Solutions and Colligative Properties PDF

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Welcome to Solutions and colligative properties  Solution By convention Constituent present in Solvent the largest amount. Constituents present in Homogeneous mixture relatively small Solute i.e., a single phase amounts. containing more than one component dispersed on a molecular scale. + = Solute Solvent Solution  Note If one of the components of a solution is water, it will always be considered as a solvent even when it is present in a very less amount. Solvent determines the physical state in which the solution exists.  Types of solutions Binary solution 2 Number of constituents 3 Ternary solution 4 Quaternary solution  Liquid solutions Gas in liquid Liquid in liquid e.g., alcohol and Solid in liquid e.g., carbonated water e.g., salt in water drinks  Composition of a solution can be described by expressing its concentration. Concentration can be expressed either qualitatively or quantitatively. Concentrated Concentration Or, Terms Dilute  Percentage Concentration Terms % w/w % w/V % V/V Amount of solute in Amount of solute in Volume of a solute grams dissolved per grams dissolved per (in mL) dissolved per 100 mL 100 g of solution. 100 mL of solution. of solution. Weight of solute (g) × 100 weight of solute (g) volume of solute (mL) % w/w = % w/V = × 100 % V/V = × 100 Weight of solution (g) Volume of solution (mL) Volume of solution (mL)  Other Concentration Terms Strength (S) Molarity (M) Molality (m) Normality (N) ppm Mole fraction ( 𝞆 )  Strength (S) Weight of solute (in gram) per litre (1000 mL) of solution. Weight of solute (g) Strength (S) = Volume of solution (L)  Molarity (M) Molality (m) Number of moles Number of moles of solute per litre of solute per 1000 g of solution. or 1 kg of solvent. No. of moles of solute (n) No. of moles of solute (n) M = Volume of solution (L) m = Mass of solvent (kg)  Normality (N) Number of gram equivalents of solute Normality = Volume of solution (L) Number of gram equivalents of solute Number of gram Mass of the species dissolved per litre of equivalents = Gram equivalent mass solution. Mass of the species = Molar mass n - factor  ‘n’ - factor For oxidising/ For acid / For reducing base salt agents For Oxidising/Reducing Agents Number of electrons involved in oxidation/ reduction half reaction per mole of oxidising/ reducing agent. ‒ ‒ 2+ 5e + 8H+ + MnO4 Mn + H2O n-factor = 5  For Acid/Base and Salts Number of moles For simple salts, of H+ ions n-factor is a total displaced/OH− ions charge on cations displaced per mole or a total charge of acid/base. on anions. Example: H2SO4 Example: NaOH Example: Al2(SO4)3 n-factor = charge on the n-factor = 2 n-factor = 1 cation =2×3=6  Parts per Million (ppm) Weight of solute (g) ppm (w/w) = Weight of solution (g) × 106 The number Weight of solute (g) of parts of solute present ppm (w/V) = Volume of solution (mL) × 106 in 1 million parts of solution. ppm Moles of solute (moles/moles) = Moles of solution × 106  Mole Fraction (𝓧) For a binary solution, Ratio of the Moles of solute n number of moles 𝓧solute = Total moles in solutions = n+N of a particular component to the total number of moles of all the Moles of solvent N components. 𝓧solvent = Total moles in solutions = n+N 𝓧solute + 𝓧solvent = 1  Based on the amount of the solute dissolved Unsaturated Saturated Supersaturated A solution which contains It is a solution in which A solution in which no more amount more amount of solute more solute can be of the dissolved solute can be dissolved at a dissolved at a than in the saturated particular temperature. particular temperature. solution at a particular temperature and pressure.  Solubility Solubility of one substance into another depends on: Solubility of a substance is its maximum amount that can be dissolved in a 1 Nature of the solute and solvent specified amount of solvent (generally 100 g of solvent) at a specified temperature 2 Temperature to form a saturated solution. 