Colligative Properties of Solutions: A Comprehensive Study, Schemes and Mind Maps of Chemistry

Download Colligative Properties of Solutions: A Comprehensive Study and more Schemes and Mind Maps Chemistry in PDF only on Docsity! Shri Swami Vivekanand Shikshan Sanstha, Kolhapur “ज्ञान, विज्ञान आणि सुसंस्कार यांसाठी शिक्षिप्रसार” - शिक्षिमहर्षी डॉ. बापूजी साळंुखे श्री स्िामी वििेकानंद शिक्षि संस्था, कोल्हापूर विद्या सशमशि E- Content Development Chapter No. 2 Solutions Marks without Option : 5 Mark with Option : 7 Shri Swami Vivekanand Shikshan Sanstha, Kolhapur SUBJECT :CHEMISTRY 2.Capacity of solutions :- 2.1 Saturated solution :- A saturated solution contains maximum amount of solute dissolved in a given amount of solvent at a given temperature. 2.2 Supersaturated solution :- A solution containing greater than the equilibrium amount of solute is said to be supersaturated solution. Example :- The capacity of 100 gm water is to dissolve 36 gms of NaCl, then the solution is saturated. If we dissolve less than 36 gms then it is unsaturated solution . If we dissolve more than 36 gms then it is supersaturated solution. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 3. Solubility :- It is amount of solute per unit volume of saturated solution at a specific temperature and is expressed as mol L-1 Factors affecting on solubility :– Nature of solute and solvent, effect of temperature and effect of pressure. a)Nature of solute and solvent :- • The compounds with similliar character are more readily soluble in each other than those of different character. • Substances having similliar intermolecular forces are likely to be soluble in each other. i.e. polar solutes dissolves in polar solvents. This is because the interactions of solute-solute, solvent-solute, solvent-solvent are all similliar magnitude. • Example - NaCl dissolves in water. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur b) Effect of temperature :- • Endothermic process :- When the temperature is increased, then solubility of substance increases by Le-chatelier’s principle. Example - KCl dissolves in Water by endothermic process. • Exothermic process:- When the temperature is increased, then solubility of substance decreases. Example - CaCl2 dissolves in water by exothermic process. • Solubilities of NaCl, KCl, NaBr changes slightly with temperature. • Solubilities of KNO3,NaNO3,KBr increases with temperature due to endothermic process. • Solubilities of Na2SO4 decreases with increase in temperature due to exothermic process. c) Effect of Pressure :- • Pressure has no effect on solubilities of solids and liquids. • However if affects on solubilities of gases in liquids, the solubility of gases increases with increase in pressure. It can be explained by Henry’s law. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 4.Vapour Pressure of solution ( liquid in liquid) :- Consider a binary solution of two volatile liquids. Both the liquids vaporize and a equilibrium is established between liquids and its vapour phase. Then the partial vapour pressure are related to their mole fractions and is given by Raoult’s law. • 1. Raoult Law :- It states that the partial vapour pressure of any volatile component of solution is equal to the vapour pressure of pure component multiplied by mole fraction of solution. Suppose a binary solution of two volatile liquids A1 & A2, P1 & P2 are the partial pressures and x1 & x2 are the mole fractions, then according to Raoult law P1 = P1 O x1 & P2 = P2 o x2 Where P1 o and P2 o are the vapour pressures of pure components. • According to Dalton’s law of partial pressure P= P1 + P2 = P1 ox1 + P2 o x2 Since x1 + x2 = 1, So x1 = 1- x2 P = P1 o(1- X2 ) + P2 ox2 = P1 o - P1 ox2 + P2 ox2 = (P2 o - P1 o) x2 + P1 o As P2 o & P1 o are constant then it is equation of straight line y=mx+c. A plot of ‘P’ vs x2 is a straight line. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur Composition of vapour phase : The composition of vapour in equilibrium with the solution can be determined by Dalton’s law of partial pressures. If we take y1 and y2 as the mole fractions of two components in the vapour, then P1 = y1P and P2 = y2P where P1 and P2 are the partial pressures of two components in the vapour and P is the total vapour pressure. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 2. Ideal and non-ideal solutions :- Ideal Solutions Non-ideal solutions 1.In this solution solvent-solvent, solute-solute and solute-solvent interactions are comparible 1.In this solution solvent-solvent, solute-solute and solute-solvent interactions are different. 2. These solutions obeys Raoult’s law over entire range of concentrations P1 = P1o.x1 2. These solutions doesnot obeys Raoult’s law over entire range of concentrations P1 ≠ P1o.x1 3.The vapour pressure of ideal solutions and its components is near about same. 3.The vapour pressure of non- ideal solutions and its components are higher or lower. 4.The volume of an ideal solutions is equal to the sum of volumes of two components. i.e. mixV = 0 4. The volumes of non-ideal solutions is not equal to its components. i.e. mixV ≠ 0 5.Enthalpy of mixing is zero. i.e. mixH = 0 5.Enthalpy of mixing is non zero. i.e. mixH ≠ 0 6. Example - Ethanol and acetone Benzene and toluene n-hexane and n-heptane 6. Example - Chloroform and acetone Ethanol and water Nitric acid and water Shri Swami Vivekanand Shikshan Sanstha, Kolhapur *Colligative Properties :- The physical properties of solutions depend on the number of solute particles in solution and not on their nature are called colligative properties. OR The properties of dilute solutions which depend on the number of particles ( ions or molecules) of the solute present in the solution are called colligative properties. The colligative properties are 1. Vapour pressure lowering (P) 2. Boiling point elevation (Tb) 3. Freezing point depression (Tf) 4. Osmotic pressure () Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 1. Vapour pressure lowering :- 1.1 : Vapour pressure : When a liquid in a closed container is in equilibrium with its vapour. Then pressure exerted by vapours on liquid surface is called vapour pressure. 1.2 : Vapour pressure lowering :- ➢ When nonvolatile, non ionizable solid is dissolved in liquid solvent, the vapour pressure of solution is lower than that of pure solvent. ➢ In other words the vapour pressure of a solvent is lowered by dissolving a non-volatile solute into it. ➢ When the solute is non-volatile it does not contribute to the vapour pressure above the solution . ➢ If P1 o is the vapour pressure of pure solvent and P1 is the vapour pressure of solution then P1 o  P1 The vapour pressure lowering is P = P1 o - P1 ---------(1) Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 1.3 : Relative lowering of V.P. :- The ratio of vapour pressure lowering of solvent divided by V.P. of pure solvent is called relative lowering of vapour pressure. Thus 𝛥𝑃 𝑃1 0 = 𝑃1 0−𝑃1 𝑃1 0 1.4 : Raoults law for non volatile solute :- It states that the vapour pressure of solvent over the solution is equal to the vapour pressure of pure solvent multiplied by mole fraction of solution. Thus P1 = P1 ox1 For a binary solution, ( x1 + x2 = 1), x1 = 1 - x2  P1 = P1 o (1 - x2) = P1 o - P1 o x2  P1 o - P1 = P1 o x2  P = P1 o x2 ----------(2) From this equation it shows that P depends upon X2. i.e. number of solute particles. Thus P is a colligative property. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 2. Boiling point elevation :- The temperature at which vapour pressure of liquid equals to atmospheric pressure is called boiling point. The boiling point of solvent is elevated by dissolving a nonvolatile solute into it. Thus the solution containing nonvolatile solute boils at higher temperature than its pure solvent. If Tbo is boiling point of pure solvent and Tb is boiling point of solution then Tb Tbo Tb = Tb - Tbo Comparision with – 1. Vapour pressure :- The vapour pressure of solution and vapour pressure of pure solvent are plotted as a function of temperature. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur Graph : Vapour pressure- temperature of pure solvent and solution • Curve AB is the vapour pressure curve of solvent. • Curve CD is the vapour pressure curve of solution. • At point ‘A’ ,the vapour pressure of solvent and atmospheric pressure is equal. Hence the temperature at point ‘A’ is called boiling point of solvent. • At point ‘C’ ,the vapour pressure of solution and atmospheric pressure is equal. Hence the temperature at point ‘C’ is called boiling point of solution. From graph it concluds that vapour pressure of solution lies below the vapour pressure of pure solvent. The difference between the two vapour pressures increases as temperature it can be predicted as P = P1 o x2 Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 2.2 Concentration of solute :- The Boiling point of elevation is directly proportional to molality of the solution. Tb  m Tb = Kb. m -----------(1) Where Kb is molal elevation constant or ebullioscopic constant. If m = 1 then, Tb = Kb, Thus molal elevation constant is the elevation in boiling point produced by 1 molal solution. Units of Kb :- Kb = Δ𝑇𝑏 𝑚 = K.kg.mol-1 2.3 : Molar mass of solute :- Eq. (1) becomes Tb = Kb. m But molality = moles of solute / mass of solvent in kg Suppose we are preparing a dilute solution having ‘W2’ gms of solute in a ‘W1’ gms of solvent. Therefore number of moles of solute becomes, n2= 𝑤2 𝑀2 molality(m) = W2/ M2 W1/1000 Tb = W2Kb1000 M2 = W2Kb1000 M2W1 Tb W1 Shri Swami Vivekanand Shikshan Sanstha, Kolhapur Graph : Vapour pressure – temperature of pure solvent, solid solvent and solution. • Consider the vapour pressure temperature diagram. The diagram consists of three curves. • Curve AB is vapour pressure curve of solid solvent • Curve CD is vapour pressure curve of liquid solvent • Curve EF is vapour pressure Curve of solution • The curve AB & CD intersects at point B, Where solid and liquid states are equilibrium having same vapour pressure Hence the temperature at point B is freezing point of solvent (Tfo) • Simillarly at point E, Curve AB & EF intersect i.e. solid solvent & solution are at equilibrium having same vapour pressure Hence the temperature at point E is freezing point of solution (Tf). • It is clear from the figure that freezing point of Solution is lower than its pure solvent. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur 3.2 Concentration of solute :- The depression of F.P. is directly proportional to molality of the solution. Tf  m Tf = Kf. M -----------(1) Where Kf is molal depression constant or cryoscopic constant. If m = 1 then, Tf = Kf. Thus molal depression constant is the depression in F.P. produced by 1 molal solution. Units of Kf :- Kf = Tf / m = K.kg.mol-1 3.3 : Molar mass of solute :- Eq. (1) becomes Tb = Kf. m But molality = moles of solute / mass of solvent in kg Suppose we are preparing a dilute solution having ‘W2’ gms of solute in a ‘W1’ gms of solvent. Therefore number of moles of solute becomes , n2=W2/M2 molality (m) = W2/M2 W1/1000 Tf = W2Kf 1000 M2 = W2Kf 1000 M2W1 Tf W1 Shri Swami Vivekanand Shikshan Sanstha, Kolhapur Problem : 1. A solution containing 18 gm of a non-volatile solute in 200 gm of water freezes at 272.07K. Calculate the molar mass of the solute. ( Tf = 273K, and Kf = 1.86 K.kg.mol⁻¹)  Given : W2 = 18 gm, Tf =273-272.07 = 0.93K , W1 = 200 gm Kf = 1.86K.kg mol⁻¹, M2 = ? Tf = Kf. W2. 1000 𝑀2𝑊1  M2 = Kf. W2. 1000 Tf.