SGCh12 - Chapter Twelve PHYSICAL PROPERTIES OF SOLUTIONS...

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Unformatted text preview: Chapter Twelve PHYSICAL PROPERTIES OF SOLUTIONS • • • • The Solution Process Concentration Units Temperature and Pressure Effects on Solubility Colligative Properties THE SOLUTION PROCESS STUDY OBJECTIVES 1. 2. 3. Describe the role of intermolecular forces in the solution process. Given a particular substance, predict what solvent will be suitable. Describe the exothermic and endothermic steps that occur during the solution process. Some Definitions. The main subject of this chapter concerns the formation and properties of liquid solutions. Recall from Chapter 4 that a s olution is a homogeneous mixture of two or more substances. The component in greater quantity is called the s olvent, and the component in lesser amount is called the s olute . In aqueous solutions water is the solvent. Solutes can be liquids, solids, or gases. Several terms are used to describe the degree to which a solute will dissolve in a solvent: 1. 2. When two liquids are completely soluble in each other in all proportions, they are said to be m iscible . For example, ethanol and water are miscible. If the liquids do not mix, they are said to be i mmiscible . Oil and water, for example, are immiscible. A solution that contains the maximum concentration of a solute at a given temperature is called a saturated solution . For example, the solubility of NaCl is 6.1 moles per liter of water. An unsaturated solution has a concentration of solute that is less than the maximum concentration. Supersaturated solutions have a concentration of solute that is greater than that of a saturated solution. The Solution Process. Dissolving is a process that takes place at the molecular level and can be discussed in molecular terms. When one substance dissolves in another, the particles of the solute disperse uniformly throughout the solvent. The solute particles occupy positions that are normally taken by solvent molecules. The ease with which a solute particle may replace a solvent molecule depends on the relative strengths of three types of interactions: • Solvent-solvent interaction • Solute-solute interaction • Solvent-solute interaction Imagine the solution process as taking place in three steps as shown in Figure 12.1. Step 1 is the separation of solvent molecules. Step 2 is the separation of solute molecules. These steps require inputs of energy to overcome attractive intermolecular forces. Step 3 is the mixing of solvent and solute molecules; it may be exothermic or endothermic. According to Hess's law (see Section 6.6 of the text) the heat of solution is given by the sum of the enthalpies of the three steps: 2 32 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Physical Properties of Solutions / 2 33 ∆ Hsoln = ∆ H1 + ∆ H2 + ∆ H3 The solute will be soluble in the solvent if the solute-solvent attraction is stronger than the solvent-solvent attraction and solute-solute attraction. Such a solution process is exothermic. Only a relatively small amount of the solute will be dissolved if the solute-solvent interaction is weaker than the solvent-solvent and solute-solute interaction; then the solution process will be endothermic. S olvation is the process in which a solute particle (an ion or molecule) is surrounded by solvent molecules due to strong solute-solvent attractive forces. When water is the solvent the process is called hydration. Step 1 Step 2 ∆ H1 ∆H2 Solute Solvent Step 3 ∆H3 Solution Figure 12.1. A molecular view of the solution process. Think of the solute molecules and solvent molecules first being spread apart, and then being mixed together. The relative strength of forces holding solvent molecules together ∆ H1 , solute particles together ∆ H2 , and the forces between solvent and solute molecules ∆ H3 in the solution are important in determining the solubility. Here we see that the solution process is assisted by an exothermic heat of solution. Yet there are a number of soluble compounds with endothermic heats of solution. In addition every substance is somewhat soluble no matter what the value of ∆ Hsoln . It turns out that all chemical processes are governed by two factors. The first of these is the energy