3 Pressure  1. Nature of Solvent & Gas Like dissolves like When a gas undergoes ionisation in a solvent, then it is highly soluble in that solvent. Polar gases dissolve in polar solvents and non-polar gases in non-polar solvents. E.g., HCl is highly soluble in water.  2. Effect of Temperature Generally, Oxygen dissolves only to a small extent in water. It is this dissolved oxygen which Dissolution of gas in sustains all aquatic life. liquid is exothermic Temperature Solubility Solubility of gases increases with decrease of temperature. It is due to this reason that aquatic species are more comfortable in cold waters rather than in warm waters.  3. Effect of Pressure (Henry’s Law) 𝓧 ∝ P P = KH 𝓧 The solubility of a gas in a liquid at a given temperature is P is the partial pressure of gas directly proportional in equilibrium with the solution. to its partial pressure, at which, it is dissolved. KH is Henry’s law constant. 𝓧 is the mole fraction of the unreacted gas in the solution.  P Increase in pressure increases solubility Movable frictionless piston ⇌ Gas P Decrease in pressure Liquid decreases solubility Gas is in equilibrium with the liquid solution  Characteristics of Henry’s Law Constant a Same unit as that of pressure: KH value increases with increase d torr or bar in temperature. b Different gases have different KH Higher the value of KH of a gas, for the same solvent. e lower will be its solubility. P c KH value of gas is different in different solvents. Since, 𝓧 = KH  Graphical Analysis (Henry’s Isotherm) Plot of P vs 𝓧 is a straight line passing through the origin with slope equal to KH. Partial pressure (P) T1 T2 T3 Partial pressure (P) Mole fraction (𝓧) Slope = KH T1 > T2 > T3 K > KH > KH H1 2 3 Mole fraction (𝓧)  Note If a mixture of gases is brought in contact with a solvent, each constituent gas dissolves in proportion to its partial pressure. Henry's law applies to each gas independent of the pressure of other gas.  Application of Henry’s Law To increase the solubility of CO2 in soft drinks and At high altitudes, the partial (a) soda water, the bottle is (b) pressure of oxygen is less sealed under high pressure. than that at the ground level. Scuba diving tanks are (c) diluted with helium. Low blood oxygen level causes anoxia.  Application of Henry’s Law The bubble blocks the capillaries and Scuba divers must cope with high creates a medical condition known as concentration of dissolved gases bends that are painful and dangerous to while, breathing air at high pressure. life. Increased pressure increases solubility of To avoid bends as well as toxic effects of atmosphere gases in the blood which are high concentration of nitrogen in blood, released when the diver comes towards the tanks used by scuba divers are filled with surface. The pressure decreases and results air diluted with helium. in formation of nitrogen bubbles in blood.  Limitations of Henry’s Law Henry’s law is valid only under the following conditions: 1 Pressure of the gas is not too high. 2 Temperature is not too low. The gas should not undergo any 3 chemical reaction with the solvent. The gas should not undergo dissociation 4 in solution.  Vapour Pressure Evaporation of a Liquid in a Closed Container Some of the more energetic particles on the surface of the liquid move fast enough to escape from the attractive forces holding the liquid together. As the gaseous particles bounce around, some of them will hit the surface of the liquid again, and be trapped there.  Evaporation and Condensation An equilibrium is set up rapidly in which, the number of particles leaving the surface is exactly balanced by the number rejoining it. Rate of evaporation At equilibrium, Rate of Rate of evaporation = condensation Rate Rate of condensation H2O (l) ⇌ H2O (g) Kp = Peq (H2O (g)) Time KP = V.P.  Vapour Pressure of Solution Vapour pressure of a liquid (a) Vapour Pressure does not depend on: Pressure exerted by the vapour 1 Amount of liquid taken of solvent 'A' and solute 'B' in equilibrium with 2 Surface area of the liquid the liquid phase. 3 Volume or shape of the container  Vapour Pressure of Solution It depends upon the (b) nature of the liquid Intermolecular Vapour Boiling attractive forces pressure point Loosely held molecules escape more easily into the vapour phase.  