Ans. 10% solution of sodium carbonate by weight (w/w) means 10g of Na 2 CO 3 are present in 100g of the solution.

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1 SOLUTIONS(5 Marks) Q1. What do you mean by 10% aqueous solution of sodium carbonate? 10% solution of sodium carbonate by weight (w/w) means 10g of Na 2 CO 3 are present in 100g of the solution. Q. Define Henry s law. Henry s law: It states that The mass of a gas dissolved in a given volume of the liquid at constant temperature is directly proportional to the pressure of the gas present in equilibrium with the liquid. Mathematically: m p or m = k H p where m = mass of the gas dissolved in a unit volume of the solvent p = pressure of the gas in equilibrium with the solvent k H = constant called Henry s law constant Q. State the law obeying the solubility of a gas in a liquid. Henry s law The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature. Expressing solubility in terms of mole fraction of the gas in the solution, for a gas, Henry s law can be written as x A = k H p A (i) or p A = k H x A (ii) Where x A is the mole faction of the gas in the solution, p A is the partial pressure of the gas above the solution and k H is Henry s constant whose value depends upon the nature of the gas, nature of the solvent and the temperature. Q. What is effect of pressure in the solubility of a gas in a liquid? The solubility of a gas in a liquid at a particular temperature increase with increase pressure. Q. Write the equation of relation between the vapour pressure and mole fraction of the gas in the gas. According to Raoult s law, p A = x A p o A and p B = x B p o B Where p A = Partial vapour pressure of component A p B = Partial vapour pressure of component B p o A= Vapour pressure of component A in the pure state p o B= Vapour pressure of component B in the pure state x A and x B are mole fractions of component A and B respectively Q. Two liquids A and B are mixed. The resulting solution is found to be cooler. What do you conclude? : The solution shows a positive deviation. Q. Define vapour pressure of a liquid. Vapour Pressure of a liquid: Vapour pressure of a liquid/ solution is the pressure exerted by the vapours in equilibrium with the liquid/ solution at a particular temperature. Q. Define vapour pressure of a liquid. What happens to the vapour pressure when (a) a volatile solute dissolves in the liquid and (b) the dissolved solute is non-volatile solute? (a) Vapour pressure of a liquid increases (b) Vapour pressure of a liquid decreases Q. Why vapour pressure of a liquid decreases when a non-volatile solute is added into it? Some liquid molecules at the surface are replaced by the molecules of the solute which are non-volatile. Q. State Raoult s law. Write expression for the Raoult s law for non-volatile solutes. [Relative lowering of vapour pressure is given by the equation Where p p p = n n + n

2 p o = vapour pressure of the pure solvent p s = vapour pressure of the solution n 2 = number of moles of the solute n 1 = number of moles of the solvent Q. What type of liquids form ideal solutions? Ideal solution is formed between liquids having similar intermolecular interactions Substances having similar structures and polarities form nearly ideal solutions. Q. Under what condition do non-ideal solutions show negative deviations? The interactions between the components are greater than in the pure components. When i. V mixing = -ve, H mixing = -ve ii. p A < x A p o A, p B < x B p o B iii. Form maximum boiling azeotropes Q. How does a non-ideal solution differ an ideal solution? When does the positive deviation occur from ideality? A non-ideal solution is that solution in which solute and solvent molecules interact with one another with a different force than the forces of interaction between the molecules of the pure components. For such solution, V mixing 0, H mixing 0 In terms of Raoult s law, a non-ideal solution is that solution in which does not obey Raoult s law under all conditions of temperatures and concentrations. i.e. p A x A p o A and p B x B p o B Non-ideal solutions showing positive deviations from ideal solution when i. The interactions between the components are less than in the pure components. ii. iii. iv. V mixing = +ve, H mixing = +ve p A >x A p o A, p B >x B p o B Form minimum boiling azeotrope Q. Why the vapour pressure of a liquid is constant at constant temperature? : Because it reaches a state of equilibrium where rate of evaporation = rate of condensation. Q. Define azeotropic mixture. Azeotropic or Constant Boiling Mixtures: A liquid mixture having a definite component and boiled like a pure liquid is called a constant boiling mixture or an azeotropic mixture. Q. Why constant boiling mixtures behave like a single component when subjected to distillation? Composition of Azeotropic mixture (liquid pairs) vapourises without change in composition, the liquid obtained by the condensation of the vapour also has same composition i.e. the mixture distils over as if it were a pure liquid. Azeotropic mixtures cannot be separated into their constituents by fractional distillation. Q. Define the term-colligative properties. Certain properties of ideal solutions depend only on the number of particles of the solute (molecules/ ions) in a definite amount of the solvent and do not depend on the nature of solute. Such properties are called colligative properties. Q. What are the colligative properties? The important colligative properties are: i. Relative lowering of vapour pressure ii. iii. iv. Elevation in boiling point Depression in freezing point Osmotic pressure