factor. In other words, does the solution process absorb energy or release energy? Disorder or randomness is the other factor that must be considered. Processes that increase randomness or disorder are also favored. In the pure state, the solvent and solute possess a fair degree of order. Here we mean the ordered arrangement of atoms, molecules, or ions in a three-dimensional crystal. Where order is high, randomness is low. The order in a crystal is lost when the solute dissolves and its molecules are dispersed in the solvent. The solution process is accompanied by an increase in disorder or randomness. It is the increase in disorder of the system that favors the solubility of any substance. A General Solubility Rule. The most general solubility rule is that "like dissolves like." In this rule, the term "like" refers to polarity. "Like dissolves like" means that substances of like polarity will mix to form solutions. And substances of different polarity will be immiscible, or will tend only slightly to form solutions. The rule predicts that two polar substances will form solutions and two nonpolar substances will form solutions, but a polar substance and a nonpolar substance will tend not to mix. Thus, water and oil, being polar and nonpolar substances, respectively, are immiscible. Water and ethanol, both being polar, are miscible. Oil will dissolve in carbon tetrachloride because both substances are nonpolar. Ionic compounds tend to be quite soluble in water (a polar solvent). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 34 / Physical Properties of Solutions _______________________________________________________________________________ EXAMPLE 12.1 Solubility Which would be a better solvent for molecular I2 (s): CCl4 or H 2 O? •Method of Solution Using the like-dissolves-like rule, first we identify I2 as a nonpolar molecule. Therefore it will be more soluble in the nonpolar solvent CCl4 than in the polar solvent H2 O. _______________________________________________________________________________ EXAMPLE 12.2 Solubility In which solvent will NaBr be more soluble, benzene or water? •Method of Solution Benzene is a nonpolar solvent and water is a polar solvent. Since NaBr is an ionic compound, it cannot dissolve in nonpolar benzene, but is very soluble in water. _______________________________________________________________________________ EXERCISES 1. 2. Explain why hexane (C6 H14 ) even though a liquid, is not miscible with water. 3. Indicate whether each compound listed is soluble or insoluble in water. a. CH 3 OH b. LiBr c. C8 H18 d . CCl 4 e. BaCl2 f. HOCH2 CH2 OH 4. Explain why ammonia gas, NH3 , is very soluble in water, but not in hexane, C6 H14 . 5. What is the difference between solvation and hydration? 6. For each of the following pairs, predict which substance will be more soluble in water. a. NaCl(s) or I2 (s) b. CH4 (g) or NH3 (g) c. CH 3 OH(l) or C6 H6 (benzene) 7. Back Isopropyl alcohol and water dissolve in each other regardless of the proportions of each. What term describes the solubilities of these liquids in each other? For each of the following pairs, predict which substance will be more soluble in CCl4 (l). a. I2 (s) or KBr(s) b. CS 2 (l) or CH 3 OH(l) c. CO 2 (g) or HCl(g) Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Physical Properties of Solutions / 2 35 CONCENTRATION UNITS STUDY OBJECTIVES 1. 2. Calculate concentrations in units of molarity, mass percent, mole fraction, and molality. Convert concentrations expressed in one unit into any of the other units. Concentration. The term concentration refers to how much of one component of a solution is present in a given amount of solution. In Chapter 4 you were introduced to the concentration unit molarity: molarity = moles of solute liters of solution Three new concentration units are introduced in this chapter. The percent by mass of solute mass of solute × 100% mass solute + mass solvent mass of solute = × 100% mass of soln Percent by mass of solute = The mole fraction of a component of the solution mole fraction of component A = X A = m oles of A sum of moles of all components The molality molality = moles of solute mass of solvent (kg) _______________________________________________________________________________ EXAMPLE 12.3 