Boiling Point Temperature at which the vapour pressure of a liquid is equal to the external pressure. At normal b.p., the vapour pressure of the pure liquid = 1 atm  Vapour Pressure of Solution (c) Temperature Temperature at which the vapour pressure of a liquid is equal to the external pressure. Vapour Temperature pressure At normal b.p., the vapour pressure of the pure liquid = 1 atm More molecules from the liquid have enough K.E. to escape from the surface of the liquid.  V.P. K.E. Temperature K.E. of particles As the temperature of a liquid increases, the kinetic energy of its molecules also increases. As the kinetic energy of the molecules increases, the number of molecules transitioning into a vapor also increases, thereby increasing the vapor pressure. In the first container (extreme left) the temperature is lowest and in the third one (extreme right) the temperature is maximum, consequently the number of molecules shown in the container is also maximum.  Temperature H2O (l) ⇌ H2O (g) ΔH > 0 According to Le Chatelier principle, increasing the temperature of a system in a dynamic equilibrium favours the endothermic change.  Raoult’s Law = o PA ∝ 𝓧A PA 𝓧APA In the solution of volatile liquids, the partial vapour pressure of each PA Partial vapour pressure of component 'A' component is directly proportional to its mole fraction. Mole fraction of 𝓧A component ‘A’ in solution Vapour pressure of pure o PA component ‘A’ at a given temperature  For Binary Solutions of A & B = = o o PA 𝓧APA PB 𝓧BPB PT = PA + PB A = o o PT 𝓧APA + 𝓧BPB B A+B 𝓧B Solutions which obey Raoult’s law 𝓧A over the entire range of concentration are called ideal solutions.  Relation between Total Pressure vs Mole Fraction PT = 𝓧APAo + 𝓧BPBo 𝓧A + 𝓧B = 1 PT = ( PAo - PoB ) 𝓧A + PBo o PT PA This represents equation o PB PA of a straight line of PT vs 𝓧A A is more o PB If PAo > PB volatile than B 𝓧A = 0 𝓧A = 1 b.p. of A < b.p. of B 𝓧B = 1 𝓧B = 0  Composition of Vapour Phase Mole fraction of A in vapour yA phase above the solution PAA = yAPT Dalton’s Law = o Mole fraction of B in vapour PAA 𝓧APA Raoult’s Law yB phase above the solution Similarly, = = o PB yBPT 𝓧BPB  PT in Terms of Composition of Vapour Phase 𝓧A + 𝓧B = 1 1 yA yB PT = o PA + o PB yAPT yBPT PAo ++ PB o = 1 yA 1 - yA = o o PA + PB 1 yA yB = o o ⇒ PT o PA + PoB PA PB PT = o o PA + (PB - PA) yA o PT in Terms of Composition of Vapour Phase o PA PT i t i o n o pos PB r com ou Vap 0 Mole fraction of A 1  Did You Know? Compositions of the liquid and vapour that are in mutual equilibrium are not necessarily the same. The vapour will be richer in the more volatile component, if the mole fraction of components in liquid phase is comparable.  Characteristics of ideal solution (i) Ideal Solutions obey Raoult's law over the entire range of concentration o PA PT = PA + PB (ii) If the forces of attraction between A—A, B—B is similar to A—B, then A and PBo B will form ideal solution. PB PA (iii) ΔmixH = 0, i.e., there should not be an enthalpy change when components of ideal solutions are mixed. 𝓧A = 1 Mole Fraction 𝓧A = 0 𝓧B = 0 𝓧B = 1 (iv) ΔmixV = 0, (1L + 1L = 2L) i.e., there should not be a change in volume on mixing.  Characteristics of ideal solution (v) (ΔmixS)sys > 0 ⇒randomness (vi) -qsys increases on mixing. (ΔmixS)surr = 0⇒ (ΔmixS)surr = =0 T (vii) (ΔmixS)univ > 0 ⇒ Mixing is a spontaneous process Probability of picking a red ball is less as it has more randomness  Examples n-Hexane and n-Heptane Benzene and Toluene Ethyl bromide and Ethyl Chlorobenzene and iodide Bromobenzene  Non-Ideal Solution Characteristics: Solutions that do not obey Raoult's law over the entire range of concentration 1 Raoult's law is not obeyed. If the forces of attraction between A—A, B—B is different from A—B, then A and 2 ΔHmix ≠ 0 B will form non-ideal solution. 3 ΔVmix ≠ 0  Non-ideal Solutions Positive deviation from Negative deviation Raoult’s Law from Raoult’s Law Non-Ideal with Positive Deviation If the forces of attraction between A—A, B—B is stronger than A—B Partial pressure of each component A and B is higher than that calculated from Raoult's law. Hence, the total pressure over the solution is also higher than the solutions, if were ideal.  