CHAPTER THREE (12) Physical Properties of Solutions

Name: CHAPTER THREE (12) Physical Properties of Solutions
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Description: CHAPTER THREE (12) Physical Properties of Solutions

CHAPTER THREE (12) Physical Properties of Solutions

Chapter 3 / Physical Properties of Solutions
Chapter Three outline Chapter Three Contains: 3.1 Types of Solutions 3.2 A Molecular View of the Solution Process 3.3 Concentration Units 3.4 The Effect of Temperature on Solubility 3.5 The Effect of Pressure on the Solubility of Gases 3.6 Colligative Properties of Nonelectrolyte Solutions 3.7 Colligative Properties of Electrolyte Solutions 3.8 Colloids

3.1 Types of Solutions A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount Our focus in this chapter will be on solutions involving at least one liquid component—that is, gas-liquid, liquid-liquid, and solid-liquid solutions

3.1 Types of Solutions There are 3 types of solutions based on their capacity to dissolve a solute : A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. (not very stable) Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate.

3.2 A Molecular View of the Solution Process Three types of interactions in the solution process: solvent-solvent interaction solute-solute interaction solvent-solute interaction DHsoln = DH1 + DH2 + DH3

3.2 A Molecular View of the Solution Process Two substances with similar intermolecular forces are likely to be soluble in each other. non-polar molecules are soluble in non-polar solvents (most organic compounds are non-polar molecules) CCl4 in C6 H6 polar molecules are soluble in polar solvents C2H5OH in H2O ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (l) Solvation is the process in which an ion or a molecule is surrounded by solvent molecules arranged in a specific manner. The process is called hydration when the solvent is water. “like dissolves like”

3.2 A Molecular View of the Solution Process What is polar Substance? and

3.2 A Molecular View of the Solution Process +

3.3 Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Chemists use several different concentration units, Let us examine the four most common units of concentration: percent by mass, mole fraction, molarity, and molality. Percent by Mass x 100% mass of solute mass of solute + mass of solvent % by mass = = x 100% mass of solute mass of solution

3.3 Concentration Units Example: A sample of g of potassium chloride (KCl) is dissolved in 54.6 g of water. What is the percent by mass of KCl in the solution? = x 100% mass of solute mass of solution % by mass = x 100% 0.892 % by mass % by mass = 1.61 %

3.3 Concentration Units Mole Fraction (X) XA = moles of A sum of moles of all components Example: Gases solution contain 2.0 g of the He and 4g of O2. What are the mole fractions of He and O2 in the solution? XHe = nHe / nHe + nO2 XO2 = nO2 / nHe + nO2 nHe = 2/4= 0.5 nO2 = 4/32= 0.125 XO2 = 0.125/ = 0.2 XHe = 0.5/ = 0.8

3.3 Concentration Units Molarity (M) M = moles of solute liters of solution Example : How would you prepare 1 Liter solution of 0.1 M CuSO4 ,starting with solid CuSO4 ? M = 0.1, V= 1 L, mass=? M = n/V ==== n= M x V = 0.1 x 1 = 0.1 mole n = mass/molar mass mass = n x molar mass = 0.1 x = g

3.3 Concentration Units Molality (m) m = moles of solute mass of solvent (kg) Example: What is the molality of CuSO4 solution when 20 g of CuSO4 dissolved in 100 g of water ? n= mass/molar mass= 20/ = mol m= n/mass (kg) = / ( 100/1000) = 1.25m

3.3 Concentration Units Example: What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is g/mL? We need to calculate the number of mole of solute (methanol) and the mass of solvent. Lets assume we have 1L of solution Thus, the number of moles of ethanol =5.86 mole. To calculate the mass of solvent Mass of solvent= mass of solution – mass of solute Mass of solute = mole of ethanol x molar mass = 5.86 x 46 =270 g Mass of solution form density 0.927 g ===== 1 ml ? g ========= 1000ml(1L) Mass of solution = x 1000 = 927 g M = moles of solute liters of solution m = moles of solute mass of solvent (kg) Mass of solvent = 927 – 270 = 657 g = kg m = moles of solute mass of solvent (kg) m = 5.86 0.657 = 8.92 m

3.3 Concentration Units Example: Calculate the molality of a 35.4 percent (by mass) aqueous solution of phosphoric acid (H3PO4). The molar mass of phosphoric acid is g. In solving this type of problem, it is convenient to assume that we start with a g of the solution. Thus, the mass of water (solvent) = =64.6 g = kg. Mole of H3PO4 = mass / molar mass = 35.4 / = mol m = moles of solute mass of solvent (kg) m = 0.361 0.0646 = 5.59 m

