Solution & Colligative Properties Exercises

Solution & Colligative Properties  Marked questions are recommended for Revision. PART - I : SUBJECTIVE QUESTIONS Section (A) : General Introduction & types of solution A-1. What are the characteristics of the supersaturated solution. A-2. What do you mean by Hygroscopic compound ? A-3. Why are some solution processes exothermic whereas others are endothermic ? Section (B) : Concentration terms (Revision of mole) Commit to memory : Molarity = Molality = no. of moles of solute ; volume of solution (in L) no. of moles of solute × 1000 W(g) of solvent molarity  1000 (where d is density of solution in g/ml) 1000d − molarity  m.wt.of solute w msolute (g) =  100 ; w msolution (g) xsolute + xsolvent = 1 % B-1. Molality = xsolute = n (where, n is moles of solute, N is moles of solvent) n+N Calculate the concentration of NaOH solution in g/ml which has the same molarity as that of a solution of HCl of concentration 0.0365 g/ml. B-2. The density of 3M solution of sodium thiosulphate (Na2S2O3) is 1.58 g/ml. Calculate (i) amount of Na2S2O3 in % w/w (ii) mole fraction of Na2S2O3 (iii) molality of Na+ and S2O32– ions. B-3. Calculate the molality and molarity of a solution made by mixing equal volumes of 30% by weight of H2SO4 (density = 1.20 g/mL) and 70% by weight of H2SO4 (density = 1.60 g/mL). Section (C) : Vapour Pressure Commit to memory : * * * The partial pressure of vapours of X in equilibrium with X at a given temperature is called as its vapour pressure. (X is given pure solid or pure liquid) A gas is said to be saturated with vapours of a liquid if the partial pressure of liquid vapours is equal to its (saturated) vapour pressure. Partial pressure of vapours of X (pure solid or liquid) will remain equal to its vapour pressure till the vapours of X are in equilibrium with X. C-1. The vapour pressure of water at 80ºC is 355 torr. A 100 ml vessel contained water−saturated oxygen at 80º C, the total gas pressure being 760 torr. The contents of the vessel were pumped into a 50.0 ml, vessel at the same temperature. What were the partial pressures of oxygen and of water vapour and the total pressure in the final equilibrium state? Neglect the volume of any water which might condense. C-2. A vessel has nitrogen gas and water vapours in equilibrium with liquid water at a total pressure of 1 atm. The partial pressure of water vapours is 0.3 atm. The volume of this vessel is reduced to one third of the original volume, at the same temperature, then find total pressure of the system. (Neglect volume occupied by liquid water) Solution & Colligative Properties Section (D) : Solutions of Solid and Gases in Liquids Commit to memory : * Henry’s law : The solubility of gas in a liquid at a given temperature is directly proportional to its partial pressure above liquid in which it is dissolved. P = KHx (where x is mole fraction of unreacted, dissolved gas and P is its partial pressure above liquid.) D-1. The partial pressure of ethane over a solution containing 6.56 × 10 –3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas ? D-2. If N2 gas is bubbled through water at 293 K, how many millimoles of N 2 gas would dissolve in 300 mole of water, if N2 exerts a partial pressure of 1 bar. Given that Henry's law constant for N 2 at 293 K is 75.00 kbar. Section (E) : Immiscible Liquids Commit to memory : * If A and B are volatile immiscible liquids, then above their mixture. PT = PºA + PºB (PT = Total pressure above mixture of A and B, PºA = vapour pressure of A, PºB = vapour pressure of B) nA PAº = (nA and nB are moles of A and B in distillate) nB PBº o w A PA MA (wA and wB are masses of A and B in distillate, MA and MB are molar masses of A and B) = w B PBoMB E-1. A mixture of an organic liquid A and water distilled under one atmospheric pressure at 99.2ºC. How many grams of steam will be condensed to obtain 1.0 g of liquid A in the distillate ? (Vapour pressure of water at 99.2ºC is 739 mm Hg. Molecular weight of A = 123) E-2. Boiling point of a mixture of water and nitrobenzene is 99°C, the vapour pressure of water is 733 mm of Hg and the atmospheric pressure is 760 mm of Hg. The molecular weight of nitrobenzene is 123. Find the ratio of weights of the components of the distillate. Section (F) : Completely miscible liquids : Raoult’s law Commit to memory : Statement of Raoult's law (for volatile liq. mixture) : In solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in solution. pA  xA pA = xAPAº ....... (1) where pA = Partial vapour pressure of component A, xA = Mole fraction of component ‘A’ in solution, PAº = Vapour pressure of pure component ‘A’ at given temperature Hence PT = xAPAº + xBPBº ....... (2) (PT = Total pressure of vapour above solution, PBº = vapour pressure of pure component 'B' at given temperature.) 