# Chapter 13. Properties of Solutions

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1 13.4 Ways of Expressing Concentration All methods involve quantifying the amount of solute per amount of solvent (or solution). Concentration may be expressed qualitatively or quantitatively. The terms dilute and concentrated are qualitative ways to describe concentration. A dilute solution has a relatively small concentration of solute. A concentrated solution has a relatively high concentration of solute. Quantitative expressions of concentration require specific information regarding such quantities as masses, moles, or liters of the solute, solvent, or solution. The most commonly used expressions for concentration are: Mass percentage Mole fraction Molarity Molality Mass percentage, ppm, and ppb Mass percentage is one of the simplest ways to express concentration. By definition: mass of component in soln Mass % of component = 100 total mass of solution Similarly, parts per million (ppm) can be expressed as the number of mg of solute per kilogram of solution. By definition: mass of component in soln 6 Parts per million (ppm) of component = 10 total mass of solution If the density of the solution is 1g/ml, then 1 ppm = 1 mg solute per liter of solution. We can extend this again! Parts per billion (ppb) can be expressed as the number of µg of solute per kilogram of solution. By definition: mass of component in soln 9 Parts per billion (ppb) of component = 10 total mass of solution If the density of the solution is 1g/ml, then 1 ppb = 1 µg solute per liter of solution

2 Sample Exercise 13.4 (p. 542) a) A solution is made by dissolving 13.5 g glucose (C 6 H 12 O 6 ) in kg of water. What is the mass percentage of solute in this solution? (11.9%) b) A 2.5-g sample of groundwater was found to contain 5.4 µg of Zn 2+. What is the concentration of Zn 2+ in parts per million? (2.2 ppm) Practice Exercise 13.4 a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (2.91%) b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? (90.5 g NaOCl) - 2 -

3 Mole Fraction, Molarity, and Molality Common expressions of concentration are based on the number of moles of one or more components. Recall that mass can be converted to moles using the molar mass. Recall: molesof component Mole fraction of component, X = total molesof all components Note: Mole fraction has no units. Note: Mole fractions range from 0 to 1. Recall: moles of solute Molarity, M = liters of solution Note: Molarity will change with a change in temperature (as the solution volume increases or decreases). We can define molality (m), yet another concentration unit: Molality, m = moles of solute kilograms of solvent Molality does not vary with temperature. Note: converting between molarity (M) and molality (m) requires density

4 Sample Exercise 13.5 (p. 544) A solution is made by dissolving 4.35 g glucose (C 6 H 12 O 6 ) in 25.0 ml of water. Calculate the molality of glucose in the solution. (0.964 m) Practice Exercise 13.5 What is the molality of a solution made by dissolving 36.5 g of naphthalene (C 10 H 8 ) in 425 g of toluene (C 7 H 8 )? (0.670 m) Sample Exercise 13.6 (p. 544) A solution of hydrochloric acid contains 36% HCl by mass. a) Calculate the mole fraction of HCl in the solution. (0.22) b) Calculate the molality of HCl in the solution. (15 m) Practice Exercise 13.6 A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate a) the mole fraction of NaOCl in the solution and (9.00 x 10-3 ) b) the molality of NaOCl in the solution (0.505 m) - 4 -

5 Sample Exercise 13.7 (p. 546) A solution with a density of g/ml contains 5.0 g of toluene (C 7 H 8 ) and 225 g of benzene. Calculate the molarity of the solution. (0.21 M) Practice Exercise 13.7 A solution containing equal masses of glycerol (C 3 H 8 O 3 ) and water has a density of 1.10 g/ml. Calculate a) the molality of glycerol; (10.9 m) b) the mole fraction of glycerol; (0.163) c) the molarity of glycerol in the solution. (5.97 M) - 5 -

6 13.5 Colligative Properties Colligative properties depend on number of solute particles. Colligative properties are physical properties. There are four colligative properties to consider: Vapor pressure lowering (Raoult's Law). Boiling point elevation. Freezing point depression. Osmotic pressure. 1. Lowering the Vapor Pressure # particles vapor pressure e.g. pure H 2 O vs. H 2 O + NaCl (salt water) higher V.P. lower V.P. Why? 1) Na + + Cl - particles (nonvolatile solutes) get surrounded by H 2 O fewer free H 2 O particles left to vaporize at surface 2) IMFs harder to get out Figure Vapor-pressure lowering. The vapor pressure over a solution formed by a volatile solvent and a nonvolatile solute (b) is lower than that of the solvent alone (a). The extent of the decrease in the vapor pressure upon addition of the solute depends on the concentration of the solute. Raoult s law quantifies the extent to which a nonvolatile solute lowers the vapor pressure of the solvent. If P A is the vapor pressure with solute, P o A is the vapor pressure of the pure solvent, and X A is the mole fraction of A, then o P = Χ P A A A - 6 -

7 Ideal solution: one that obeys Raoult s law. Real solutions show approximately ideal behavior when: The solute concentration is low. The solute and solvent have similarly sized molecules. The solute and solvent have similar types of intermolecular attractions. Raoult s law breaks down when the solvent-solvent and solute-solute intermolecular forces are >> or << solute-solvent IMFs. NEW - optional Sample Exercise 13.8 (p. 547) Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.25 g/ml at 25 o C. Calculate the vapor pressure at 25 o C of a solution made by adding 50.0 ml of glycerin to ml of water. The vapor pressure of pure water at 25 o C is 23.8 torr. (23.2 torr) Practice Exercise 13.8 The vapor pressure of pure water at 110 o C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 o C. Assuming that Raoult s law is obeyed, what is the mole fraction of ethylene glycol in the solution? (0.290) 2. Boiling Point Elevation # particles of nonvolatile solute VP boiling point Since boiling point is when VP = atmospheric pressure, if the V.P. is lower, then more energy must be used to overcome atmospheric pressure and reach 1 atm = higher T at boiling It takes more energy to break apart solvent/solute bonds so solvent can vaporize. H 2 O: 1 mole particles = o C 1 kg H 2 O molal boiling-point-elevation constant, K b, expresses how much T b changes with molality, m: T b = K b m The boiling-point elevation is proportional to [solute particles]. A 1 m solution of NaCl is 2 m in total solute particles