Concentration Units The dehydrated form of Epsom salt is magnesium sulfate. a. What is the percent MgSO 4 by mass in a solution made from 16.0 g MgSO4 and 100 mL of H 2 O at 25°C? The density of water at 25°C is 0.997 g/mL. b. What is the mole fraction of each component? c. Calculate the molality of the solution. •Method of Solution for (a) Write the equation for percent by mass. percent MgSO4 = m ass MgSO 4 × 100% mass MgSO 4 + m ass water •Calculation for (a) 0 .997 g = 99.7 g H2 O 1 mL 1 6.0 g 16.0 g percent MgSO4 = × 100% = × 100% 16.0 g + 99.7 g 115.7 g mass H2 O = 100 mL × percent MgSO4 = 13.8% Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 36 / Physical Properties of Solutions •Method of Solution for (b) Write the formula for the mole fraction of MgSO 4 . XMgSO4 = m ol MgSO 4 mol MgSO 4 + m o l H 2 O • Calculation for (b) Convert the masses in part (a) into moles to substitute into the equation. 16.0 g MgSO4 × 99.7 g H2 O × 1 m ol MgSO 4 = 0.133 mol MgSO4 120.4 g MgSO 4 1 m ol H 2 O = 5.54 mol H2 O 18.0 g H 2 O The mole fraction of MgSO 4 is XMgSO4 = 0 .133 mol = 0.0235 0.133 mol + 5.54 mol The mole fraction of H2 O is XH2 O = 5 .54 mol = 0.977 5.67 mol Notice that the sum of the two mole fractions is 1.000. The sum of the mole fractions of all solution components is always 1.00. •Method of Solution for (c) Write the formula for molality: molality = moles of MgSO 4 kilograms of H 2 O Substitute into this equation the quantities previously calculated in parts (a) and (b). •Calculation for (c) 0.133 mol MgSO 4 103 g × = 1.33 m 99.7 g 1 kg _______________________________________________________________________________ EXAMPLE 12.4 Molality and Molarity of the Same Solution molality = Concentrated hydrochloric acid is 36.5 percent HCl by mass. Its density is 1.18 g/mL. Calculate: a. The molality. b. The molarity of HCl. •Method of Solution for (a) Write the formula for molality. molality = mol HCl kg H 2 O Next try to find the number of moles of HCl per kilogram of H2 O. We take 100 g of solution, and determine how many moles of HCl and how many kilograms of the solvent it contains. A solution that is 36.5 percent Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Physical Properties of Solutions / 2 37 HCl by mass corresponds to 36.5 g HCl/100 g solution. Since 100 g of solution contains 36.5 g HCl, the difference 100 g – 36.5 g must equal the mass of water which is 63.5 g H2 O. We have two ratios: 36.5 g HCl 100 g soln and 36.5 g HCl 63.5 g H 2 O • Calculation for (a) The moles of HCl is given by: moles HCl = 36.5 g HCl × 1 m ol HCl = 1.00 mol HCl 36.5 g HCl The kilograms of water is given by: kg H2 O = 63.5 g H2 O × 1 kg = 0.0635 kg H2 O 103 g Then we calculate molality: molality = 1.00 mol HCl = 15.7 m 0.0635 kg H 2 O •Method of Solution for (b). Write the formula for molarity: molarity = moles HCl liters soln Take 100 g of solution and determine how many moles of HCl and how many liters of solution are present. 36.5 g HCl 100 g soln • Calculation for (b). We know that 100 g of solution contains 36.5 g HCl, which is 1.00 mol HCl. Convert 100 g of solution to volume of solution by using the density. Volume of soln = (100 g soln) × 1 mL 10–3 L × = 0.0847 L 1.18 g soln 1 mL 1.00 mol HCl = 11.8 M 0.0847 L soln _______________________________________________________________________________ molarity = EXERCISES 8. Calculate the percent by mass of the solute in each of the following aqueous solutions. a. 6.50 g NaCl in 75.2 g of water. b. 27.2 g ethanol in 250 g of solution. c. 2.0 g I2 in 125 g methanol. 9. Calculate the molality of each of the following solutions. a. 6.50 g NaCl in 75.2 g of water. b. 27.5 g glucose (C6 H12 O6 ) in 425 g of water. 10. Calculate the molarity of a 2.44 m NaCl solution given that its density is 1.089 g/mL. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 38 / Physical Properties of Solutions 11. Calculate the percent AgNO3 by mass in a 0.650 m AgNO3 solution. 12. Calculate the molality of a 5.5% AgNO3 solution. TEMPERATURE AND PRESSURE EFFECTS ON SOLUBILITY STUDY OBJECTIVES 1. 