Expressing Solution Composition Mass fraction = Mass Solute Total Mass of Solution Weight percent = Mass fraction x 100% Example Saline solutions, NaCl(aq), are often used in medicine. What is the weight percent of NaCl in a solution of 4.6 g of NaCl in 500. g of water? 4.6 g 500. g + 4.6 g mass fraction = = 0.0091 weight percent = 0.0091 x 100% = 0.91 % 13-תכונות קוליגטיביות

Parts per Million, Billion & Trillion mass solute mass solution % = Parts per hundred = x 102 Very dilute = low mass fraction. Units for high dilution (trace solute): x 106 Parts per million (ppm) = mass solute mass solution x 109 Parts per billion (ppb) = mass solute mass solution x 1012 Parts per trillion (ppt) = mass solute mass solution 13-תכונות קוליגטיביות

Parts per Million, Billion & Trillion 1 ppm of solute in water = 1 mg / 1000 g of solution Since 1 L of water has a mass ≈ 1000 g. 1 ppm ≈ 1 mg/L Similarly 1 ppb ≈ 1 μg/L 1 ppt ≈ 1 ng/L 13-תכונות קוליגטיביות

Parts per Million, Billion & Trillion Sea water is 10,600 ppm Na+. Calculate the mass fraction and mass percent of sodium ions in sea water. Mass fraction 10,600 ppm = 10,600 g Na+ in 106 g of solution Mass fraction = 10,600 g Na+ 106 g of solution = 0.0106 Mass percent = 1.06 % 13-תכונות קוליגטיביות

Molarity Molarity = M = = moles of solute liters of solution n V Because V varies with T, molarity is T dependent! 13-תכונות קוליגטיביות

Molality Molality is another concentration scale: moles of solute kilograms of solvent Molality is another concentration scale: A mass-based unit. Uses solvent mass (not solution). It is T independent (unlike molarity). 13-תכונות קוליגטיביות

Molarity and Molality Calculate the molarity and molality of an 93.0 % by mass sulfuric acid solution. Its density = 1.835 g/mL. 93 % = 93.0 g H2SO4 + 7.0 g H2O (in 100.0 g soln.) nH2SO4 = 93.0 g 98.09 g/mol = 0.9481 mol Vsoln = 100.0 g 1.835 g/mL = 54.50 mL [H2SO4] = 0.9481 mol 0.05450 L = 17.4 M 0.9481 mol 0.0070 kg = 1.4 x 102 mol/kg (m) mH2SO4 = 13-תכונות קוליגטיביות

Mole fraction is the moles of component over the total moles of solution. It is abbreviated as X. 13-תכונות קוליגטיביות

Mole Fraction Determine the mole fractions of glucose and water in a solution containing 5.67 g of glucose dissolved in 25.2 g of water. 13-תכונות קוליגטיביות

Mole Fraction 5.67 g of glucose = 0.0315 mol of glucose. Calculate the total moles of solution. The total moles of solution are: Therefore, 13-תכונות קוליגטיביות

Colligative Properties of Solutions Colligative properties are those properties that: Only depend upon the number of “particles” in solution. each ion or molecules has the same effect; the particle type is unimportant. Solvent vapor P drops if non-volatile solute is added. Lower purity solvent = lower vapor P. 13-תכונות קוליגטיביות

Vapor Pressure Lowering over pure solvent Raoult’s law: Psolvent = Xsolvent P°solvent over the solution mole fraction moles of 1 n1 total moles (n1+n2+…) X1 = = 13-תכונות קוליגטיביות

Raoult’s law To obtain an explicit expression for the vapor-pressure lowering of a solvent in a solution, assume that Raoult’s law holds and that the solute is a nonvolatile electrolyte. Therefore, Substituting Raoult’s law gives: A: solvent B: nonelectrolyte solute P = PAXB 13-תכונות קוליגטיביות