Non-Ideal with Positive Deviation Vapour pressure of solution o Dashed lines represent vapour PA pressures and total pressure corresponding to ideal solution o PB PB PA V.P. B.P. 𝛘A = 1 Mole fraction 𝛘A = 0 𝛘B = 0 𝛘B = 1 b.p. of solution < b.p. of both A and B  Characteristics 1 Raoult's law is not obeyed. 3 ΔHmix > 0 2 ΔVmix > 0 (1L + 1L > 2L) When the liquids are mixed, less heat is evolved when the new attractions are set up Intermolecular forces between than was absorbed to break the original molecules of A and B are weaker ones. than those in the pure liquids.  Characteristics Examples (ΔmixS)sys > 0 ⇒ randomness 4 increases on mixing. Water and ethanol -qsys-q Chloroform and water 5 sys (ΔmixS)surr = 0⇒ (ΔmixS)surr T= 0 ⇒ Mixing is a 6 spontaneous process Methanol and chloroform Non-Ideal with Negative Deviation If the forces of attraction between A—A, B—B are weaker than A—B Partial pressure of each component A and B is lower than that calculated from Raoult's law. Hence, the total pressure over the solution is also lower than the solutions, if were ideal.  Non-Ideal with Negative Deviation Vapour pressure o PA of solution Dashed lines represent vapour pressures and total pressure corresponding to ideal solution. o PB PA PB V.P. B.P. 𝛘A = 1 Mole fraction 𝛘A = 0 𝛘B = 0 𝛘B = 1 B.P. of solution > B.P. of both A and B  Characteristics 1 Raoult's law is not obeyed. 3 ΔHmix < 0 2 ΔVmix < 0 (1L + 1L < 2L) When the liquids are mixed, more heat is evolved when the new attractions are set up than was absorbed to break the original ones. Intermolecular forces between molecules of A and B are stronger than those in the pure liquids.  Characteristics Examples Chloroform and 4 (ΔmixS)sys > 0 ⇒ randomness acetone increases on mixing. Chloroform and methyl acetate -qsys H2O and HCI 5 (ΔmixS)surr = 0 ⇒ (ΔmixS)surr = T >0 H2O and HNO3 Acetic acid and 6 (ΔmixS)univ > 0 ⇒ Mixing is a pyridine spontaneous process Phenol and aniline  Comparing Ideal & Non-Ideal Solutions Non-ideal solutions Ideal solutions Positive deviation Negative deviation o o o o o o PT = 𝓧APA + 𝓧BPB PT > 𝓧APA + 𝓧BPB PT < 𝓧APA + 𝓧BPB A-A & B-B A-A & B-B A-A & B-B molecular molecular molecular interaction are interaction are interaction are similar as A-B stronger than A-B weaker than A-B ∆mixH = 0 ∆mixH > 0 ∆mixH < 0 ∆mixV = 0 ∆mixV > 0 ∆mixV < 0  Non-ideal solutions Ideal solutions Positive Negative deviation deviation (∆mixS)sys > 0 (∆mixS)sys > 0 (∆mixS)sys > 0 (∆mixS)surr = 0 (∆mixS)surr < 0 (∆mixS)surr > 0 (∆mixS)univ > 0 (∆mixS)univ > 0 (∆mixS)univ > 0 (∆mixG)sys < 0 (∆mixG)sys < 0 (∆mixG)sys < 0  Azeotropic Mixtures Liquid mixtures which boils at a constant Very large deviations from ideality temperature and can be distilled without lead to a special class of mixtures any change in the composition. known as azeotropes, azeotropic mixtures, or constant-boiling mixtures. A boiling liquid mixture at the azeotropic composition produces vapours of exactly the same composition as that of the liquid. Composition of liquid mixtures at which, distillation cannot separate the two liquids because the Azeotropes are of two types: condensate has the same composition as that of the 1. Maximum boiling azeotropes azeotropic liquid. 2. Minimum boiling azeotropes  Minimum Boiling Azeotropes Minimum Boiling Azeotropes Non-ideal solutions showing large Non-ideal solutions showing large positive deviation from Raoult's law, form negative deviation from Raoult's law, minimum boiling azeotropes that boil form maximum boiling azeotropes that boil at a temperature lower than the boiling at a temperature higher than the boiling points of its components 'A' and 'B'. points of its components 'A' and 'B'. Ethanol - water mixture containing A mixture of HNO3 and H2O containing ≈ 95% by volume of ethanol ≈68% HNO3 and 32% water by mass Chloroform - methanol mixture Formic acid - water mixture containing containing 87.4% chloroform and 77.6% formic acid and 22.4% water by 12.6% methanol by weight mass.  