2. Discuss the effects of temperature on the solubility of salts in water. Discuss the effects of temperature and pressure on the solubilities of gases. Temperature Effect on the Solubility of Solids. Temperature changes strongly affect the solublilty of most solid solutes. The effects of temperature on the solubilities of some common salts is shown in Figure 12.3 (text). In most cases the solubility of a solid in water increases with increasing temperature. However, this is not always true. Table 12.1 list several solids whose solubility decreases with increasing temperature. In general, it is not practical to predict just how the temperature will affect the solubility of a given substance. Temperature effects must be determined experimentally. TABLE 12.1 Compounds Whose Solubility Decreases with Increasing Temperature ___________________________________ CaSO4 Ca(OH)2 Na2 SO4 . 10H2 O Ca(C2 H3 O2 )2 Ce2 (SO4 )3 Li2 SO4 ____________________________________ Pressure and the Solubility of Gases. The solubility of gases in liquids is directly proportional to the gas pressure. H enry's law relates gas concentration c, in moles per liter, to the gas pressure p in atmospheres. c = kP where k is a constant for each gas and has units of mol/(L . atm). The greater the value of k, the greater the solubility of the gas. A few of Henry's law constants for gases dissolved in water at 25°C are listed in Table 12.2 TABLE 12.2 Some Henry's Law Constants ____________________________ Gas k(mol/L. atm) ____________________________ O2 1.28 × 10 –3 N2 6.5 × 10 –4 CO2 3.38 × 10 –2 ___________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Physical Properties of Solutions / 2 39 Gas Solubility and Temperature. A common observation is that a glass of cold water when warmed to room temperature shows the formation of many small air bubbles. This phenomenon results in part from the decreased solubility of gases in water with increasing temperature. The solubility of an unreactive gaseous solute is due to intermolecular attractive forces between solute molecules and solvent molecules. As the temperature of such a solution is increased, more solute molecules attain sufficient kinetic energy to break away from these attractive forces and enter the gas phase, and so the solubility decreases. _______________________________________________________________________________ EXAMPLE 12.5 Henry's Law What is the concentration of O2 at 25°C in water that is saturated with air at an atmospheric pressure of 645 mm Hg? Assume the mole fraction of oxygen in air is 0.209. •Method of Solution Henry's law states that the concentration of dissolved O2 (C O2 ) is proportional to its partial pressure (PO2 ) in atm. C O2 = k PO2 The partial pressure of O2 in air is found by using Dalton's law of partial pressures. 1 a tm P O2 = XO2 P T = 0.209 (645 mm Hg) × 760 mm Hg = 0.177 atm Taking k from Table 12.2, the concentration of dissolved oxygen is: C O2 = kP O2 = (1.28 × 10 –3 mol/L·atm) (0.177 atm) = 2.27 × 10 –4 mol/L _______________________________________________________________________________ EXERCISES 13. The solubility of KNO3 at 70°C is 135 g per 100 g of water. At 10°C the solubility is 21 g per 100 g of water. What mass of KNO3 will crystallize out of solution if exactly 100 g of its saturated solution at 70°C is cooled to 25°C? 14. As temperature increases the solubility of all gases in water increases / decreases. 15. The Henry's law constant for argon is 1.5 × 10 –3 mol/L·atm at 20°C. Calculate its solubility in water at 20°C and 7.6 mm Hg. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 40 / Physical Properties of Solutions COLLIGATIVE PROPERTIES STUDY OBJECTIVES 1. 2. 3. Calculate for a given solution the magnitude of the colligative properties: vapor-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure. Determine the molar mass of a solute from freezing-point depression and osmotic pressure measurements. Calculate the van't Hoff factor (i) for electrolyte solutions Colligative Properties. The properties of solutions that depend on the number of solute particles in solution are called c olligative properties . The four colligative properties of interest here are: 1. The v apor-pressure lowering ∆ P is given by o ∆ P = X2 P 1 o where X2 is the mole fraction of the solute, and P1 is the vapor pressure of the pure solvent. Recall that the vapor pressure of a liquid is the pressure exerted by a vapor in equilibrium with its liquid phase. For a solution containing a nonvolatile solute the vapor pressure due to the solvent is less than it is for the pure solvent. The amount of lowering of the vapor pressure can be seen to depend on X2 , the mole fraction of the solute. 