1 y y Thus, = A + Bo (for complete derivation, refer page no. 12 of sheet) o PT PA PB (yA = mole fraction of A in vapour phase above the solution and y B = mole fraction of B in vapour phase above the solution) F-1. Two liquids A and B form an ideal solution. At 300 K, the vapour pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm of Hg. At the same temperature, if one mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of A and B in their pure states. F-2. Two liquids, A and B, form an ideal solution. At the specified temperature, the vapour pressure of pure A is 200 mm Hg while that of pure B is 75 mm Hg. If the vapour over the mixture consists of 50 mol percent A, what is the mole percent A in the liquid ? F-3. Two solutions of A and B are available. The first is known to contain 1 mole of A and 3 moles of B and its total vapour pressure is 1.0 atm. The second is known to contain 2 moles of A and 2 moles of B; its vapour pressure is greater than 1 atm, but it is found that this total vapour pressure may be reduced to Solution & Colligative Properties 1 atm by the addition of 6 moles of C. The vapour pressure of pure C is 0.80 atm. Assuming ideal solutions and that all these data refer to 25°C, calculate the vapour pressure of pure A and of pure B. F-4. At 80oC, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg. Calculate the composition of a liquid in mole per cent which at 80oC is in equilibrium with the vapour containing 30 mole per cent of benzene. F-5. Vapour pressure of C6H6 and C7H8 mixture at 50ºC is given P (mm Hg) = 180X B + 90, where XB is the mole fraction of C6H6. A solution is prepared by mixing 12 mol benzene and 8 mol toluene and if vapours over this solution are removed and condensed into liquid and again brought to the temperature 50ºC, what would be mole fraction of C6H6 in the vapour state. (At. wt. of C = 12, H = 1) Section (G) : Non-ideal Solutions Commit to memory : * +ve deviation –ve deviation PT.exp > (xAPºA + xBPºB) PT.exp < (xAPºA + xBPºB) (where PT.exp is experimental total pressure above mixture of volatile liquids A and B). Hmix = +ve Hmix = –ve Vmix = +ve Vmix = –ve Smix = +ve Smix = +ve Gmix = –ve Gmix = –ve G-1. A non ideal solution was prepared by mixing 30 ml chloroform and 50 ml acetone. Comment on volume of mixture. G-2. Total vapour pressure of mixture of 1 mole of volatile component A (PºA = 100 mm Hg) and 3 mole of volatile component B (PºB = 80 mm Hg) is 90 mm Hg. Find out nature of solution and sign of entropy of solution. Section (H) : Degree of Ionisation/Dissociation for Weak Electrolytes Commit to memory : For dissociation/association i = 1 + (n – 1) (where n is total number of particles produced per solute particle after association/dissociation, i = vant Hoff factor,  = degree of dissociation / association. Moles of solute particles after association / dissociation of X Observed value of colligative property i= = Theoretical value of that colligative property Moles of X without association / dissociation * i = Theoretical molar mass of solute Observed molar mass of solute H-1. Complete the following table. Dissociation / association Solute reaction KCl H2SO4 CH3COOH (in water) CH3COOH (in benzene) Urea NaBr A 3A → A3 Degree of dissociation / association 1 1 0.2 0.5 n i 0.8 1 H-2. Calculate the percentage degree of dissociation of an electrolyte XY2 (Normal molar mass = 164) in water if the observed molar mass by measuring elevation in boiling point is 65.6. Solution & Colligative Properties Section (I) : Relative lowering of vapour pressure Commit to memory : * Pº − PS Pº − Ps n = xsolute; = Pº N Ps (where Pº = vapour pressure of pure solvent, Ps = partial pressure of vapour above solution, n = dissolved moles of solute, N = moles of solvent.) Pº − Ps msolute(g) Msolvent = × (where m = given mass in solution, M = molar mass.) Ps msolvent(g) Msolute RLVP = Pº − Ps M = ( molality ) × solvent Ps 1000 If solution is of single solute and it gets associated or dissociated; Pº − Ps i.n Pº − Ps M = ; that is = i × (molality) × solvent N Ps Ps 1000 I-1. Twenty grams of a solute are added