3.3 Concentration Units Example: a) How many grams of concentrated nitric acid should be used to prepare 250 ml of 2.0 M HNO3 ,The concentrated HNO3 is 70 % HNO3 by mass b) If the density of the concentrated nitric acid is1.42 g/ml, what volume should be used? a) M = n/V ==== n= M x V = 2 x (250/1000) = 0.5 mole n = mass/molar mass mass = n x molar mass = 0.5 x 63 = 31.5 g From the given percentage there are 70 g of HNO3 in 100g of solution 70g HNO3 ---- 100g solution 31.5g HNO  ? Solution = 31.5 x 100 / 70 = 45g. b) d= mass / volume === volume = mass / density volume = 45 / 1.42 = 31.7 ml Stopped here

3.3 Concentration Units Example: What are the mole fractions of solute and solvent in 1.00 m aqueous solution ? 1.00 m its mean 1 mole of solute in 1 kg (1000g) of H2O nH2O = mass/ molar mass = 1000g / 18 = 55.6 mol m = moles of solute mass of solvent (kg) Xsolute = nsolute / (nsolute + nH2O) XH2O = nH2O / (nsolute + nH2O) Xsolute = 1 / ( ) = 0.018 XH2O = 55.6 / ( ) = 0.982

3.4 The Effect of Temperature on Solubility Temperature affects the solubility of most substances. In this section we will consider the effects of temperature on the solubility of solids and gases. solubility increases with increasing temperature solubility decreases with increasing temperature Solid solubility and temperature 1) If H is –ve value an increase in temperature ,well decrease solubility. 2) If H is +ve value an increase in temperature, well increase solubility.

3.4 The Effect of Temperature on Solubility Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO3 contaminated with 10 g NaCl. Fractional crystallization: Dissolve sample in 100 mL of water at 600C Cool solution to 00C All NaCl will stay in solution (s = 34.2g/100g) 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g

3.4 The Effect of Temperature on Solubility O2 gas solubility and temperature solubility usually decreases with increasing temperature

3.5 The Effect of Pressure on the Solubility of Gases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c is the molar concentration (M) of the dissolved gas c = kP P is the pressure of the gas over the solution k is a constant for each gas (mol/L•atm) that depends only on temperature low P low c high P high c

3.5 The Effect of Pressure on the Solubility of Gases Example: The solubility of nitrogen gas at 25°C and 1 atm is 6.8 x10-4 mol/L. What is the concentration (in molarity) of nitrogen dissolved in water under atmospheric conditions? The partial pressure of nitrogen gas in the atmosphere is 0.78 atm. First we determine the value of k c=kP 6.8 x10-4 = k (1) k = 6.8 x10-4 mol/L.atm Thus c= 6.8 x10-4 x 0.78 = 5.3 x 10-4 M.

3.6 Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. The colligative properties are vapor-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure. For our discussion of colligative properties of nonelectrolyte solutions it is important to keep in mind that we are talking about relatively dilute solutions, that is, solutions whose concentra-tions are ≤ M . Vapor-Pressure Lowering P1 = X1 P 1 P 1 = vapor pressure of pure solvent HEEEER X1 = mole fraction of the solvent Raoult’s law If the solution contains only one solute: X1 = 1 – X2 P 1 - P1 = DP = X2 X2 = mole fraction of the solute

3.6 Colligative Properties of Nonelectrolyte Solutions Example: Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (molar mass =180.2 g/mol) in 460 mL of water at 30°C. What is the vapor-pressure lowering? The vapor pressure of pure water at 30°C = mmHg. Assume the density of the solution is 1.00 g/mL ? We need to calculate y the molar fraction of the solvent (water): Mass = density x volume (ml) Mass = 1 x 460 = 460 g n1 = mass/ molar mass = 460 / 18 = 25.5 mol Mole of solute (glucose ) n2= mass /molar mass = 218 / = 1.21 mol X1 = n1 / (n1 + n2) = 25.5 / ( ) = 0.955 P1 = X1 P 1 = x = 30.4 mmHg P1 = X1 P1

3.6 Colligative Properties of Nonelectrolyte Solutions PA = XA P A Ideal Solution PB = XB P B PT = PA + PB PT = XA P A + XB P B

3.6 Colligative Properties of Nonelectrolyte Solutions Example: Heptane (C7H16) and octane (C8H18 ) form ideal solution .What is the vapor pressure at 40C of a solution that contains 3.0 mol of heptane and 5 mol of octane, At 40C pheptan = atm and poctane = atm ? Xn = 3 / (3+5) = 0.375 Xo = 5 / (3+5) = Pt = Xn pn + Xo Po = x x 0.041 = = atm

3.6 Colligative Properties of Nonelectrolyte Solutions PT is greater than predicted by Raoults’s law PT is less than predicted by Raoults’s law Force A-B A-A B-B < & Force A-B A-A B-B > &