2. The b oiling-point elevation . The boiling point of a solution is higher than that of pure solvent because the vapor pressure of a solution is always less than the vapor pressure of pure solvent. Thus, a solution must be hotter than a pure solvent, if both vapor pressures are to be 1 atm. Figure 12.11 in the text shows the effect that lowering the vapor pressure has on the boiling point. The boiling-point elevation ∆ Tb of a solution of molality m is given by ∆ Tb = Kb m where Kb is a constant called the molal boiling-point elevation constant and has the units °C/m. There is a Kb for each solvent, and Table 12.1 (text) lists Kb values for six solvents. The magnitude of ∆ Tb , the increase in boiling point of the solution over the boiling point of the pure solvent, is proportional to the solute concentration m. The preceding equation is independent of the solute, but requires that the solute be nonvolatile. 3. The f reezing-point depression ∆ Tf is given by: ∆ Tf = Kf m where Kf is the molal freezing-point depression constant and has the units °C/m. As is the case of Kb , K f is different for each solvent, and its value must be found in a table such as Table 12.2 (textbook). The equation applies to all solutes, even volatile ones, because freezing points are rather insensitive to vapor pressure. 4. The o smotic pressure π is given by: π = MRT Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Physical Properties of Solutions / 2 41 where M is the molar concentration, R is the ideal gas constant (0.0821 L. atm/K. mol), and T is the absolute temperature. Osmosis is the net flow of solvent molecules through a semipermeable membrane from a more dilute solution to a more concentrated solution. Note that in this process the solvent flows from a region of greater solvent concentration to a region of lesser solvent concentration. The osmotic pressure of a solution is the pressure required to stop osmosis. Molar Mass Determination. Colligative properties provide a means to determine the molar mass of the solute. For instance, the freezing-point depression equation is ∆ Tf = Kf m = K f × m ol solute kg solvent Since ∆ Tf can be measured and Kf is known, then the molality, m, of the solution can be calculated. Recall that molality is molality = mol solute m ass solute/molar mass of solute = kg solvent kg solvent When a known mass of solute is dissolved in a known mass of solvent, and you already know the molality of the solution, this leaves the molar mass of the solute as the only unknown quantity. Freezing-point depression is more sensitive to the number of moles of solute than the boiling-point elevation because for the same solvent Kf is larger then K b . The most sensitive method for the determination of the molar mass of a solute is to measure the osmotic pressure. The equation for osmotic pressure then becomes useful in the following form: π = MRT = (g solute/molar mass of solute)RT L soln Colligative Properties of Electrolyte Solutions. For the same solute concentrations solutions of electrolytes always have greater colligative properties than those of nonelectrolyte solutions. For a 0.1 m methanol solution, ∆ Tf is the same as the calculated value. For 0.1 m NaCl, ∆ Tf is 1.9 times greater than calculated. And for Na2 SO4 , ∆ Tf is 2.7 times greater than the calculated value. Table 12.3 compares values of ∆ Tf(obs)/∆ Tf(calc)for several solutes. For solutions of electrolytes the ratio is always greater than one. Since colligative properties depend on the number of solute particles, the most likely explanation of this behavior is that NaCl and Na2 SO4 dissociate into ions. This gives a greater concentration of solute particles. Therefore, the effective concentration of solute particles in 0.10 m NaCl is essentially 2 × 0.10 m, and in Na2 SO4 it is almost 3 × 0.10 m. The freezing-point depression depends on the total concentration of all solute particles. Each ion acts as an independent particle. TABLE 12.3 van't Hoff Factor i for Several Solutes ______________________________________________________________ Solution ∆ Tf ∆ Tf(calc) ∆ Tf/∆ Tf(calc) ______________________________________________________________ 0.10 m CH3 OH 0.186°C 0.186°C 1.0 0.10 m NaCl 0.353°C 0.186°C 1.9 0.10 m Na2 SO4 0.502°C 0.186°C 2.7 ______________________________________________________________ The ratio ∆ Tf/∆ Tf(calc) is called the v an't Hoff factor i . For 1:1 electrolytes, such as NaCl, KCl, and MgSO 4 , i ~ 2; for 2:1 electrolytes, such as MgCl2 and K2 SO4 , i is a little less than 3. The deviation from a whole number results from the temporary formation of some ion pairs such as "NaCl" in the solution. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 42 / Physical Properties of Solutions _______________________________________________________________________________ EXAMPLE 12.6 Boiling Point Elevation What is the boiling point of an "antifreeze/coolant" solution made from a 50-50 mixture (by volume) of ethylene glycol, C 2 H6 O2 (density 1.11 g/mL), and water? •Method of Solution The boiling point depends on the molality of the 50-50 mixture. ∆ T = Kb m. For simplicity assume 100 mL of the solution. Then the mass of 50 mL (50 vol %) of H2 O is 50 g. The mass of 50 mL (50 vol %) of ethylene glycol is 55.5 g. •Calculation m= mol C 2 H6 O2 kg H 2 O 1 m ol mol C2 H6 O2 = 55.5 g × = 0.895 mol 62.0 g 0.895 mol C 2 H6 O2 m= = 17.9 mol/kg = 17.9 m 0.050 kg ∆ Tb = Kb m = (0.52°C/m) × 17.9 m = 9.3°C Therefore the boiling point of the solution is 9.3°C above the normal boiling point of water. Boiling point = 109.3°C. _______________________________________________________________________________ EXAMPLE 12.7 Vapor Pressure Lowering Calculate the vapor pressure P of an aqueous solution at 30°C made from 100 g of sucrose (C12 H22 O11 ), and 100 g of water. The vapor pressure of water at 30°C is 31.8 mm Hg. •Method of Solution Sucrose is a nonvolatile solute, so the vapor pressure over the solution will be due to H2 O molecules. The problem can be worked in two ways. a. The vapor-pressure lowering is proportional to the mole fraction of sucrose X2 , and P o, the vapor pressure of pure water at 30°C. o ∆ P = X2 P 1 First let's calculate the mole fraction of sucrose, X2 . X2 = n2 n1 + n 2 1 m ol = 0.292 mol 342 g 1 m ol n1 = 100 g × = 5.55 mol 18.0 g 0.292 0.292 X2 = = 0.292 + 5.55 5.84 n2 = 100 g × X2 = 0.050 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Physical Properties of Solutions / 2 43 The vapor pressure lowering is ∆ P = (0.050) (31.8 mm Hg) = 1.6 mm Hg o The vapor pressure, P1 , i s P 1 – ∆ P. P 1 = 31.8 – 1.6 = 30.2 mm Hg b. Alternatively o Raoult's law can be used: P 1 = X1 P 1 o where P1 and P 1 are the vapor pressures of pure solvent and of the solvent in solution, respectively. From part a, X2 = 0.050, therefore X1 = 1.00 – X2 = 0.95, and P 1 = (0.95)(31.8 mm Hg) = 30.2 mm Hg _______________________________________________________________________________ EXAMPLE 12.8 Finding the Molar Mass of a Solute Benzene has a normal freezing point of 5.51°C. The addition of 1.25 g of an unknown compound to 85.0 g of benzene produces a solution with a freezing point of 4.52°C. What is the molecular mass of the unknown compound? •Method of Solution We need to find the number of moles in 1.25 g of unknown compound X. The freezing-point depression is proportional to the number of moles of X per kilogram of solvent. •Calculation The freezing-point depression is: ∆ Tf = 5.51°C – 4.52°C = 0.99°C We can calculate the molality of the solution because ∆ Tf is given above. m= ∆ Tf Kf where Kf is the freezing-point depression constant for benzene (Table 12.2 textbook). m= 0.99°C = 0.19 m 5.12°C/m This means that there are 0.19 mol of X per kg of benzene. The number of moles of X in 0.085 kg of the solvent benzene (the given amount) is 0.085 kg × 0 .19 mol X = 1.6 × 10 –2 mol X 1 kg benzene Therefore 1.25 g X = 1.6 × 10 –2 mol X, and molar mass of X = Back Forward Main Menu 1.25 g 1.6 × 1 0 –2 m o l TOC = 77.4 g/mol Study Guide TOC Textbook Website MHHE Website 2 44 / Physical Properties of Solutions _______________________________________________________________________________ EXAMPLE 12.9 Colligative Propereties of Electrolytes and Nonelectrolytes List the following aqueous solutions in the order of increasing boiling points: 0.10 m glucose (C6 H12 O6 ); 0.10 m Ca(NO 