to 100 g of water at 25ºC. The vapour pressure of pure water is 23.76 mmHg; the vapour pressure of the solution is 22.41 Torr. (a) Calculate the molar mass of the solute. (b) What mass of this solute is required in 100 g of water to reduce the vapour pressure to one-half the value for pure water ? l-2. The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution containing 7 g salt per 100 g of water at 100ºC is 70%. If the vapour pressure of water at 100ºC is 760 mm of Hg, calculate the vapour pressure of the solution. Section (J) : Elevation of Boiling Point & Depression of Freezing Point Commit to memory : * Tb = i × Kb × m (where Tb = elevation in boiling point of solution, Kb = ebullioscopic constant, m = molality of single solute, Tb = boiling point of solvent (in K). 2 Kb = 2 RTb M RTb = 1000  Hvap 1000  L vap (M = Molar mass of solvent) Tf = i × Kf × m (where Tf = depression in freezing point of solution, Kf = cryoscopic constant, Tf = freezing point of solvent (in K). 2 Kf = 2 RTf M RTf = 1000  Hfusion 1000  L fusion J-1. (a) A solution containing 0.5 g of naphthalene in 50 g CCl 4 yield a boiling point elevation of 0.4 K, while a solution of 0.6 g of an unknown solute in the same mass of the solvent gives a boiling point elevation of 0.65 K. Find the molar mass of the unknown solute. (b) The boiling point of a solution of 0.1 g of a substance in 16 g of ether was found to be 0.100ºC higher that of pure ether. What is the molecular mass of the substance. Kb(ether) = 2.16 K kg mol–1. J-2. The amount of benzene that will separate out (in grams) if a solution containing 7.32 g of triphenylmethane in 1000 g of benzene is cooled to a temperature which is 0.2°C below the freezing point of benzene ? (Kf = 5.12 K-Kg/mol) J-3. The boiling point of a solution of 5 g of sulphur in 100 g of carbon disulphide is 0.474ºC above that of pure solvent. Determine the molecular formula of sulphur in this solvent. The boiling point of pure carbon disulphide is 47ºC and its heat of vaporisation is 84 calories per gram. [Hint : Kb = RTb2 2  (320)2 = = 2.438] 1000  84 1000Lv J-4. A 0.01 molal solution of ammonia freezes at –0.02ºC. Calculate the van’t Hoff factor, i and the percentage dissociation of ammonia in water. (K f(H2O) ) = 1.86 deg molal–1. Solution & Colligative Properties Section (K) : Osmotic Pressure Commit to memory : *  = CRT = n RT (n = Total moles of solute particles in solution, V = Total volume of solution in L) V K-1. (a) Predict the osmotic pressure order for the following (assume salts are 100% dissociated). I 0.1 M urea II 0.1 M NaCl III 0.1 M Na2SO4 IV 0.1 M Na3PO4 (b) If equal volumes of all these solutions are mixed then calculate the osmotic pressure of the net resultant solution obtained at 300K. K-2. A solution containing 3.00 g of calcium nitrate in 100 c.c. of solution had an osmotic pressure of 11.2 atmosphere at 12°C. Calculate the degree of ionisation of calcium nitrate at this dilution and temperature. K-3. 17.4% (w/v) K2SO4 solution at 27ºC is isotonic with 5.85% (w/v) NaCl solution at 27ºC. If NaCl is 100% ionised, what is % ionisation of K2SO4 in aq. solution ? K-4. At 2ºC the osmotic pressure of a urea solution is found to be 500 mm of Hg. The solution is diluted and the temperature is raised to 27ºC, when the osmotic pressure is found to be 109.09 mm of Hg. Determine the extent of dilution. PART - II : ONLY ONE OPTION CORRECT TYPE Section (A) : General Introduction & types of solution A-1. Which statement best explains the meaning of the phrase “like dissolves like “ ? (A) A Solute will easily dissolve a solute of similar mass (B) A solvent and solute with similar intermolecular forces will readily form a solution (C) The only true solutions are formed when water dissolves a non-polar solute (D) The only true solutions are formed when water dissolves a polar solute A-2. An ionic compound that attracts atmospheric water so strongly that a hydrate is formed is said to be : (A) Dilute (B) Hygroscopic (C) Immiscible (D) Miscible Section (B) : Concentration terms (Revision of mole) B-1. Persons are medically considered to have lead poisoning if they have a concentration greater than 10 micrograms of lead per decilitre of blood. Concentration in parts per billion is : (A) 1000 (B) 100 (C) 10 (D) 1 Section (C) : Vapour Pressure C-1. A liquid is kept in a closed vessel. If a glass plate (negligible mass) with a small hole is kept on top of the liquid surface, then the