Minimum Boiling Azeotropes Maximum Boiling Azeotropes Constant pressure Constant Pressure P = 101.325 kPa P = 101.325 kPa Vapour composition Vapour only (Dew point curve) Temperature [K] Temperature [K] s Vapour se Azeotrope ha Co composition tp ex en iste (Dew point ist nt ex ph curve) Co as Liquid composition Liquid es (Bubble point curve) composition (Bubble point curve) Liquid only Azeotrope 0 Mole fraction of chloroform 1 1 Pure Pure 0 Pure Mole fraction of formic acid Pure methanol chloroform water formic acid  Minimum Boiling Azeotropes Maximum Boiling Azeotropes Constant Temperature Azeotrope Constant Temperature T = 298.15 K T = 298.15 K Liquid Liquid only composition (Bubble Pressure [kPa] Pressure [kPa] point curve) s es a h Liquid composition t p n (Bubble point curve) es e ist s x Vapour only ha e Co tp en ist ex Vapour Azeotrope Co composition (Dew point curve) Vapour composition (Dew point curve) 0 0 1 Mole fraction of chloroform 0 Pure Mole fraction of formic acid 1 Pure Pure Pure water formic acid methanol chloroform  Dissolution & Crystallisation When a solid solute is added to the At equilibrium solvent, some solute dissolve and its concentration increases in the solution. Rate of Rate of dissolution = crystallisation Some solute particles in the solution Solute + Solvent Solution collide with the other solid solute particles and get separated out of the solution.  Equilibrium At this stage, the concentration of solute in solution will remain constant under the given temperature and pressure conditions. Such a solution is said to be saturated with the given solute.  Factors affecting Solubility of Solid in liquid Nature of the solvent Temperature Pressure & the solute  1. Nature of Solvent & Solute Sodium chloride dissolves readily Like dissolves like in water, whereas naphthalene and anthracene do not. Polar solutes dissolve in polar solvents and non-polar solutes dissolve in non-polar solvents. Naphthalene and anthracene dissolve readily in benzene but sodium chloride do not.  2. Effect of Temperature Solute + Solvent Solution Solute + Solvent Solution If dissolution If dissolution ΔH < 0 ΔH > 0 is exothermic is endothermic By Le Chatelier's principle, By Le Chatelier's principle, T Solubility T Solubility  3. Effect Of Pressure Pressure does not have any significant effect on the solubility of solids in liquids. Solids and liquids are highly incompressible, and practically remain unaffected by changes in the pressure.  Colligative Properties Relative lowering The properties of the solution of the vapour pressure that are dependent only on the total number of solute particles relative to Elevation in the boiling solvent/solution are known as point colligative properties. Colligative Properties Depression in the freezing point They are not dependent on the nature of particle i.e., shape, size, charge, etc. Osmotic pressure  Abnormal Colligative Property For electrolytic solutes, If solute gets associated or dissociated in the number of particles solution, then experimental/observed/actual would be different from the value of colligative property will be different number of particles actually from the theoretically predicted value. added due to dissociation or association of the solute. Abnormality in colligative NaCl (s) + H2O (l ) Na+ (aq) + Cl‒ (aq) property can be calculated in terms of van’t-Hoff factor. Example: On adding 1 mole of NaCl in excess water gives 1 mole of Na+ and 1 mole of Cl‒ ions i.e. 2 moles of solute.  van’t Hoff Factor (i) The actual extent of dissociation/ Observed colligative property association can i = Calculated colligative property be expressed with a correction factor known as van’t Hoff factor (i). Moles of solute particles in solution after dissociation/ association i = Moles of solute particles before association/dissociation  van’t–Hoff Factor (i) Dissociation >1 i No dissociation/ =1 association PS, where P0 is V.P. of the pure solvent and Solvent PS is the V.P. of the solution. Solution  1. Relative Lowering of Vapour Pressure Lowering in V.P. = Po - PS = ΔP Relative lowering of the vapour pressure is a colligative property, Relative lowering ΔP whereas, lowering in the in vapour pressure = Po vapour pressure is not.  