8 3. Freezing Point Depression # particles freezing point When a solution freezes, crystals of almost pure solvent are formed first. Solute molecules are usually not soluble in the solid phase of the solvent. the triple point occurs at a lower T because of the lower VP for the solution. The melting-point (freezing-point) curve is a vertical line from the triple point. the solution freezes at a lower temperature ( T f ) than the pure solvent. The in freezing point ( T f ) is directly proportional to molality, i.e. [solute particles]. K f = molal freezing-point-depression constant. T f = K f m Boiling point: T b = K b m i Boiling Point/Freezing Point Calculations Change in specific molal concentration i = # particles upon boiling T for each of solution dissociation of solute solvent Freezing point: T f = K f m i Change in freezing T - 8 -

9 For H 2 O: K b = o C/m K f = 1.86 o C/m e.g. What is the boiling point of a solution that contains 1.25 mol CaCl 2 in g of water? mol CaCl 2 = 1.25 mol CaCl 2 x g = m x 3 = m g H 2 O g H 2 O 1 kg T b = o C x 2.679m = 1.37 o C o C = o C m i = # particles when CaCl 2 dissociates in H 2 O Sample Exercise 13.9 (p. 550) Automotive antifreeze consists of ethylene glycol (C 2 H 6 O 2 ), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water. (102.7 o C; o C) Practice Exercise 13.9 Calculate the freezing point of a solution containing kg of chloroform (CHCl 3 ) and 42.0 g of eucalyptol (C 10 H 18 O), a fragrant substance found in the leaves of eucalyptus trees. (See Table 13.4) (-65.6 o C) - 9 -

10 Sample Exercise (p. 551) List the following aqueous solutions in order of their expected freezing points: m CaCl 2 ; 0.15 m HCl; m HC 2 H 3 O 2 ; 0.10 m C 12 H 22 O 11. Practice Exercise Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO 3 ) 2, 2 mol of KCl, 3 mol of ethylene glycol (C 2 H 6 O 2 )? (2 mol KCl) Osmosis Osmosis = the net movement of a solvent from an area of low [solute] to an area of high [solute] across a semipermeable membrane, e.g. cell membranes and cellophane

11 Osmotic pressure, π, is the pressure required to prevent osmosis. Osmotic pressure obeys a law similar in form to the ideal-gas law. For n moles, V= volume, M= molarity, R= the ideal gas constant, and an absolute temperature, T, the osmotic pressure is: πv = nrt π = n V RT = MRT Two solutions are said to be isotonic if they have the same osmotic pressure. Hypotonic solutions have a lower π, relative to a more concentrated solution. Hypertonic solutions have a higher π, relative to a more dilute solution. Sample Exercise (p. 553) The average osmotic pressure of blood is 7.7 atm at 25 o C. What concentration of glucose (C 6 H 12 O 6 ) will be isotonic with blood? (0.31 M) Practice Exercise What is the osmotic pressure at 20 o C of a M sucrose (C 12 H 22 O 11 ) solution? (0.048 atm or 37 torr)

12 Determination of Molar Mass Any of the four colligative properties may be used to determine molar mass. Sample Exercise (p. 555) A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving g of the substance in 40.0 g of CCl 4. The boiling point of the resultant solution was o C higher than that of the pure solvent. Calculate the molar mass of the solute. (88.0 g/mol) Practice Exercise Camphor (C 10 H 16 O) melts at o C, and it has a particularly large freezing-point-depression constant, K f = 40.0 o C/m. When g of an organic substance of unknown molar mass is dissolved in g of liquid camphor, the freezing point of the mixture is found to be o C. What is the molar mass of the solute? (110 g/mol) Sample Exercise (p. 556) The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine its molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 ml of solution. The osmotic pressure of the solution at 25 o C was found to be 1.54 torr. Calculate the molar mass of the protein. (8.45 x 10 3 g/mol)

13 Practice Exercise A sample of 2.05 g of the plastic polystyrene was dissolved in enough toluene to form L of solution. The osmotic pressure of this solution was found to be 1.21 kpa at 25 o C. Calculate the molar mass of the polystyrene. (4.20 x 10 4 g/mol) 13.6 Colloids Colloids or colloidal dispersions are suspensions in which the suspended particles are larger than molecules but too small to separate out of the suspension due to gravity. Particle size: 5 to 1000 nm. Tyndall effect: ability of colloidal particles to scatter light. The path of a beam of light projected through a colloidal suspension can be seen through the suspension

14 Sample Integrative Exercise 13 (p. 560) A L solution is made by dissolving g of CaCl 2(s) in water. a) Calculate the osmotic pressure of this solution at 27 o C, assuming that it is completely dissociated into its component ions. (2.93 atm) b) The measured osmotic pressure of this solution is 2.56 atm at 27 o C. Explain why it is less than the value calculated in (a), and calculate the van t Hoff factor, i, for the solute in this solution. (2.62) c) The enthalpy of solution for CaCl 2 is H = kj/mol. If the final temperature of the solution was 27.0 o C, what was its initial temperature? (Assume that the density of the solution is 1.00 g/ml, that its specific heat is 4.18 J/g-K, and that the solution loses no heat to its surroundings.) (26.2 o C) (don t worry if you have trouble with this one we will work on thermodynamics next semester)

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