vapour pressure of the liquid in the vessel is : (A) More than what would be if the glass plate were removed (B) Same as what would be if the glass plate were removed (C) Less than what would be if the glass plate were removed (D) Cannot be predicted C-2. The vapour pressure of water depends upon : (A) Surface area of container (C) Temperature C-3. (B) Volume of container (D) All Among the following substances, the lowest vapour pressure is exerted by : (A) Water (B) Mercury (C) Acetone (D) Ethanol C-4. At higher altitudes, water boils at temperature < 100ºC because (A) temperature of higher altitudes is low (B) atmospheric pressure is low (C) the proportion of heavy water increases (D) atmospheric pressure becomes more. Solution & Colligative Properties Section (D) : Solutions of Solid and Gases in Liquids D-1. The solubility of gases in liquids : (A) increases with increase in pressure and temperature (B) decreases with increase in pressure and temperature (C) Increases with increase in pressure and decrease in temperature (D) decreases with increase in pressure and increase in temperature D-2. Which of the following curves represents theHenry’slaw? (A) (B) (C) (D) D-3. According to Henry’s law, the solubility of a gas in a given volume of liquid increases with increase in : (A) Temperature (B) Pressure (C) Both (A) and (B) (D) None of these D-4. Some of the following gases are soluble in water due to formation of their ions : I : CO2 ; II : NH3 ; III : HCl ; IV : CH4 ; V : H2 Water insoluble gases can be : (A) I, IV , V (B) I, V (C) I, II, III (D) IV, V D-5. The solubility of N2(g) in water exposed to the atmosphere, when its partial pressure is 593 mm is 5.3 × 10–4 M. Its solubility at 760 mm and at the same temperature is : (A) 4.1 × 10–4 M (B) 6.8 × 10–4 M (C) 1500 M (D) 2400 M Section (E) : Immiscible Liquids E-1. When a liquid that is immiscible with water was steam distilled at 95.2ºC at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at 95.2ºC, what is the molar mass of liquid? (A) 7.975 g/mol (B) 166 g/mol (C) 145.8 g/mol (D) None of these Section (F) : Completely miscible liquids : Raoult’s law F-1. For a binary ideal liquid solution, the total pressure of the solution is given as : (A) Ptotal = PºA + (PºA – PºB) XB (B) Ptotal = PºB + (PºA – PºB) XA (C) Ptotal = PºB + (PºB – PºA) XA (D) Ptotal = PºB + (PºB – PºA) XB F-2. An ideal solution contains two volatile liquids A (pº = 100 torr) and B (pº = 200 torr). If mixture contain 1 mole of A and 4 mole of B then total vapour pressure of the distillate is: (A) 150 (B) 180 (C) 188.88 (D) 198.88 F-3. At 323 K, the vapour pressure in millimeters of mercury of a methanol-ethanol solution is represented by the equation p = 120 XA + 140, where XA is the mole fraction of methanol.Then the value of p lim A is x A →1 X A (A) 250 mm (B) 140 mm (C) 260 mm (D) 20 mm F-4. Given at 350 K pA° = 300 torr and pB° = 800 torr, the composition of the mixture having a normal boiling point of 350 K is : (A) XA = 0.08 (B) XA = 0.06 (C) XA = 0.04 (D) XA = 0.02 F-5. Two liquids A and B have PºA and PºB in the ratio of 1 : 3 and the ratio of number of moles of A and B in liquid phase are 1 : 3 then mole fraction of ‘A’ in vapour phase in equilibrium with the solution is equal to: (A) 0.1 (B) 0.2 (C) 0.5 (D) 1.0 Solution & Colligative Properties Section (G) : Non-ideal Solutions G-1. The vapour pressure of the solution of two liquids A(pº = 80 mm) and B(pº = 120 mm) is found to be 100 mm when xA = 0.4. The result shows that (A) solution exhibits ideal behaviour (B) solution shows positive deviations (C) solution shows negative deviations (D) solution will show positive deviations for lower concentration and negative deviations for higher concentrations. G-2. Consider a binary mixture of volatile liquids. If at XA = 0.4 the vapour pressure of solution is 580 torr then the mixture could be (pAº = 300 torr, pB° = 800 torr) : (A) CHCl3 – CH3COCH3 (B) C6H5Cl – C6H5Br (C) C6H6 – C6H5CH3 (D) nC6H14 – nC7H16 G-3. A solution of sulphuric acid in water exhibits : (A) Negative deviations from Raoult’s law (C) Ideal properties (B) Positive deviations from Raoult’s law (D) The applicability of Henry’s law G-4. When KCl dissolves in water (assume endothermic dissolution), then : (A) H = + ve, S = + ve, G = + ve (B) H = + ve, S = – ve, G = – ve (C) H = + ve, S = + ve, G = – ve (D) H = – ve, S = – ve, G = + ve G-5. The dissolving process is exothermic when : (A) The energy