Note Raoult’s Law (For non–volatile solutes) Vapour pressure The vapour pressure of a solution of a of pure solvent non-volatile solute is equal to the vapour pressure of the pure solvent at that Vapour pressure temperature multiplied by its mole fraction. PS = 𝓧solvent Po = ( 1 - 𝓧solute) Po Po - PS n RLVP = o P = 𝓧solute = n+N 1 0 Mole fraction of solvent n = Number of moles of non-volatile solute 𝓧solvent N = Number of moles of solvent in the solution  1. Relative Lowering of Vapour Pressure Po n+N N Po - P S o P - PS = n = 1+ n PS = wsolute x Msolvent Msolute wsolvent N Po PS wsolute n = Po - PS ‒ 1 = Po ‒ PS = Msolvent 1000 x w x Msolute solvent 1000 Po ‒ PS n PS = N w = Weight of species M = Molar mass of species  1. Relative Lowering of Vapour Pressure Po - PS wsolute 1000 Msolvent If the solute gets associated or PS = Msolute xw solvent x 1000 dissociated Po - PS ixn Po = ixn+N Po - PS Msolvent PS = Molality x 1000 Po - PS i xn Msolvent PS = N = i x Molality x 1000  Boiling Point Temperature at which the V.P. of a liquid is equal to the external pressure present at the surface of the liquid. External pressure Vapour pressure At normal B.P., V.P. of the pure liquid = 1 atm 2. Elevation in Boiling Point When a non-volatile solute is added into a volatile liquid to form solution, V.P. decreases. The solution need to be heated to a higher temperature to boil it, so that V.P. becomes equal to external pressure.  2. Elevation in Boiling Point For example, Vapour pressure of an aqueous solution of sucrose is less than 1.013 bar at 373.15 K. Atmospheric Atmospheric pressure pressure To make this solution boil, its vapour pressure must be increased to 1.013 bar by raising the temperature above the boiling temperature Lower vapour of the pure solvent (water). pressure Thus, The boiling point The boiling point of Pure Solution of a solution > the pure solvent (with solvent solute)  2. Elevation in Boiling Point T = B.P. of solution (K) o b Tb = B.P. of pure solvent (K) V.P. of solution < V.P. of pure solvent 1 atm t l ven Vapour pressure So n tio Hence, to make the V.P. equal to Pext, l u So we have to heat the solution by a greater amount in comparison to pure solvent. ΔTb Tbo Tb Temperature (K)  2. Elevation in Boiling Point o ΔTb = Tb - Tb ΔTb = Kb m ΔTb ∝ m Kb = B.P. elevation constant or, Molal elevation constant or, Ebullioscopic constant m Molality  Kb o2 RTb M It is equal to elevation in Kb = 1000 x ΔHvap boiling point of 1 molal solution. o Tb is B.P. of pure solvent (K) Units K/m or °C/m M is the molar mass of solvent in g/mol or K kg mol–1 ΔHvap is the molar enthalpy of vapourisation of the solvent (J/mol) R = 8.314 J/mol-K  Note If solute gets associated/dissociated 1 then ΔTb = i × Kb × Molality 2 Kb is the property of solvent Elevation in B.P. is proportional to 3 the lowering of vapour pressure. ΔTb ∝ ΔP  Freezing Point (F.P.) Temperature at which, the vapour pressure of a solid becomes equal to the vapour pressure of liquid at 1 atm is called normal freezing point.  3. Depression in Freezing Point When a non-volatile solute is dissolved in the solvent, the V.P. of the solvent in the solution decreases. Liquid n t ve ol Solid res Pu Pressure 1 atm n i o o l ut S V.P. of solid and liquid solvent Gas ΔTf will become equal at a lower temperature, i.e., F.P. of solution is lower Temperature (K) than that of a pure solvent.  3. Depression in Freezing Point ΔTf o ΔTf = Tf - Tf The difference between o F.P. of a pure solvent Tf ∝ ΔTf m and F.P. of its solution Tf. ΔTf = Kf m m Molality Kf is equal to the depression in Kf = F.P. depression constant or molal freezing point of 1 molal solution. depression constant or cryoscopic constant.  