3 )2 ; 0.10 m NaCl. •Method of Solution Ethanol is a nonelectrolyte, whereas NaCl and Ca(NO3 )2 are electrolytes. NaCl dissociates into 2 ions per formula unit, and Ca(NO3 )2 dissociates into three ions per formula unit. The effective molalities are approximately: glucose 0.10 m NaCl ≅ 0.20 m Ca(NO3 )2 ≅ 0.30 m Therefore, the boiling-point elevation and the boiling point are greatest for a solution of Ca(NO 3 )2 , second highest for NaCl and lowest for glucose. •Comment If you included the van't Hoff factors, the effective molalities would be affected slightly, but not enough to change the predicted results. _______________________________________________________________________________ EXERCISES 16. What is the freezing point of a solution made from 1.00 g of C6 H12 O6 (glucose) and 100.0 g of H 2 O? 17. Calculate the approximate osmotic pressure at 25°C of a solution that has a freezing point of –18°C. 18. Calculate the molar mass of naphthalene given that a solution of 2.11 g of naphthalene in 100 g of benzene has a freezing-point depression of 0.85°C. 19. A solution contains 1.00 g of a compound (a nonelectrolyte) dissolved in 100.0 g of water. The freezing point of the solution is –0.103°C. What is the molar mass of the compound? 20. Calculate the freezing point of an aqueous solution that boils at 100.5°C. 21. How many solute particles does each basic unit of the following compounds give in aqueous solution? a. C 2 H6 O2 (ethylene glycol) b. (H2 N)2 CO (urea) c. HBr d. (NH4 )3 PO4 e. NaOH f. Ca(OH)2 22. Arrange the following aqueous solutions in order of increasing freezing point. 0.10 m MgCl2 , 0.20 m C 6 H12 O6 , 0.15 m NaCl, 0.10 m KI. 23. Which of following aqueous solutions has the lower freezing point? 0.10 m HCl or 0.10 m acetic acid _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Physical Properties of Solutions / 2 45 CONCEPTUAL QUESTIONS 1. 2 3. 4. List three things that happen when a salt dissolves in water. Give the sign of ∆ H for each step and identify the forces involved in each. Many salts with ∆ Hsoln near zero still dissolve in appreciable amounts. Explain. When a "seed" crystal is placed in a supersaturated solution at constant temperature, precipitation is induced. When precipitation stops is the solution saturated, unsaturated, or superaturated? What are ion pairs? What effect does ion-pair formation have on the osmotic pressure? PRACTICE TEST 1. For each of the following pairs predict which substance will be more soluble in water: a. HCl(g) or H2 (g) b. CCl 4 or AlCl3 c. CH3 Cl or CCl4 d. CH3 OH or CH 3 —O—CH3 2. For each of the following pairs predict which substance will be more soluble in CCl4 (l). a. H2 O or oil b. C 6 H6 (benzene) or CH3 OH c. I2 or NaCl 3. Calculate the molality of a solution prepared by dissolving 28.0 g of urea, CO(NH2 )2 , in 450 g of H2 O. 4. How many grams of KI must be dissolved in 300 g of water to make a 0.500 m solution? 5. What is the molarity of a 3.0 percent hydrogen peroxide (H2 O2 ) aqueous solution? The density is essentially 1.0 g/cm3 . 6. Concentrated sulfuric acid is 96 percent H2 SO4 by weight. Its density is 1.83 g/mL. Calculate the molarity of concentrated H2 SO4 . 7. What are the mole fractions of methanol and water in a solution that contains 40.0 g CH3 OH and 40.0 g H 2 O? 8. An aqueous solution contains 167 g CuSO4 in 820 mL of solution. Calculate the following: a. Molarity b. Percent by mass c. Mole fraction d. Molality of the solution The density of the solution is 1.195 g/mL. 9. The Henry's law constant for CO is 9.73 × 10 –4 mol/L·atm at 25°C. What is the concentration of dissolved CO in water if the partial pressure of CO in the air is 0.015 mm Hg? 10. How many grams of isopropyl alcohol, C3 H7 OH, should be added to 1.0 L of water to give a solution that will not freeze above –16°C. 11. What is the boiling point of an aqueous solution of a nonvolatile solute that freezes at –3.0°C? 12. When 48 g of glucose (a nonelectrolyte) is dissolved in 500 g of H 2 O, the solution has a freezing point of –0.94°C. What is the molar mass of glucose? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 46 / Physical Properties of Solutions 13. 7.85 g of a compound with an unknown formula is dissolved in 300 g of benzene. The freezing point of the solution is 2.10°C below that of pure benzene. What is the molar mass of the compound? 