released in solvation exceeds the energy used in breaking up solute-solute and solventsolvent interactions. (B) The energy used in solvation exceeds the energy released in breaking up solute-solute and solventsolvent interactions. (C) The energy released in solvation is about the same as the energy used in breaking up solute-solute and solvent-solvent interactions. (D) The energy used in solvation is about the same as the energy used in breaking up solute-solute and solvent-solvent interactions. G-6. Which of the following is less than zero for ideal solutions ? (A) Hmix (B) Vmix (C) Gmix (D) Smix Section (H) : Degree of Ionisation/Dissociation for Weak Electrolytes H-1. One mole of a solute A is dissolved in a given volume of solvent. The association of the solute take place as follows : nA An If  is the degree of association of A, the van’t Hoff factor i is expressed as :  1–  +  n (A) i = 1–  (B) i = 1 + (C) i = (D) i = 1 n 1 H-2. The degree of dissociation of an electrolyte is  and its van’t Hoff factor is i. The number of ions obtained by complete dissociation of 1 molecule of the electrolyte is : i+  –1 i + 1+  i–1 (A) (B) i –  – 1 (C) (D)  1–   H-3. If Mnormal is the normal molecular mass and  is the degree of ionization of K3[Fe(CN)6], then the abnormal molecular mass of the complex in the solution will be : (A) Mnormal (1 + 2)–1 (B) Mnormal (1 + 3)–1 (C) Mnormal (1 + )–1 (D) equal to Mnormal H-4. A complex containing K+, Pt(IV) and Cl– is 100% ionised giving i = 3. Thus, complex is : (A) K2[PtCl4] (B) K2[PtCl6] (C) K3[PtCl5] (D) K[PtCl3] Solution & Colligative Properties Section (I) : Relative lowering of vapour pressure I-1. If P0 and P are the vapour pressures of a solvent and its solution respectively and N 1 and N2 are the mole fractions of the solvent and non-volatile solute respectively, then correct relation is : (A) P = P0N2 (B) P = P0N1 (C) P0 = PN1 (D) P = P0(N1/N2) I-2. Relative decrease in vapour pressure of an aqueous NaCl is 0.167. Number of moles of NaCl present in 180g of H2O is : (A) 2 mol (B) 1 mol (C) 3 mol (D) 4 mol I-3. The vapour pressure of pure benzene, C6H6 at 50°C is 268 Torr. How many moles of non-volatile solute per mol of benzene is required to prepare a solution of benzene having a vapour pressure of 167 Torr at 50°C? (A) 0.377 (B) 0.605 (C) 0.623 (D) 0.395 I-4. If relative decrease in vapour pressure is 0.4 for a solution containing 1 mol NaCl in 3 mol H 2O, NaCl is .... % ionised. (A) 60% (B) 50% (C) 100% (D) 40% I-5. The vapour pressure of a solution of a non-volatile solute B in a solvent A is 95% of the vapour pressure of the solvent at the same temperature. If the molecular weight of the solvent is 0.3 times the molecular weight of the solute, what is the ratio of weight of solvent to solute. (A) 0.15 (B) 5.7 (C) 0.2 (D) none of these I-6. Relative decrease in vapour pressure of an aqueous solution containing 2 moles [Cu(NH 3)3Cl]Cl in 3 moles H2O is 0.50. On reaction with AgNO3, this solution will form(assuming no change in degree of ionisation of substance on adding AgNO3) (A) 1 mol AgCl (B) 0.25 mol AgCl (C) 0.5 mol AgCl (D) 0.40 mol AgCl Section (J) : Elevation of Boiling Point & Depression of Freezing Point J-1. An aqueous solution containing 1g of urea boils at 100.25°C. The aqueous solution containing 3 g of glucose in the same volume will boil at (A) 100.75 °C (B) 100.5 °C J-2. (C) 100°C (D) 100.25°C Elevation in boiling point was 0.52 °C when 6 g of a compound x was dissolved in 100 g of water. Molecular weight of x is : (K = 0.52 kg mol–1) (A) 120 (B) 60 (C) 100 (D) 342 J-3. A solute’S’ undergoes a reversible trimerization when dissolved in a certain solvent. The boiling point elevation of its 0.1 molal solution was found to be identical to the boiling point elevation in case of a 0.08 molal solution of a solute which neither undergoes association nor dissociation. To what percent had the solute ‘S’ undergone trimerization? (A) 30% (B) 40% (C) 50% (D) 60% J-4. A complex of iron and cyanide ions is 100% ionised at 1m (molal). If its elevation in b.p. is 2.08. Then the complex is (Kb = 0.52° mol–1 kg) : (A) K3[Fe(CN)6] (B) Fe(CN)2 (C) K4[Fe(CN)6] (D) Fe(CN)4 J-5. PtCl4.6H2O can exist as a hydrated complex 1 molal aq. solution has depression in freezing point of 3.72°. Assume 100% ionisation and Kf(H2O) = 1.86° mol–1 kg, then complex is (A) [Pt(H2O)6]Cl4 (B) [Pt(H2O)4Cl2]Cl2 . 2H2O (C) [Pt(H2O)3Cl3]Cl . 3H2O (D) [Pt(H2O)2Cl4] . 