmotic pressure in atmospheres • M is the molarity of the solution • R is the gas law constant, 0.08206 L*atm/mol*K • T is the temperature in Kelvin Osmotic pressure

18. An aqueous solution of 10.00 g of catalase, an enzyme found in the liver, has a volume of 1.00 L at 27*C. The solution’s osmotic pressure at 27*C is 0.745 torr. Calculate the molar mass of the catalase. Example 6 M = π = 0.745 torr x (1 atm/760 torr) = 3.98 x 10-5 M [0.08206 x atm/mol*K] 300K 1.00 L x (3.98 x 10-5 mol/L) = 3.98 x 10-5 mol Molar mass = 10.0 g = 2.51 x 105 g/mol 3.98 x 10-5 mol

19. The end