Vapor Pressure Lowering A solution of urea in water has a vapor P of 291.2 torr. The vapor P of pure water is 355.1 torr. Calculate the mole fraction of urea in the solution. Urea is non-volatile. Water obeys Raoult’s law: Pwater = XwaterP°water 291.2 torr = Xwater(355.1 torr) Xwater= 291.2/355.1 = 0.820 Xurea = 1 – 0.820 = 0.180 (Remember: X1 + X2 + … = 1) 13-תכונות קוליגטיביות

Raoult’s Law for Volatile Solute When both the solvent and the solute can evaporate, both molecules will be found in the vapor phase. The total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent. For an ideal solution Ptotal = Psolute + Psolvent The solvent decreases the solute vapor pressure in the same way the solute decreased the solvent’s. Psolute = χsolute∙P°solute and Psolvent = χsolvent ∙P°solvent 13-תכונות קוליגטיביות

Such solutions occur when the substances are chemically similar. An ideal solution for substances A and B is one in which both substances follow Raoult’s law for all values of mole fractions. Such solutions occur when the substances are chemically similar. 13-תכונות קוליגטיביות

Vapor-Pressure Lowering of Solutions: Raoult’s Law Ptotal = PA + PB = (PA XA) + (PB XB) 13-תכונות קוליגטיביות

Vapor Pressure of a Nonideal Solution When the solute–solvent interactions are stronger than the solute–solute  solvent–solvent, the total vapor pressure of the solution will be less than predicted by Raoult’s law, because the vapor pressures of the solute and solvent are lower than ideal. When the solute–solvent interactions are weaker than the solute–solute  solvent–solvent, the total vapor pressure of the solution will be more than predicted by Raoult’s law. 13-תכונות קוליגטיביות

Deviations from Raoult’s Law Benzene- toluene Acetone- water Ethanol- hexane 13-תכונות קוליגטיביות

Boiling Point Elevation Non-volatile solutes increase the b.p. of a solvent. Why? Solvent vapor P is lowered. Higher T is needed to get the vapor P = external P. Quantitatively ΔTb = Kb msolute Note: the Kb value only depends on the solvent, the molality, m depends on the solute. 13-תכונות קוליגטיביות

Freezing-Point Lowering A non-volatile solute lowers the f.p. of a solvent: Quantitatively ΔTf = Kf msolute Kf is a constant for the solvent. Ethylene glycol (antifreeze) is added to car radiators to lower the freezing point of water (stops freezing). 13-תכונות קוליגטיביות

Boiling-Point Elevation and Freezing-Point Depression of Solutions 13-תכונות קוליגטיביות

Boiling-Point Elevation and Freezing-Point Depression of Solutions Nonelectrolytes Electrolytes ΔTb = Kbm ΔTb = Kbmi ΔTf = Kfm ΔTf = Kfmi 13-תכונות קוליגטיביות

Freezing Point Lowering Calculate the f.p. of an aqueous 30.0% ethylene glycol mixture. For water Kf = -1.86°C kg mol-1. 100.0 g of 30% mix: 30.0 g C2H2(OH)2 + 70.0 g H2O nglycol = 30.0 g / 62.07 g mol-1 = 0.4833 mol mglycol = (0.4833 mol / 0.070 kg) = 6.904 molal ΔTf = -1.86°C kg mol-1(6.904 mol/kg) = -12.8 °C Freezing point = 0.00°C – 12.8°C = -12.8°C normal freezing point of water 13-תכונות קוליגטיביות

Molar Mass By Boiling Point Elevation Benzyl acetate is one of the active components of oil of jasmine. If 0.125 g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is 61.82 C. What is the molar mass of benzyl acetate? Given: bp (chloroform) = 61.70 C Kb (chloroform) = 3.63C/m Solution: The molality of the solution can be found from the freezing point depression. Knowing molality and mass of solvent, one can find the moles of solute. Knowing mass and moles of solute, the molar mass of the compound can be determined. 13-תכונות קוליגטיביות