4H2O J-6. How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 104°C between boiling point and freezing point. (Kf = 1.86 K Kg mol–1, Kb = 0.52 K Kg mol–1) (A) 1.68 (B) 3.36 (C) 8.40 (D) 0.840 Solution & Colligative Properties J-7. Which of the following has been arranged in order of decreasing freezing point? (A) 0.05 M KNO3 > 0.04 M CaCl2 > 0.140 M sugar > 0.075 M CuSO4 (B) 0.04 M BaCl2 > 0.140 M sucrose > 0.075 M CuSO4 > 0.05 M KNO3 (C) 0.075 M CuSO4 > 0.140 M sucrose > 0.04 M BaCl2 > 0.05 M KNO3 (D) 0.075 M CuSO4 > 0.05 M NaNO3 > 0.140 M sucrose > 0.04 M BaCl2 J-8. A solution of x moles of sucrose in 100 grams of water freezes at −0.2ºC. As ice separates the freezing point goes down to – 0.25ºC. How many grams of ice would have separated? (A) 18 grams (B) 20 grams (C) 25 grams (D) 23 grams J-9. Sea water is found to contain 5.85 % NaCl and 9.50% MgCl 2 by weight of solution. Calculate its normal boiling point assuming 80% ionisation for NaCl and 50% ionisation of MgCl 2 (Kb(H2O) = 0.51 kgmol–1 K). (A) Tb = 101.9°C (B) Tb = 102.3°C (C) Tb = 108.5°C (D) Tb = 110.3°C Section (K) : Osmotic Pressure K-1. The relationship between osmotic pressure at 273 K when 10 g glucose (P 1), 10 g urea (P2) and 10 g sucrose (P3) are dissolved in 250 mL of water is : (A) P1 > P2 > P3 (B) P3 > P1 > P2 (C) P2 > P1 > P3 (D) P2 > P3 > P1 K-2. Osmotic pressure of blood is 7.40 atm at 27°C. Number of moles of glucose to be used per litre for an intravenous injection that is to have the same osmotic pressure as blood is : (A) 0.3 (B) 0.2 (C) 0.1 (D) 0.4 K-3. The total concentration of dissolved particles inside red blood cells is approximately 0.30 M and the membrane surrounding the cells is semipermeable. What would the osmotic pressure (in atmosphere) inside the cells become if the cells were removed from the blood plasma and placed in pure water at 298 K ? (A) 7.34 atm (B) 1.78 atm (C) 2.34 atm (D) 0.74 atm K-4. Consider following cases : I : 2M CH3COOH solution in benzene at 27°C where there is dimer formation to the extent of 100% II : 0.5 M KCl aq. solution at 27°C, which ionises 100% Which is/are true statements(s) : (A) both are isotonic (B) I is hypertonic (C) II is hypotonic (D) none is correct PART - III : MATCH THE COLUMN 1. If at a particular temperature, the density of 18 M H2SO4 is 1.8 g cm–3. Then : Column – I Column – II (A) Molality (p) 0.1 (B) % concentration by wt. of solute (q) 0.9 (C) mole fraction of H2SO4 (r) 500 (D) mole fraction of H2O (s) 98 2. Match the following columnColumn – I (A) Acetone + CHCl3 (B) Ethanol + Water (C) C2H5Br + C2H5I (D) Acetone + Benzene (p) (q) (r) (s) (t) Column – II Smix. > 0 Vmix. > 0 Hmix. < 0 Maximum boiling azeotropes Minimum boiling azeotropes Solution & Colligative Properties  Marked questions are recommended for Revision. PART - I : ONLY ONE OPTION CORRECT TYPE 1. All of the water in a 0.20 M solution of NaCl was evaporated and a 0.150 mol of NaCl was obtained. What was the original volume of the sample ? (A) 30 mL (B) 333 mL (C) 750 mL (D) 1000 mL 2. A 20.0 mL sample of CuSO4 solution was evaporated to dryness, leaving 0.967 g of residue. What was the molarity of the original solution ? (Cu = 63.5) (A) 48.4 M (B) 0.0207 M (C) 0.0484 M (D) 0.303 M 3. The vapour pressure of water at 20°C is 17.54 mmHg. What will be the vapour pressure of the water in the apparatus shown after the piston is lowered, decreasing the volume of the gas above the liquid to one half of its initial volume (assume temperature constant). (A) 8.77 mmHg (B) 17.54 mmHg (C) 35.08 mmHg (D) between 8.77 and 17.54 mmHg 4. A sample of air is saturated with benzene (vapor pressure = 100 mm Hg at 298 K) at 298K, 750mm Hg pressure. If it is isothermally compressed to one third of its initial volume, the final pressure of the system is (A) 2250 torr (B) 2150 torr (C) 2050 torr (D) 1950 torr 5. Water and chlorobenzene are immiscible liquids. Their mixture boils at 89ºC under a reduced pressure of 7.7 × 104 Pa. The vapour pressure of pure water at 89ºC is 7 × 10 4 Pa. Weight percent of chlorobenzene in the distillate is: (A) 50 (B) 60 (C) 78.3 (D) 38.46 6. If two liquids A (PºA =100 torr) and B (PºB = 200 torr) are completely immiscible with each other, each one will behave independently of the other, are present in a closed vessel. The total vapour pressure of the system will be: (A) less than 100 torr (B) less than 200 torr (C) between 100 to 200 torr (D) 300 torr 7. Given P-x curve for a non-ideal