14. The dart poison in root extracts used by the Peruvian Indians is called curare. It is a nonelectrolyte. The osmotic pressure at 20°C of an aqueous solution containing 0.200 g of curare in 100 mL is 56.2 mm Hg. Calculate the molar mass of curare. 15. What is the concentration of a NaCl solution with an osmotic pressure of 10 atm at 25°C? 16. The walls of red blood cells are semipermeable membranes, and the solution of NaCl within those walls exerts an osmotic pressure of 7.82 atm at 37°C. What concentration of NaCl must a surrounding solution have so that this pressure is balanced and cell rupture (hemolysis) is prevented? 17. Arrange the following aqueous solutions in order of increasing freezing points, lowest to highest. 0.100 m ethanol 0.050 m Ca(NO 3 )2 0.100 m NaBr 0.050 m HCl 18. Calculate the value of i for a 1:1 electrolyte, if a 1.0 m aqueous solution of the electrolyte freezes at – 3.28°C. General Problem 19. In the course of research a chemist isolates a new compound. An elemental analysis gives the following: C 50.7%, H 4.25%, O 45.1%. When 5.01 g of the compound is dissolved in 100 g of water, it produces a solution with a freezing point of –0.65°C. What is the molecular formula of the compound? ANSWERS Exercises 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. Back miscible Hexane is a nonpolar liquid and water is a polar liquid. a. soluble b. soluble c. insoluble d. insoluble e. soluble f. soluble Ammonia NH3 can form hydrogen bonds to water, but hexane cannot form hydrogen bonds.. Solvation involves surrounding a solute molecule or ion with solvent molecules. The process is called hydration when the solvent is water. Solvation is the more general term. a. NaCl(s) b. NH 3 (g) c. CH 3 OH(l) a. I2 (s) b. CS 2 (l) c. CO 2 (g) a. 7.96% b. 9.81% c. 1.6% a. 1.47 m b. 0.359 m 2.32 M NaCl 9.94% AgNO3 0.34 M AgNO 3 115 g KNO 3 decreases 1.5 × 10 –5 mol/L -0.103°C Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Physical Properties of Solutions / 2 47 17. 18. 19. 20. 21. 22. 23. 237 atm 127 g/mol 181 g/mol –1.8°C a. 1 b. 1 c. 2 d. 4 e. 2 f. 3 0.20 m C6 H12 O6 = 0.10 m KI < 0.15 m NaCl = 0.10 m MgCl2 The Tf of 0.10 m HCl (a strong acid) is lower than that of 0.10 m acetic acid (a weak acid). Conceptual Questions 1. 2. 3. 4. First, the water molecules in the solvent must spread apart a bit to make room for the solute particle. This requires a positive ∆ H. Energy must be added in order to break hydrogen bonds between water molecules. Second, the ions in the lattice of the solid salt must be separated from all their neighbors. The energy for this step is related to the lattice energy. The forces holding the crystal together are ionic bonds. ∆ H is positive. Next, water molecules and the anions and cations from the salt are mixed. Hydration occurs and heat is released (∆ H is negative) as ion-dipole forces bring water molecules and ions together. Energy is only one factor affecting the solution process. A second factor which affects all natural events is a tendency to produce more disordered systems. In the pure state the solute and solvent molecules are arranged in a fairly ordered manner. Much of this order disappears when the solute dissolves in the solvent. The molecules become mixed together. It is the increase in disorder of the system that favors the solubility of any substance. Saturated Ion pairs can form in solutions of electrolytes. The formation of an ion pair will reduce the number of solute particles is the solution. Therefore the osmotic pressure would decrease compared to a solution in which there were no ion pairs. Practice Test 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. Back a. HCl b. AlCl3 c. CH 3 Cl d. CH3 OH a. oil b. C 6 H6 c. I2 1.03 m urea 24.9 g KI 0.88 M H2 O2 18 M XCH3 OH = 0.360, XH2 O = 0.640 a. 1.28 M b. 17.0% CuSO4 , 83.0% H2 O c. X CuSO4 = 0.023, XH2 O = 0.977 d. 1.29 m 1.9 × 10 –8 M CO 520 g 100.84°C 190 g/mol 63.8 g/mol 650 g/mol 0.20 M NaCl 0.15 M NaCl NaBr; Ca(NO3 )2 ; HCl = ethanol i = 1.76 C 6 H6 O4 ___________________________________________________________________________________ Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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