3 Q. Why vapour pressure of a liquid decreases when a non-volatile solute is added into it? : Because some liquid molecules at the surface are replaced by the molecules of the solute which are nonvolatile. Q. 2g each of the solutes A and B (Mol. Mass of A > B) are dissolved separately in 20g each of the same solvent C. Which will show greater lowering of vapour pressure and why? : B will show greater lowering of vapour pressure because = Here, lowering of vapour pressure (p 0 -p s ) is directly related with amount in gram and inversely related with molar mass of non-volatile solute. Q. Show that the relative lowering of vapour pressure is a colligative property. From the equation, = The relative lowering of vapour pressure is a colligative property depends upon the number of moles of the solute dissolved in a definite number of moles of the solvent and is independent of the nature of the solute and depends upon molality of the solution. Q. Define Boiling point of a liquid. The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. Q. What do you mean by elevation in boiling point? Show that it is a colligative property. When a non-volatile solute is added to a liquid, the boiling point of the solution is higher than that of the pure solvent. The increase is called the elevation in boiling point. i.e. Boiling point of the solution, T > Boiling point of the pure solvent, T o The difference between two boiling points, called the elevation in boiling point, T b is given by T b = T T o From the expression, T b = K b x m, it is clear that the elevation in boiling point depends on molality; the number of moles of the solute dissolved in 1000g of the solvent and not upon the nature of the solute. Hence, it is a colligative property. Q. Why the boiling point of a liquid gets raised on dissolution of non-volatile solute into it? The elevation in boiling point is due to the vapour pressure of the solution is lower than that of the pure solvent and increase with increase temperature. Hence the solution has to be heated more to make the vapour pressure equal to the atmospheric pressure Q. Define Molal elevation constant or ebullioscopic constant? The elevation in boiling point depends upon concentration of the solution in terms of molality as bellow: T b = K b x m Where K b is called the molal elevation constant or ebullioscopic constant and m is the molality of the solution. If m = 1 then T b = K b Hence Molal elevation constant may be defined as the elevation in boiling point when the molality of the solution is unity i.e. 1 mole of the solute is dissolved 1kg (1000g) of the solvent. Q. How is lowering in vapour pressure related to the elevation in boiling point of the solution? Greater the lowering in vapour pressure ( P), higher is the elevation in boiling point ( T b ) i.e. T b α P Q. Define freezing point of a liquid and depression in freezing point. The Freezing point of a liquid is the temperature at which the solid and the liquid forms of the substance are in equilibrium i.e. the solid and the liquid forms of substance have the same vapour pressure. When a non-volatile solute is added to a liquid, the freezing point of the solution is always lower than that of the pure solvent. The decrease is called the depression in freezing point.