liquid mixture (Fig.). Identify the correct T-x curve for the same mixture. (A) (B) (C) (D) 8. If vapour pressures of pure liquids ‘A’ & ‘B’ are 300 and 800 torr respectively at 25°C. When these two liquids are mixed at this temperature to form a solution in which mole percentage of ‘B’ is 92, then the total vapour pressure is observed to be 0.95 atm. Which of the following is true for this solution. (A) Vmix > 0 (B) Hmix < 0 (C) Vmix = 0 (D) Hmix = 0 9. Barium ions, CN¯ and Co2+ form an ionic complex. If that complex is supposed to be 75% ionised in water with vant Hoff factor ‘i’ equal to four, then the coordination number of Co2+ in the complex can be: (A) Six (B) Five (C) Four (D) Six and Four both Solution & Colligative Properties 10. In the following aqueous solutions (a) 1 m sucrose, (b) 1 m potassium ferricyanide and (c) 1 m potassium sulphate Maximum value of vapour pressure of solution is that of : (A) a (B) b (C) c (D) equal 11. When only a little quantity of HgCl2(s) is added to excess KI(aq) to obtain a clear solution, which of the following is true for this solution? (no volume change on mixing). The reaction is 4KI(aq.) + HgCl2(s) —→ K2[HgI4] (aq.) + 2KCl (aq.) (A) Its boiling and freezing points remain same (B) Its boiling point is lowered (C) Its vapour pressure become lower (D) Its boiling point is raised (E) Its freezing point is lowered. 12. x mole of KCI and y mole of BaCl2 are both dissolved in 1 kg of water. Given that x + y = 0.1 and Kf for water is 1.85 K/molal, what is the observed range of Tf, if the ratio of x to y is varied ? (A) 0.37º to 0.555º (B) 0.185º to 0.93º (C) 0.56º to 0.93º (D) 0.37º to 0.93º 13. FeCl3 on reaction with K4[Fe(CN)6] in aq. solution gives blue colour. These are separated by a semipermeable membrane PQ as shown. Due to osmosis there is(A) blue colour formation in side X (B) blue colour formation in side Y (C) blue colour formation in both of the sides X and Y (D) no blue colour formation 14. Two beakers, one containing 20 ml of a 0.05 M aqueous solution of a non volatile, non electrolyte and the other, the same volume of 0.03 M aqueous solution of NaCl, are placed side by side in a closed enclosure. What are the volumes in the two beakers when equilibrium is attained ? Volume of the solution in the first and second beaker are respectively. (A) 21.8 ml and 18.2 mL (B) 18.2 mL and 21.8 mL (C) 20 mL and 20 mL (D) 17.1 mL and 22.9 mL PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE 1. What volume of 98% sulphuric acid (in ml) should be mixed with water to obtain 200 mL of 15% solution of sulphuric acid by weight ? Given density of H 2O = 1.00 g cm–3, sulphuric acid (98%) = 1.88 g cm–3 and sulphuric acid (15%) = 1.12 g cm–3. 2. At 300 K, 40 mL of O3 (g) dissolves in 100 g of water at 1.0 atm. What mass of ozone (in gram) dissolved in 1600 g of water at a pressure of 4.0 atm at 300 K? 3. An ideal aqueous solution containing liquid A(M.Wt. = 128) 64% by weight has a vapour pressure of 145 mm Hg. If the vapour pressure of A is x mm of Hg and that of water is 155 mm Hg at the same temperature. Then find x/5. The solutions is ideal. 4. A and B form ideal solutions; at 50ºC, PAº is half PBº. A solution containing 0.2 mole of A and 0.8 mole of B has a normal bolling point of 50ºC. Find 18 × PBº. (PBº is in atm) 5. The vapour pressure of pure liquid A at 300 K is 577 Torr and that of pure liquid B is 390 Torr. These two compounds form ideal liquid and gaseous mixtures. Consider the equilibrium composition of a mixture in which the mole fraction of A in the vapour is 0.35. Find the mole % of A in liquid. 6. How many facts related to CHCl3 + ethylmethylketone solution are correct ? (a) It shows negative derivation. (b) It forms maximum boiling azeotropic mixture (c) S > 0 (d) G < 0 (e) Components can be separated by fractional distillation. 7. How many of the following solutions show negative deviation from Raoult's Law ? Liquid A + Liquid B (i) (CH3)2CO + CS2 (ii) CCl4 + C6H6 (iii) CCl4 + CHCl3 (iv) H2O + C2H5OH Solution & Colligative Properties (v) (vi) (vii) (viii) (ix) (C2H5)2O CH3COOH C6H5NH2 C6H5Cl Cyclohexane + + + + + (CH3)2CO C5H5N (pyridine) (CH3)2CO C6H5Br Ethanol 8. 0.1 mole XY2 is dissolved in 2L water, where it ionizes to give X 2+ and Y22–. Observed osmotic pressure is 3 atm. Molar mass of X is 24 and Y is 32. Find M observed + 2i (where Mobserved is observed molar mass of XY2). (Use R = 1/12 L-atm/mol.K and temperature is 87°C) 9. How many grams of sucrose (C12H22O11) must be dissolved in 90 g of water to produce a solution over which the relative humidity is 80%? Assume the solution is ideal. Give your answer after dividing by 10. 