𝑊1  M2 = 1.86181000 0.93200 = 167.4 / 0.93  M2 = 180 g/mol Shri Swami Vivekanand Shikshan Sanstha, Kolhapur Osmotic pressure :- 4.1 Osmotic pressure & Concentration Of solution :- • For a dilute solution osmotic pressure follows, •  = CRT ----------(1) •  = 𝑛2𝑅𝑇 𝑣 ----------(2) • Where V is the volume of solution in dm3, n2 is moles of solute, R is a gas / solution constant 0.082 dm3 atm K-1 mol-1,  is osmotic pressure. •   = MRT Where M = 𝑛2 𝑣 4.2 : Molar mass of solute and osmotic pressure :- From eq. (2) = 𝑛2𝑅𝑇 𝑣 If ‘W2’ gms of solute is dissolved in ‘V’ liters of solution having the molar mass ‘M2’ then, n2= 𝑊2 𝑀2  = 𝑊2𝑅𝑇 𝑀2𝑉 M2 = 𝑊2𝑅𝑇 𝑉 -----(3) From equation (3) , the molar mass of solute can be calculated. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur • Reverse osmosis :- The direction of osmosis can be reversed by applying a pressure larger than the osmotic pressure. The pure solvent then flows from solution into pure solvent through semipermeable membrane. This phenomenon is called reverse osmosis. Example :- Separation of salt from sea water. In figure fresh water and salty water are separated by a semipermeable membrane. When the pressure larger then osmotic pressure is applied to solution, then pure water from salty water passes into pure water through semipermeable membrane. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur Problem : 1. Osmotic pressure of a solution containing 7 gm of dissolved protein per 100 cm3 of solution is 20 mm of Hg at 310K. Calculate the molar mass of protein? ( R= 0.0821atm.L mol-1K-1)  Given : W2 = 7 gm, V = 100 cm3 = 0.1 dm3, T = 310K,  = 20 mm of Hg = 20 / 760 = 0.026 atm, R = 0.0821 atm.L mol-1K-1 M2 = ?  = 𝑊2𝑅𝑇 𝑀2𝑉  M2 = 𝑊2𝑅𝑇  𝑉 = 70.0821310 0.0260.1 = 178.157 / 0.0026  M2 = 68521.9 g.  M2 = 68.52 kg. Shri Swami Vivekanand Shikshan Sanstha, Kolhapur • Modification of expressions of colligative properties : The expressions of colligative properties mentioned earlier for nonelectrolytes are to be modified so as to make them applicable for electrolyte solutions. The modified equations are • Van’t Hoff factor and degree of dissociation :- Consider an electrolyte solution AxBy and dissociates in aq. Solution as AxBy xA + yB Initial 1 mole 0 0 At aq. (1-) x y If  is degree of dissociation of electrolyte then x moles of A and y moles of B dissociated at equilibrium. Initially we dissolved 1 moles of electrolyte out of  moles are dissociated and (1-) moles are undissociated. Total moles after dissociation = (1-) + x+ y = 1+ (x+y-1) = 1+  (n-1) (n=x+y) Thus Van’t Hoff factor is given by i = 1+  (n-1) / 1 i = 1+  (n-1)   = ⅈ−1 𝑛−1 Shri Swami Vivekanand Shikshan Sanstha, Kolhapur Problem : 1) 0.01 m aqueous formic acid solution freezes at - 0.021oC. Calculate its degree of dissociation. ( Kf = 1.86 K.Kg/mol)  Given :- m = 0.01 m, Kf = 1.86 K.Kg/mol, Tf (observed) = ( 0-0.021) =0.021 oC ,  = ? Now, Tf (Therotical) = Kf. m = 1.86  0.01 = 0.0186 i = Tf (observed) / Tf (Therotical) = 0.021 / 0.0186 i = 1.13 But  = ⅈ−1 𝑛−1  = i-1 ( because n=2) = 1.13-1  = 0.13  = 13 % Shri Swami Vivekanand Shikshan Sanstha, Kolhapur Previously asked Questions in Board examination :- Feb -19 1. Derive Van’t Hoff general solution equation? ( 2 mark) 2. Define - 1) Isotonic solution, ii) Hypertonic solution, iii) Hypotonic solution. ( 3 mark) 3. What is a freezing point of a liquid . The freezing point of pure benzene is 278.4 K. Calculate the freezing point of solution when 2.0 gm of solute having molecular weight 100 gm/mole is added to 100 gm of benzene?( Kf for benzene = 5.12 K.Kg/mol) ( 3 mark) July-19 1.Define osmotic pressure . Write mathematical expression between cryoscopic constant and molar mass of solute. ( 2 mark) 2. Derive = CRT ( 2 mark) Feb- 20 1.What is the concentration of dissolved oxygen at 50C under pressure of one atmosphere, if partial pressure of oxygen at 50C is 0.14 atm. (1 mark) 2. Define Osmotic pressure . Write mathematical expression between cryoscopic constant and molar mass of solute? ( 3 mark) Shri Swami Vivekanand Shikshan Sanstha, Kolhapur