4 i.e. Freezing point of the solution, T < Freezing point of the pure solvent, T o The difference between two freezing points, called the depression in freezing point, T f is given by T f =T o T Q. In which concentration term does depression in freezing point depends? Molality i.e., The depression in freezing point depends upon concentration of the solution in terms of molality as bellow: T f = K f x m Where K f is called the molal depression constant or cryoscopic constant and m is the molality of the solution. Q. Explain why the freezing point of a solvent is lowered on dissolving a non-volatile solute into it? The Depression in freezing point is due to the vapour pressure of the solution is lower than that of the pure solvent. Hence the solution has to be cool to make the vapour pressure of the liquid equal to the vapour pressure of the solid form. Q. Define molal depression constant or cryscopic constant. The depression in freezing point depends upon concentration of the solution in terms of molality as bellow: T f = K f x m Where K f is called the molal depression constant or cryoscopic constant and m is the molality of the solution. If m = 1 then T f = K f Hence Molal depression constant may be defined as the depression in freezing point when the molality of the solution is unity i.e. 1 mole of the solute is dissolved 1kg (1000g) of the solvent. Q. Why is depression in freezing point considered to be a colligative property? From the expression, T f = K f x m, it is clear that the depression in freezing point depends on molality; the number of moles of the solute dissolved in 1000g of the solvent and not upon the nature of the solute. Hence, it is a colligative property. Q. What is osmotic pressure? The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semi-permeable membrane is called the osmotic pressure. Q. Define osmosis and osmotic pressure. Osmosis: The net spontaneous flow of the solvent molecules form the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semi-permeable membrane is called osmosis. Osmotic Pressure: The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semi-permeable membrane is called the osmotic pressure. Q. What is van t Hoff equation for dilute solution? The Relation, P = RT or PV = nrt This equation is called Van t Hoff equation for dilute solutions. Q. Define semi-permeable membrane. A membrane which allows the flow the solvent molecules to pass through but not the solute particles is called semi-permeable membrane. Q. What is osmotic pressure? How does it depend on temperature and atmospheric pressure? Osmotic pressure: The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semi-permeable membrane is called the osmotic pressure. Osmotic pressure (P) of a solution is found to be directly proportional to the molar concentration (C) of the solution and its temperature (T). Mathematically, P C T P C T P = R C T Where R is a constant (called solution constant) and its value is found to be same as that of the Gas constant. R = litre atm per degree per mole P = CRT

5 Q. What concentration term does osmotic pressure related to? Molarity Q. What are isotonic solutions? Solutions which have the same osmotic pressure at the same temperature are called isotonic solutions. Thus two isotonic solutions are also isosmotic. Q. Define osmotic pressure. Show that it is a colligative property. Osmotic pressure: The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semi-permeable membrane is called the osmotic pressure. From the equation, P = RT It is clear that osmotic pressure depends upon the number n of moles of the solute per litre of the solution irrespective of the nature of the solute. Hence osmotic pressure is a colligative property. Q. How is the colligative property of solution changed when a solute in a solution undergoes (i) association (ii) Dissociation? As the value of any colligative property depends upon the number of particles of the solute, therefore the experimentally observed value of colligative property comes out to be higher than the theoretically expected value. Further as molecular mass is inversely proportional to the colligative property, therefore the observed molecular mass comes out to be less than the theoretically expected value. Association: Under association, the observed value of colligative property will be half and the molecular mass will be double than the expected value. Dissociation: Under dissociation, the experimentally observed value of colligative property comes out to be higher than the theoretically expected value but the observed molecular mass comes out to be less than the theoretically expected value. Q. Why do electrolyte show abnormal molecular masses? Name the Factors responsible for abnormality. When the molecular mass of a substance determined by studying any of the colligative properties comes out to be different than the theoretically expected value, the substance is said to show abnormal molecular mass. Such abnormality molecular mass and colligative property is due to the following two ways: a) Electrolytic Dissociation b) Association Q. What is van t Hoff factor? What possible values can it have if the solute molecules undergo (i) Association and (ii) Dissociation, in solution Van t Hoff factor: It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property. i.e. Van t Hoff factor, i = = Further, as the colligative property is inversely proportional to the molecular mass of the solute, we can also write i = () ( ) () ( )

6 i = For substance undergoing association or dissociation in the solution, the various expressions for the colligative properties are modified as follows: Elevation in boiling point, T b = i K b x m Depression in freezing point, T f = i K f x m Osmotic pressure (P or π) = i RT Q. Why NaCl solution freezes at lower temperature than water but boils at higher temperature than water? : Because of depression in freezing point whereas of an elevation in boiling point. Q. How much molecular mass of NaCl is obtained experimentally using colligative properties? Since NaCl undergoes dissociation in solution, gives two ions I,e, Na+ and Cl- ions. Colligative properties are depend upon the number of ions whereas inversely depends on mol. mass. Hence mol. mass of NaCl becomes half i.e.. = Q. How is the colligative property of solution changed when a solute in a solution undergoes (i) association (ii) dissociation? (i) The value of colligative property decreases (ii) The value of colligative property increases. Q. Which of the following solution will have the highest and which will have the lowest freezing point and why? (i) 0.1M NaCl solution (ii) 0.1 M Glucose solution (iii) 0.1M BaCl 2 solution [Hints: Higher the Colligative Properties, Higher the Relative lowering of vapour pressure, lesser is the vapour pressure of solution, Greater the Elevation in boiling point, Higher is boiling point of solution. Greater the Depression in Freezing point, lesser is freezing point of solution. BaCl 2 solution will have the lowest while Glucose solution will have the highest freezing point because NaCl dissociates to give two ions, BaCl 2 dissociates to give three ions but glucose does not. Depression in freezing point will be maximum for BaCl 2 solution and minimum for glucose solution. MOST SUCCESSFUL QUESTIONS-ANSWERS Q. Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0g of urea (Mol. Mass =60) per 250 g of water. Hints: Molality =. Calculation of mole fraction of the solute: Number of moles of solute, n = Number of moles of solute, n = mole fraction of the solute = (). (). () =...