3.6 Colligative Properties of Nonelectrolyte Solutions Fractional Distillation Apparatus

3.6 Colligative Properties of Nonelectrolyte Solutions Boiling-Point Elevation DTb = Tb – T b T b is the boiling point of the pure solvent T b is the boiling point of the solution Tb > T b DTb > 0 DTb = Kb m m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) for a given solvent

3.6 Colligative Properties of Nonelectrolyte Solutions Freezing-Point Depression DTf = T f – Tf T f is the freezing point of the pure solvent T f is the freezing point of the solution T f > Tf DTf > 0 DTf = Kf m m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) for a given solvent

3.6 Colligative Properties of Nonelectrolyte Solutions

3.6 Colligative Properties of Nonelectrolyte Solutions Example: What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is g. Mole of ethylene glycol = 478 / = 7.71 mol DTf = Kf m Kf water = C/m m = moles of solute mass of solvent (kg) m = 7.71 3.202 = 2.41 m DTf = Kf m = 1.86 x = C DTf = T f – Tf Tf = T f – DTf = – = C

3.6 Colligative Properties of Nonelectrolyte Solutions Example: What are the boiling point and freeing point of a solution prepared by dissolving 2.48 g of biphenyl ( C12H10) in 75 g of benzene, kb and kf for benzene are 2.53C/m and 5.12 C/m respectively , the boiling point and freezing point of benzene are 80.1 and 5.5C respectively . n = 2.4 / 159 = mol m = mol / Kg = m DTb = Kb m Tb = 2.53 x 0.208 = 0.526C Boiling point Of solution = boiling point of pure solvent + Tb = = C˚ DTf = Kf m ∆Tf = 5.12 x = 1.06C˚ Freezing point of solution = freezing point of pure solvent - ∆Tf = 5.5 – 1.06 = 4.4C˚

3.6 Colligative Properties of Nonelectrolyte Solutions Example: A 7.85-g sample of a compound is dissolved in 301 g of benzene. The freezing point of the solution is 1.05°C below that of pure benzene. What are the molar mass of this compound? DTf = Kf m Kf benzene = C/m m = DTf / Kf = 1.05 / 5.12 = m = mol/kg Mole = molality x mass of solvent (kg) = x = mol Mole = mass (g) / molar mass Molar mass = mass / mole = 7.85 / = 127 g/mol m = moles of solute mass of solvent (kg)

3.6 Colligative Properties of Nonelectrolyte Solutions Osmotic Pressure (p) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (p) is the pressure required to stop osmosis. more concentrated dilute

3.6 Colligative Properties of Nonelectrolyte Solutions Osmotic Pressure (p) p = MRT Stopped here High P Low P M is the molarity of the solution R is the gas constant T is the temperature (in K)

3.6 Colligative Properties of Nonelectrolyte Solutions A cell in an: isotonic solution hypotonic hypertonic If two solutions are of equal concentration and, hence, have the same osmotic pressure, they are said to be isotonic. If two solutions are of unequal osmotic pressures, the more concentrated solution is said to be hypertonic and the more dilute solution is described as hypotonic

3.6 Colligative Properties of Nonelectrolyte Solutions Example: The average osmotic pressure of seawater is about 30.0 atm at 25°C. Calculate the molar concentration of an aqueous solution of sucrose (C12H22O11) that is isotonic with seawater? p = MRT M = p / (RT) = 30 / ( X 298) = 1.23 M

3.6 Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P1 = X1 P 1 Boiling-Point Elevation DTb = Kb m Freezing-Point Depression DTf = Kf m Osmotic Pressure (p) p = MRT

3.7 Colligative Properties of Electrolyte Solutions 0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution 0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be nonelectrolytes 1 NaCl 2 CaCl2 3

3.7 Colligative Properties of Electrolyte Solutions Boiling-Point Elevation DTb = i Kb m Freezing-Point Depression DTf = i Kf m Osmotic Pressure (p) p = iMRT

3.7 Colligative Properties of Electrolyte Solutions Example: The osmotic pressure of a M potassium iodide (KI) solution at 25°C is atm. Calculate the van’t Hoff factor for KI at this concentration? p = iMRT i = p / (MRT) = / ( 0.01 X X 298) = 1.90

3.8 Colloids A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. Colloid versus solution collodial particles are much larger than solute molecules collodial suspension is not as homogeneous as a solution

3.8 Colloids Among the most important colloids are those in which the dispersing medium is water. Such colloids are divided into two categories called hydrophilic, or water-loving, and hydrophobic ,or water-fearing. The Cleansing Action of Soap

Chapter 2 / Entropy, Free Energy, and Equilibrium
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CHAPTER THREE (12) Physical Properties of Solutions

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