Molar Mass By Boiling Point Elevation, continued 13-תכונות קוליגטיביות

Osmotic Pressure of Solutions Semipermeable membrane Allows passage of small “particles”. Stops large “particles”. Ion + H2O coordination sphere – too large to pass through net solvent flow Large molecule cannot pass. e.g. animal bladders and cell membranes. Osmosis Movement of solvent through a semipermeable membrane from dilute to more concentrated solution. 13-תכונות קוליגטיביות

An Osmosis Cell 13-תכונות קוליגטיביות

Osmosis at the Particulate Level 13-תכונות קוליגטיביות

Osmosis 13-תכונות קוליגטיביות

Osmotic Pressure of Solutions Osmotic pressure = P that must be applied to stop osmosis. Π = c R T gas constant molarity of solution absolute T R = 0.08206 (atm∙L)/(mol∙K) Height of the column of solution is a measure of Π. Pure water Water enters the bag, increasing the P… semipermeable bag of 5% sugar water 13-תכונות קוליגטיביות

Osmotic Pressure Π = c R T n R T P = = c R T V Easy to remember. Very similar to the ideal gas law: c R T n R T V P = = 13-תכונות קוליגטיביות

Osmotic Pressure A cell can be exposed to 3 kinds of solution: Isotonic [solute]out=[solute]in No Net flow. Hypertonic [solute]out >[solute]in Net flow out. Hypotonic [solute]out <[solute]in Net flow in. 13-תכונות קוליגטיביות

Reverse Osmosis Used to purify water: Normal Osmosis Reverse Osmosis Water molecules cross the membrane diluting the brine. Flow stops when P = Π is applied. Reverse Osmosis Flow is reversed if P > Π is applied. Pure water can be separated from brine. 13-תכונות קוליגטיביות

Example A solution made up of 2.00 g of an unknown solute and 275 mL of water (d = 1.00 g/mL) has an osmotic pressure of 0.872 atm at 25°C What is the molar mass of the unknown solute? 13-תכונות קוליגטיביות

Example Analysis: Information given Mass of solute (2.00 g) Volume of solvent (275 mL) π (0.872 atm); T (25°C) Density of solvent (1.00 g/mL) Density of solution 13-תכונות קוליגטיביות

Example Strategy: Find the molarity M, by substituting in Determine the volume of the solution by first finding its mass Mass of solution = mass of solute + mass of water Volume of solution = mass/density Find the moles of solute using the defining equation for molarity M = moles solute/volume of solution (L) Find the molar mass Molar mass = mass/moles 13-תכונות קוליגטיביות

Example Solution: M Volume of solution 13-תכונות קוליגטיביות

Example Moles of solute Molar mass 13-תכונות קוליגטיביות

Colligative Properties of Electrolytes Ionic solutes dissolve to form more than one particle per formula unit. We alter the colligative property equations to account for this fact by including i, the number of ions per formula unit: DP = iPAXB DTb = iKbm DTf = iKfm p = iMRT 13-תכונות קוליגטיביות

Colligative Properties of Electrolytes In aqueous solution: 1 mol sucrose → 1 mol particles 1 mol NaCl → 2 mol particles (Na+ and Cl-) 1 mol CaCl2 → 3 mol (Ca2+ and 2 Cl-) Modify the formulas by replacing msolute with isolutemsolute where i = the number of particles per formula unit. = van’t Hoff factor 13-תכונות קוליגטיביות

Colligative Properties of Electrolytes A simple correction, but it only works at low [ion]. Ions attract each other in solution. Only act as separate units at very low [ion]. i is smaller than expected in many solutions: [MgSO4] iexpected iobserved 0.0005 M 2 2.00 0.005 M 1.72 0.50 M 1.07 13-תכונות קוליגטיביות

Colligative Properties of Strong Electrolyte Solutions 13-תכונות קוליגטיביות

Boiling Point Elevation At what T will 0.100 molal NaCl(aq) boil? Kb = 0.51°C kg mol-1 The b.p. of 0.100 molal NaCl: ΔTb = Kb isolutemsolute = 0.51°C kg mol-1(2)(0.100 mol/kg) = 0.10 °C b.p. = 100.00°C + 0.10°C = 100.10 °C 13-תכונות קוליגטיביות