7 Q. Calculate the molarity and normality of a solution containing 9.8g of H 2 SO 4 in 250 cm 3 of the solution. Hints: Molarity =.. From relation between normality and molarity Normality of an acid = Molarity x basicity Basicity of H 2 SO 4 =2 Q. The vapour pressure of an aqueous solution of cane sugar (mol. mass = 342) is 756 at 373K. How many grams of sugar are present in 1000g of water? Given, vapour pressure of pure water = 760 mm Hg [ 100.5g] We have, by relation = or w = ( ) Q. Vapour pressure of an aqueous solution of glucose is 750 mm of Hg at 373 K. Calculate the molality and mole fraction of solution. [ x 2 = 0.013, molality = 0.73] From the relative lowering in vapour pressure, = m = p p 1000 p M Mole fraction of glucose, x 2 x = (for dilute solution) [n 1 = number of moles of water= = =55.5] [n 2 =number of moles of glucose =0.73] x = = Q. The osmotic pressure of blood is 8.21 atm at 37 0 C. How much glucose would be used for an injection that is at the same osmotic pressure as blood? From the eqn. Osmotic pressure,p P = w = [V=1litre, R =0.0821L atm K -1 mol -1 ]

8 Q. A solution containing 10.2 g glycerine per litre of a solution is found to be isotonic with 2.0% solution of glucose (molar mass=180). Calculate the molecular mass of glycerine.. Since the two solutions are isotonic, they must have same concentration in moles/litre. For glucose solution, concentration = (Given, 2.0% solution) = in litre = [Mol. mass of Glucose, C 6 H 12 O 6 = 180]. For glycerine substance, (x), concentration = (Given 10.2g in litre) =. For isotonic solution, =. =. [Mol. Mass of Unknown Substance, M =? ]. M = M = 91.8 g mol -1 Q. A solution contains 3.5g of a non volatile solute in 125 g of water and it boils at K. Calculate the molecular mass of solute (K b for water = 0.52K/m). From the equation, T b = K b M = [ T = (T T )] Q. The vapour pressure of an aqueous solution of cane sugar (mol. mass = 342) is 732 mm at C. Calculate the boiling point of the solution (K b for water= C) To be calculating molality, m from relative lowering of vapour pressure: = or m = ( ) Again, from the equation, T b = K b m [ T = (T T )] [Mass of solvent (water)=1kg] Q. A solution of 1.25g of a certain non-electrolyte in 20.0g water freezes at K. Calculate the molecular mass of the solute. K f = 1.86 Km -1 From the equation, T f = K f or M = [ T = (T T)] Q. Calculate the freezing point of a solution containing 0.52g glucose in 80.2g of water. K f for water = 1.86 K/m Ans From the equation, T f = K f Depression in freezing point, T f = (T 0 -T s )

