1. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 1 Chapter Twelve Physical Properties of Solutions

2. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 2 Solution: Solute dispersed in a solvent. Some Types of Solutions

3. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 3 Molarity: moles of solute/liter of solution Molality: moles of solute/kg of solvent Percent by mass: grams of solute/grams of solution (then multiplied by 100%) Percent by volume: milliliters of solute/milliliters of solution (then multiplied by 100%) Mass/volume percent: grams of solute/milliliters of solution (then multiplied by 100%) Most concentration units are expressed as: Solution Concentration Amount of solvent or solution Amount of solute

4. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 4 Example 12.1 How would you prepare 750 g of an aqueous solution that is 2.5% NaOH by mass? Example 12.2 At 20°C, pure ethanol has a density of 0.789 g/mL and USP ethanol has a density of 0.813 g/mL. What is the mass percent ethanol in USP ethanol?

5. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 5 Example 12.1 How would you prepare 750 g of an aqueous solution that is 2.5% NaOH by mass? ? g NaOH = 750 g solution x 2.5% = 19 g NaOH ? g H 2 O = 750 g solution – 19 g NaOH = 731 g H 2 O Example 12.2 At 20 °C, pure ethanol has a density of 0.789 g/mL and USP ethanol (= 95% ethanol by volume) has a density of 0.813 g/mL. What is the mass percent ethanol in USP ethanol? Assume you a working with 100 ml sample of the USP ethanol, then: 95.0 ml ethanol x 78.9 g ethanol= 75.0 g ethanol 100 ml solution Mass percent ethanol = 75.0 g ethanolx 100 % = 92.3 % 81.3 g solution

6. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 6 Parts per million (ppm): grams of solute/grams of solution (then multiplied by 10 6 or 1 million) Parts per billion (ppb): grams of solute/grams of solution (then multiplied by 10 9 or 1 billion) Parts per trillion (ppt): grams of solute/grams of solution (then multiplied by 10 12 or 1 trillion) Amount of solution Amount of solute ppm, ppb, ppt ordinarily are used when expressing extremely low concentrations (a liter of water that is 1 ppm fluoride contains only 0.001 g F – !) Solution Concentration (cont’d) Most concentration units are expressed as:

7. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 7 PPM and PPB are Tiny Amounts......................................................................................................................................................

8. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 8 Example 12.3 The maximum allowable level of nitrates in drinking water in the United States is 45 mg NO 3 – /L. What is this level expressed in parts per million (ppm)?

9. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 9 Example 12.3 The maximum allowable level of nitrates in drinking water in the United States is 45 mg NO 3 – /L. What is this level expressed in parts per million (ppm)? The density of water with only small amounts of dissolved substances is essentially 1.0 g/ml. So one liter of dilute aqueous nitrate solution has a mass of 1000 g. 45 mg NO 3 – 1 g water45 mg NO 3 – 1000 g water1000 mg1,000,000 mg water X = 45 ppm NO 3 – =

10. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 10 Molality (m): moles of solute/kilograms of solvent. Molarity varies with temperature (expansion or contraction of solution). Molality is based on mass of solvent (not solution!) and is independent of temperature. We use molality in describing certain properties of solutions when temperature changes are occurring. Solution Concentration (cont’d) Most concentration units are expressed as: Amount of solvent or solution Amount of solute

11. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 11 Example 12.4 What is the molality of a solution prepared by dissolving 5.05 g naphthalene [C 10 H 8 (s)] in 75.0 mL of benzene, C 6 H 6 (d = 0.879 g/mL)? Example 12.5 How many grams of benzoic acid, C 6 H 5 COOH, must be dissolved in 50.0 mL of benzene, C 6 H 6 (d = 0.879 g/mL), to produce 0.150 m C 6 H 5 COOH?

12. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 12 Mole fraction (x i ): moles of component i per moles of all components (the solution). The sum of the mole fractions of all components of a solution is ____. Mole percent: mole fraction times 100%. Concentration expressed as: Solution Concentration (cont’d) Amount of solvent or solution Amount of solute

13. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 13 Example 12.6 An aqueous solution of ethylene glycol HOCH 2 CH 2 OH used as an automobile engine coolant is 40.0% HOCH 2 CH 2 OH by mass and has a density of 1.05 g/mL. What are the (a) molarity, (b) molality, and (c) mole fraction of HOCH 2 CH 2 OH in this solution? Example 12.7 An Estimation Example Without doing detailed calculations, determine which aqueous solution has the greatest mole fraction of CH 3 OH: (a) 1.0 m CH 3 OH, (b)10.0% CH 3 OH by mass, or (c) x CH3OH = 0.10.

14. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 14 Solution formation can be considered to take place in 3 steps: 1.Move solvent molecules apart (breaking solvent-solvent bonds) to make room for the solute molecules.  H 1 > 0 (endothermic) 2.Separate the molecules/ions of solute (breaking solute- solute bonds).  H 2 > 0 (endothermic) 3.Allow the separated solute and solvent molecules to mix randomly (form solute-solvent bonds).  H 3 < 0 (exothermic)  H soln =  H 1 +  H 2 +  H 3 Enthalpy of Solution

15. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 15 For dissolving to occur, the magnitudes of  H 1 +  H 2 and of  H 3 must be roughly comparable. Visualizing Enthalpy of Solution

16. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 16 An ideal solution exists when all intermolecular forces are of comparable strength,  H soln = 0. When solute–solvent intermolecular forces are somewhat stronger than other intermolecular forces,  H soln < 0. When solute–solvent intermolecular forces are somewhat weaker than other intermolecular forces,  H soln > 0. When solute–solvent intermolecular forces are much weaker than other intermolecular forces, the solute does not dissolve in the solvent. –Energy released by solute–solvent interactions is insufficient to separate solute particles or solvent particles. Intermolecular Forces in Solution Formation

17. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 17 Intermolecular Forces in Solution For a solute to dissolve, solvent– solvent forces … … and solute– solute forces … … must be comparable to solute– solvent forces.

18. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 18 Non-Ideal Solutions 50 mL of ethanol … … and 50 mL of water … … when mixed, give less than 100 mL of solution. In this solution, forces between ethanol and water are > / < / = other intermolecular forces.

19. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 19 The forces causing an ionic solid to dissolve in water are ion–dipole forces, the attraction of water dipoles for cations and anions. The attractions of water dipoles for ions pulls the ions out of the crystalline lattice and into aqueous solution. The extent to which an ionic solid dissolves in water is determined largely by the competition between: –interionic attractions that hold ions in a crystal, and –ion–dipole attractions that pull ions into solution. Aqueous Solutions of Ionic Compounds

20. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 20 Ion–Dipole Forces in Dissolution Positive ends of dipoles attracted to anions. Negative ends of dipoles attracted to cations.

21. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 21 Example 12.8 Predict whether each combination is likely to be a solution or a heterogeneous mixture: (a) methanol, CH 3 OH, and water, HOH (b) pentane, CH 3 (CH 2 ) 3 CH 3, and octane, CH 3 (CH 2 ) 6 CH 3 (c) sodium chloride, NaCl, and carbon tetrachloride, CCl 4 (d) 1-decanol, CH 3 (CH 2 ) 8 CH 2 OH, and water, HOH Remember rule of thumb: “like dissolved like”

22. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 22 Liquids that mix in all proportions are called miscible. –Ones that don’t = immiscible When there is a dynamic equilibrium between an undissolved solute and a solution, the solution is saturated. The concentration of the solute in a saturated solution is the solubility of the solute. A solution which contains less solute than can be held at equilibrium is unsaturated. Some Solubility Terms

23. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 23 Formation of a Saturated Solution Solid begins to dissolve. As solid dissolves, some dissolved solute begins to crystallize. Eventually, the rates of dissolving and of crystallization are equal; no more solute appears to dissolve. Longer standing does not change the amount of dissolved solute.

24. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 24 Most ionic compounds have aqueous solubilities that increase significantly with increasing temperature A few have solubilities that change little with temperature A very few have solubilities that decrease with increasing temperature (see following slide) If solubility increases with temperature, a hot, saturated solution may be cooled carefully without precipitation of the excess solute. This creates a supersaturated solution. Supersaturated solutions ordinarily are unstable … Solubility as Function of Temperature

25. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 25 Some Solubility Curves What is the (approx.) solubility of KNO 3 per 100 g water at 90 °C? At 20 °C?

26. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 26 A single “seed crystal” of solute is added. Solute immediately begins to crystallize … … until all of the excess solute has precipitated. A Supersaturated Solution

27. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 27 Most gases become less soluble in liquids as the temperature increases. (Explanation**)Explanation At a constant temperature, the solubility (S) of a gas is directly proportional to the pressure of the gas (P gas ) in equilibrium with the solution. S = k P gas The value of k depends on the particular gas and the solvent. The effect of pressure on the solubility of a gas is known as Henry’s law. ** http://antoine.frostburg.edu/chem/senese/101/solutions/faq/print-temperature-gas- solubility.shtml The Solubilities of Gases

28. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 28 Effect of Temperature on Solubility of Gases Eco-Connection Thermal pollution: as river/lake water is warmed (when used by industry for cooling), less oxygen dissolves, and fish no longer thrive.

29. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 29 Pressure and Solubility of Gases (Henry’s Law) … thus more frequent collisions of gas molecules with the surface … … giving a higher concentration of dissolved gas. Higher partial pressure means more molecules of gas per unit volume …

30. Example 12.9 A 225-g sample of pure water is shaken with air under a pressure of 0.95 atm at 20°C. How many milligrams of Ar(g) will be present in the water when solubility equilibrium is reached? Use data from Figure 12.14 and the fact that the mole fraction of Ar in air is 0.00934. According to the graph, argon has a solubility in water of 6 mg/atm100 g H 2 O. 225 g H 2 O x 0089 atm x 6 mg/atm100 g H 2 O = 0.12 mg Ar

31. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 31 Colligative properties of a solution depend only on the concentration of solute particles, and not on the nature of the solute. Non-colligative properties include: color, odor, density, viscosity, toxicity, reactivity, etc. We will examine four colligative properties of solutions: –Vapor pressure (of the solvent) –Freezing point depression –Boiling point elevation –Osmotic pressure Colligative Properties of Solutions

32. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 32 Review: Vapor Pressure The pressure of the vapor phase of a substance above a liquid sample of the substance after a dynamic equilibrium has been established in a closed system.

33. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 33 The vapor pressure of solvent above a solution is less than the vapor pressure above the pure solvent. Raoult’s law: the vapor pressure of the solvent above a solution (P solv ) is the product of the vapor pressure of the pure solvent (P° solv ) and the mole fraction of the solvent in the solution (x solv ): P solv = x solv ·P° solv The vapor in equilibrium with an ideal solution of two volatile components has a higher mole fraction of the more volatile component than is found in the liquid. Vapor Pressure of a Solution

34. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 34 Example 12.10 The vapor pressure of pure water at 20.0 °C is 17.5 mmHg. What is the vapor pressure at 20.0°C above a solution that has 0.250 mol sucrose (C 12 H 22 O 11 ) and 75.0 g urea [CO(NH 2 ) 2 ] dissolved per kilogram of water? Moles of urea = 75.0 g urea x 1 mol urea / 60 g urea = 1.25 mol urea Moles of solute =.250 mole sucrose + 1.25 mole urea = 1.50 mole solute Moles solvent = 1000 g H 2 0 x 1 mole H 2 0 / 18.0 g H 2 0 = 55.6 moles Mole fraction of solvent = 55.6 moles solvent/57.1 moles solution = 0.974 P solv = x solv ·P° solv = 0.974 · 17.5 mmHg P solv = 17.0 mmHg

35. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 35 Example 12.11 At 25°C, the vapor pressures of pure benzene (C 6 H 6 ) and pure toluene (C 7 H 8 ) are 95.1 and 28.4 mmHg, respectively. A solution is prepared that has equal mole fractions of C 7 H 8 and C 6 H 6. Determine the vapor pressures of C 7 H 8 and C 6 H 6 and the total vapor pressure above this solution. Consider the solution to be ideal. Since there are equal mole fractions, the vapor pressure of each substance will be ½ of the vapor pressure of the pure substance. The total vapor pressure will be the sum of the partial pressures of the two vapors (Dalton’s law of partial pressures). Example 12.12 What is the composition, expressed as mole fractions, of the vapor in equilibrium with the benzene–toluene solution of Example 12.11? At constant temperature and number of moles, the pressures of gases are proportional to the molar amounts. So we can use the partial pressure values like moles to produce mole fractions of the two gases.

36. Fractional Distillation The vapor here … … is richer in the more volatile component than the original liquid here … … so the liquid that condenses here will also be richer in the more volatile component.

37. Fractional Distillation of Crude Oil Fractional distillation on a very large scale

38. Example 12.13 A Conceptual Example Figure 12.16 (below) shows two different aqueous solutions placed in the same enclosure. After a time, the solution level has risen in container A and dropped in container B. Explain how and why this happens.

39. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 39 Vapor Pressure Lowering by a Nonvolatile Solute Raoult’s Law: the vapor pressure from a solution (nonvolatile solute) is lower than … … the vapor pressure from the pure solvent Result: the boiling point of the solution increases by  T b.

40. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 40  T f = –K f × m × i  T b = K b × m × i Freezing Point Depression and Boiling Point Elevation

41. Example 12.14 What is the freezing point of an aqueous sucrose solution that has 25.0 g C 12 H 22 O 11 per 100.0 g H 2 O? 25.0 g x 1 mol sucrose / 342 g = 0.0731 mol C 12 H 22 O 11 `  T f = −1.86 ºC/m x m x 1 m = 0.0731 mol / 0.100 kg = 0.731 m  T f = − 1.86 ∙ 0.731 m ∙ 1  T f = −1.36ºCSo, the freezing point is −1.36ºC Example 12.15 Sorbitol is a sweet substance found in fruits and berries and sometimes used as a sugar substitute. An aqueous solution containing 1.00 g sorbitol in 100.0 g water is found to have a freezing point of –0.102 °C. Elemental analysis indicates that sorbitol consists of 39.56% C, 7.75% H, and 52.70% O by mass. What are the (a) molar mass and (b) molecular formula of sorbitol? Use freezing point depression equation to calculate molality of solution With m, moles of solute known. Molar mass = g/mol, and you’re given 1.00 g of the solute, so 1.00 g sorbitol/moles from the calculation gives molar mass.  T f = –K f × m × i  T b = K b × m × i

42. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 42 A semipermeable membrane has microscopic pores, through which small solvent molecules can pass but larger solute molecules cannot. During osmosis, there is a net flow of solvent molecules through a semipermeable membrane, from a region of lower concentration to a region of higher concentration. The pressure required to stop osmosis is called the osmotic pressure (  of the solution.  = (nRT/V) = (n/V)RT = M RT Osmotic Pressure This equation should look familiar …

43. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 43 Osmosis and Osmotic Pressure Net flow of water from the outside (pure H 2 O) to the solution. The solution increases in volume until … … the height of solution exerts the osmotic pressure (π) of the solution.

44. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 44 Example 12.16 An aqueous solution is prepared by dissolving 1.50 g of hemocyanin, a protein obtained from crabs, in 0.250 L of water. The solution has an osmotic pressure of 0.00342 atm at 277 K. (a) What is the molar mass of hemocyanin? (b) What should the freezing point of the solution be?

45. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 45 Ordinarily a patient must be given intravenous fluids that are isotonic—have the same osmotic pressure as blood. Practical Applications of Osmosis External solution is hypertonic; produces osmotic pressure > π int. Net flow of water out of the cell. External solution is hypotonic; produces osmotic pressure < π int. Net flow of water into the cell. Red blood cell in isotonic solution remains the same size.

46. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 46 Reverse osmosis (RO): reversing the normal net flow of solvent molecules through a semipermeable membrane. Pressure that exceeds the osmotic pressure is applied to the solution. RO is used for water purification. Pressure greater than π is applied here … … water flows from the more concentrated solution, through the membrane. Practical Applications of Osmosis (cont’d)

47. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 47 Whereas electrolytes dissociate, the number of solute particles ordinarily is greater than the number of formula units dissolved. One mole of NaCl dissolved in water produces more than one mole of solute particles. The van’t Hoff factor (i) is used to modify the colligative- property equations for electrolytes:  T f = i × (–K f ) × m  T b = i × K b × m  = i × M RT For nonelectrolyte solutes, i = 1. For electrolytes, we expect i to be equal to the number of ions into which a substance dissociates into in solution. Solutions of Electrolytes

48. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 48 At very low concentrations, the “theoretical” values of i are reached. At higher concentrations, the values of i are significantly lower than the theoretical values; ion pairs form in solution.

49. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 49 Example 12.17 A Conceptual Example Without doing detailed calculations, place the following solutions in order of decreasing osmotic pressure: (a)0.01 M C 12 H 22 O 11 (aq) at 25 °C (b)0.01 M CH 3 CH 2 COOH(aq) at 37 °C (c)0.01 m KNO 3 (aq) at 25 °C (d)a solution of 1.00 g polystyrene (molar mass: 3.5 × 10 5 g/mol) in 100 mL of benzene at 25 °C

50. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 50 In a solution, dispersed particles are molecules, atoms, or ions (roughly 0.1 nm in size). Solute particles do not “settle out” of solution. In a suspension (e.g., sand in water) the dispersed particles are relatively large, and will settle from suspension. In a colloid, the dispersed particles are on the order of 1–1000 nm in size. Although they are larger than molecules/atoms/ions, colloidal particles are small enough to remain dispersed indefinitely. Colloids

51. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 51 Why are there no gas- in-gas colloids?

52. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 52 The Tyndall Effect The dissolved Fe 3+ ions are not large enough to scatter light; the beam is virtually invisible. Light scattered by the (larger) colloidal particles of Fe 2 O 3 makes the beam visible.

53. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 53 Formation and Coagulation of a Colloid When a strong electrolyte is added to colloidal iron oxide, the charge on the surface of each particle is partially neutralized … … and the colloidal particles coalesce into a suspension that quickly settles.

54. Chapter Twelve General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 54 Cumulative Example A 375-mL sample of hexane vapor in equilibrium with liquid hexane C 6 H 14 (d = 0.6548 g/mL), at 25.0 °C is dissolved in 50.0 mL of liquid cyclohexane, C 6 H 12, at 25.0 °C (d = 0.7739 g/mL, vp = 97.58 Torr). Use information found elsewhere in the text (such as Example 11.3) to calculate the total vapor pressure above the solution at 25.0 °C. How reliable is this calculation?