Cryoscopic Constant (Kf) o2 RTf M K/m or °C/m Kf = 1000 x ΔHfus Units or K kg mol–1 o Tf is F.P. of pure solvent (K) M is the molar mass of solvent in g/mol ΔHfus is the molar enthalpy of fusion of the solvent (J/mol) R = 8.314 J/mol-K  Note Depression in freezing point is proportional 1 to the lowering of vapour pressure i.e. ΔTf ∝ ΔP If solute gets associated/dissociated 2 then ΔTf = i × Kf × Molality At freezing point or below it, 3 only solvent molecules will freeze & not solute molecules.  Osmosis The spontaneous flow of solvent particles from solvent side to solution side, or from solution of low concentration side to solution of high concentration side through a semi-permeable membrane (SPM). Solution Solvent C1 SPM  Semi-permeable Membrane (SPM) Natural A membrane that allows only solvent particles SPM to move across it. Artificial  Semi-permeable Membrane (SPM) Natural SPM Artificial SPM Animal/plant cell Cu2[Fe(CN)6] membrane formed just & Silicate of Ni, below the outer skins. Fe, Co can act as SPM.  Example of Osmosis People taking a lot of salt, experience A raw mango placed in a water retention in tissue cells. This i concentrated salt solution loses water ii results in puffiness or swelling called & shrivel into pickle. edema.  4. Osmotic Pressure (𝚷) Pext The external pressure that Solution Solvent must be applied on the solution side to just stop the process of C1 SPM osmosis. Pext = 𝚷  4. Osmotic Pressure (𝚷) 𝚷 ∝ Concentration 𝚷 = CRT (Molarity) 𝚷 = Osmotic pressure 𝚷 ∝ T C = Concentration (mol/L) R= Universal gas constant T = Temperature (K)  4. Osmotic Pressure (𝚷) If more than one type of solute particles are present. 𝚷 = CRT CT = C1 + C2 + C3 + …….. n 𝚷 = V RT n1 + n2 + n3 + …….. = V 𝚷 = CTRT  4. Osmotic Pressure (𝚷) If two solutions of concentration C1 & C2 are kept separated by SPM, and C1 > C2, then the solvent particle movement take place Solution 1 Solution 2 from lower to higher concentration. C1 SPM C2 C1 > C2 So, an extra pressure is applied on the higher concentration side to stop osmosis. Pext = 𝚷1 – 𝚷2  Note Osmotic pressure of very dilute solutions is also quite If solute gets associated significant. So, its or dissociated then 𝚷 = i x CRT measurement in lab is very easy.  Reverse Osmosis If the pressure applied on the solution side is more than the osmotic pressure of the solution, then the solvent particles Used in will move from solution to solvent side. desalination of sea-water Pext > 𝚷  Based on the difference in osmotic pressure 1 Isotonic solution 2 Hypotonic solution 3 Hypertonic solution  Isotonic Solution Two solutions having same osmotic Solution 1 Solution 2 pressure are considered as isotonic solution. C1 SPM C2 𝚷1 = 𝚷2  Hypotonic & Hypertonic Solutions Hypotonic Solution 1 Solution 2 If two solutions Hypertonic ‘1’ and ‘2’ are such that 𝚷2 > 𝚷1 , then ‘2’ is C1 SPM C2 called hypertonic solution and ‘1’ is called hypotonic solution. 𝚷1 < 𝚷2  Hypotonic & Hypertonic Solutions Pext Hypotonic Solution 1 Solution 2 Hypertonic C1 SPM C2 𝚷1 < 𝚷2 Pext = 𝚷2 - 𝚷1  Types of Solutions Hypotonic Isotonic Hypertonic Solution Solution Solution Net water gain No net Net water loss Cell Swells loss or gain Cell Shrinks  Plasmolysis When the cell is placed in a solution having its osmotic pressure greater than that of the cell sap, water passes out of the cell due to osmosis. Consequently, the cell material shrinks gradually. The gradual shrinking of the cell material is called plasmolysis.  Application of Osmotic Pressure a. Determination of molecular mass of the solute. Widely used to determine molar masses of proteins and other biomolecules, as they are generally not stable at higher temperatures wRT Msolute = 𝚷V w is mass of solute This method has the advantage over other methods as pressure V is volume of the solution measurement 𝚷 is osmotic pressure of the solution is around the room temperature. T is temperature R is universal gas constant  Note If solute gets associated or dissociated then theoretical molar mass of solute will be different from experimentally calculated molar mass Theoretical molar mass of substance i = Experimental molar mass of the substance  Application of Osmotic Pressure b. Desalination of water by reverse osmosis c. Dialysis Pext Pext > 𝚷 Artificial kidney removes waste products from blood through osmosis. Sea water Pure water SPM