9 Q. A solution of urea in water has a boiling point of C. Calculate the freezing point of the same solution. (Kf and Kb for water are C and C respectively) From Relation between Depression in freezing point, T f =K f m and Elevation in boiling point, T b =K b m, where m is the molality. From Elevation in boiling point, T b =K b m [ T b = T T 0 ) T b = T T 0 = K b m Again, Depression in freezing point, T f =K f m or T f = (T 0 - T)=K f m Q. Two aqueous solutions, containing respectively 7.5g of urea (mol.wt.=60) and 42.75g substance X in 100 g of water freeze at the same temperature. Calculate the molecular weight of X. Depression in freezing point, T = K m As T is same for the two solutions, their molalities will be equal i. e T Soln 1 = T Soln2 or K = K [Let Mol. weight of Unknown, X of soln 2.] Yearly Qs Analysis [COHSEM-2010] Q. (i) A solution contains 75% water, and 25% ethanol by mass. Calculate the mole fractions of ethanol and molality of the solution. (ii) Calculate the osmotic pressure at 273K of a 5% solution of urea (molecular mass =60). Given: R=0.0821litre.atm/deg/mole (iii) Will equimolar solutions of sodium chloride and glucose be isotonic or not? Give reason in support of your answer =5 (i) Amount of water =75g Amount of ethanol=25g Number of moles of water = = 4.16 [Mol.mass of water=18] Number of moles of ethanol = = 0.54 [Mol. mass of C 2H 5 OH, ethanol =46] Mole fraction of ethanol =. =. = Molality of the solution= = = 7.2m (ii) Since 5% solution of urea means that amount of urea,nh 2 CONH 2,w =5g and Volume of solution,v =100ml By relation, Osmotic pressure, P P = P =. [ Mol. mass of urea,m =60] P =. = atm (iii) Not, because NaCl dissociates to give two ions, Na + and Cl - while glucose does not dissociate, osmotic pressure of NaCl solution is higher than that of Glucose solution. [COHSEM-2009] Q. What are isotonic solutions? 1 mark : Solutions which have the same osmotic pressure at the same temperature are called isotonic solutions.

10 Q. The vapour pressure of a 5% aqueous solution of a non- volatile substance at 373K is 745mm. Calculate the molecular mass of the solute. 3 marks 5% aqueous solution of the solute means that 5g of the solute are present in 100g of the solution i.e. Weight of solute (w 2 ) = 5g Weight of solution= 100g Weight of solvent (w 1 ) = = 95g Further as the solution is aqueous, it means that the solvent is water and we know that Vapour pressure of pure water at 373 K, (p o ) = 760 mm Vapour pressure of solution at 373 K, (p s ) = 745 mm Molecular mass of solvent (water), M 1 = 18 Molecular mass of solute, M 2 = to be calculated Using the formula for dilute solution, M = ( ) ( ) M = ( ) M = =48 [COHSEM-2008] Q. State Henry s Law. 1 mark It states that the mass of a gas dissolved in a given of the liquid at constant temperature is directly proportional to the pressure of the gas present in equilibrium with the liquid. Mathematically: m = k H p where k H = constant called Henry s law constant. Q. 1M FeCl 3 solution has lower freezing point than 1M glucose solution. Give reasons. 1mark Because glucose does not dissociate, FeCl 3 dissociates to give four ions i.e. Fe 3+ and 3Cl - ions. Depression in freezing point will be maximum for FeCl 3 and minimum for glucose solution. Hence FeCl 3 solution will have the lower freezing point than that of glucose. Q. Why does the boiling point of a solution of a solution containing a non-volatile solute is higher and the freezing point is lower than that of the pure solvent? 2marks. : When a solute is dissolved in a solvent, the vapour pressure decreases. As a result, there is an elevation in boiling point whereas there is a depression in freezing point. [COHSEM-2007] Q. When a pressure higher than the osmotic pressure is applied on the surface of a solution separated from a solvent by a semi permeable membrane the level of the solution in the container decreases? What is the name of this process? 1mark : Reverse Osmosis. Q. 0.6g of urea is added to 500g of water (i) Calculate the molality of the solution. (ii) What changes will be observed on the vapour pressure and freezing point of water? molal depression constant for water is 1.86 K kg mol -1, what is the freezing point of the solution? 3marks (i) Here, Weight of urea (w 2 ) = 0.6g Weight of solvent (w 1 ) =500g Molecular mass of solvent (water), M 1 = 18 (iii) If the

11 Molecular mass of urea, M 2 = 60g Molality of the solution = =. = 0.02m (ii) Lowering in vapour pressure and depression in freezing point. (iii) Depression in freezing point, T = (T T) = K m T = = [Molal depression constant for water,k f =1.86] T = T T = T = T [T=Freezing point of solution=? T o =Freezing point of water=273k T = = 272.9K [COHSEM-2006] Q. How is the colligative property of a dilute solution changed when the solute undergoes association in the solution? 1 mark The experimentally observed value of the colligative property is lower and the molecular mass is higher than the expected valued when the solute undergoes association in the solution. Q. The freezing point depression of 0.1 molal solution of acetic acid in benzene is of 0256K. K f for benzene is 5.12K kg mol -1. What conclusion can you draw about the molecular state of acetic acid in benzene? 2marks Here, m=0.1m K f = 5.12K kg mol -1 Theoretically calculated valued of T f will be T f = K f m T = = 0.512K Given, experimental value of T = 0.256K Van t Hoff factor, i = i =.. = Also i = [ molecular mass of CH COOH = 60 (Calculated)] Observed Mol. Mass =. Observed mol. mass = 60 2 = 120 Thus observed molecular mass is double of the theoretical value. Hence acetic acid exists as doubly associated i.e. (CH 3 COOH) 2 [COHSEM-2005] Q. Draw a neat and properly lebelled diagram for demonstration of Osmosis between distilled water and molar urea solution. 2marks. Q. What is the molarity of 0.05N oxalic acid solution? 1 mark : Here, Normality = 0.05 N Molarity =? (To be calculated) Basicity of oxalic acid, ( H 2 C 2 O 4 ) = 2 From the relationship between normality and molarity of acid, Normality of an acid = Molarity x Basicity Molarity of an acid = =. = 0.025M

12 Q. What is an ideal solution? 1mark An ideal solution may be defined as the solution in which each component obeys Raoult s law under all conditions of temperatures and concentrations. Q. 1g each of the solutes A and B (Mol. Mass of A > B) are dissolved separately in 50g each of the same solvent. Which one o the two will show higher boiling point and why? 3 marks B will show higher boiling point because T = 1000K w w M Here, Elevation of Boiling point ( T ) directly related with amount (w B ) in gram and inversely related with molar mass (M B ) of non-volatile solute. Q. Defined Osmosis pressure. [COHSEM-2004] State two van t Hoff s laws of osmotic pressure and give the van t Hoff equation for dilute solutions. Three litres aqueous solution of urea (molar mass=60g) weighing 3018g is found to record an osmotic pressure of atmospheres at 27 o C. Calculate the molality of the urea solution.(r= atm.mol -1 K -1 ) ½+2+1/2+2 =5 marks Osmotic pressure may be defined as the minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semi permeable membrane. van t Hoff s law of osmotic pressure of a solution is state that osmotic pressure of a solution is directly proportional to the molar concentration (C) of the solution and its temperature (T). Mathematical, P = CRT or or P = RT is called van t Hoff equation for dilute solution. Where V = Volume of the solution in litres n= number of moles of solutes. C = Given, Wt. of urea in three litres of aq. solution of urea = 3018 g Wt. of urea in one litre of aq. solution of urea = = 1006g Osmotic pressure, P= atm. Temperature, T = K = 298K Molar mass, M = 60 g We have, P = CRT or C = =,. =.. = 0.1M in 1006 g Wt. of solvent = ( x 60) = 1000g or 1Kg Hence, Molality = 0.1/1Kg = 0.1m. [COHSEM-2003] Q. A 9% solution of glucose is isotonic with a 3% solution of solute (x). Calculate the molecular mass of x. : Since the two solutions are isotonic, they must have same concentration in moles/litre. For glucose solution, concentration= (Given) = in litre = [Mol. mass of Glucose, C 6 H 12 O 6 = 180] For unknown substance, (x), concentration = (Given) = = [Mol. Mass of Unknown Substance, M =? ]

13 For isotonic solution, = M = M = 60 g mol -1 Q. State the law which governs the solubility of gases in liquids. 1mark [COHSEM-2002] Henry s law: It states that the mass of a gas dissolved in a given of the liquid at constant temperature is directly proportional to the pressure of the gas present in equilibrium with the liquid. Mathematically: m = k H p where k H = constant called Henry s law constant. Q. What are colligative properties? Name them. 1+2=3marks Certain properties of ideal solution depend only on the number of particles of the solute in a definite amount of the solvent and do not depend on the nature of solute. Such properties are called colligative properties. They are i. Relative lowering of vapour pressure ii. Osmotic pressure iii. Elevation in boiling point iv. Depression in freezing point

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