10. 1.22 g of a monobasic acid is dissolved in 100 g of benzene. Boiling point of solution increases by 0.13ºC with respect to pure benzene. Find the molecular mass of acid in benzene solvent (in u). Report your answer after dividing it by 100 and Round it off to nearest integer. (Kb of benzene = 2.6 K kg mol–1). 11. 1 g of a monobasic acid dissolved in 200 g of water lowers the freezing point by 0.186ºC. On the other hand when 1 g of the same acid is dissolved in water so as to make the solution 200 mL, this solution requires 125 mL of 0.1 M NaOH for complete neutralization. Calculate % dissociation of acid ? (K f K − kg =1.86 ) mol 12. At 27ºC, a 1.2% solution (wt./vol.) of glucose is isotonic with 4.0 g/litre of solution of solute X. Find the molar mass of X, if the molar mass of glucose is 180. (R = 0.082 L atm mol –1 K–1, Molar mass of glucose = 180 g/mole) 13. 10 g of solute A and 20 g of solute B both are dissolved in 500 ml. of water. The solution has the same osmotic pressure as 6.67 g of A and 30 g of B are dissolved in the same volume of water at the same temperature. If the ratio of molar masses of A and B is x/y, find x + y. PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE 1. We have 100 mL of 0.1 M KCl solution. To make it 0.2 M, (A) evaporate 50 mL water (B) evaporate 50 mL solution (C) add 0.1 mol KCl (D) add 0.01 mol KCl 2. Which of the following concentration factors can be calculated if the mole fraction and density of an aqueous solution of HCl are known ? (A) Molality (B) Molarity (C) Percent by mass (D) Normality 3. The vapour pressure of a dilute solution of a solute is influenced by : (A) Temperature of solution (B) Mole fraction of solute (C) M.pt. of solute (D) Degree of dissociation of solute 4. According to Henry’s law, the partial pressure of gas (Pg) is directly proportional to mole fraction of gas in dissolved state , i.e., Pgas = KH . Xgas where KH is Henry’s constant. Which are correct ? (A) KH is characteristic constant for a given gas–solvent system (B) Higher is the value of KH , lower is solubility of gas for a given partial pressure of gas (C) KH has temperature dependence (D) KH increases with temperature 5. Select correct statements : (A) Gases which have high value of van der Waals constant ‘a’ are easily liquefied (B) Easily liquefiable gases are water soluble (C) Gases which forms ions in a solvent are soluble in that solvent (D) Under same conditions, NH3 has low solubility in water than that of CO2. 6. Two liquids X and Y are perfectly immiscible. If X and Y have molecular masses in ratio 1 : 2, the total vapour pressure of a mixture of X and Y prepared in weight ratio 2 : 3 should be (P x0 = 400 torr, Py0 = 200 torr) (A) 600 torr (B) 400 torr (C) 800 torr (D) 1000 torr Solution & Colligative Properties 7. Which is/are true about ideal solutions ? (A) The volume change on mixing is zero (C) The entropy of mixing is zero (B) The enthalpy of mixing is zero (D) The enthalpy of mixing is negative 8. At 40ºC, vapour pressure in Torr of methanol and ethanol solution is P = 119x + 135 where x is the mole fraction of methanol. Hence (A) vapour pressure of pure methanol is 119 Torr (B) vapour pressure of pure ethanol is 135 Torr (C) vapour pressure of equimolar mixture of each is 127 Torr (D) mixture is completely immiscible 9. Which is/are correct statement(s) ? (A) When mixture is more volatile than both pure components, there is positive deviation from Raoult’s law. (B) When mixture is less volatile than both pure components, there is negative deviation from Raoult’s law. (C) Ethanol and water form ideal solution (D) CHCl3 and water form ideal solution 10. At 35°C, the vapour pressure of CS2 is 512 mm Hg, and of acetone is 344 mm Hg. A solution of CS2 and acetone in which the mole fraction of CS2 is 0.25, has a total vapour pressure of 600 mm Hg. Which of the following statements is/are correct ? (A) A mixture of 100 mL of acetone and 100 mL of CS2 has a volume of 200 mL (B) When acetone and CS2 are mixed at 35°C, heat must be absorbed in order to produce a solution at 35°C. (C) Process of mixing is exothermic (D) Entropy of mixing is zero 11. For the given electrolyte AxBy, the degree of dissociation '' can be given as i −1 1− i (A)  = (B) i = (1 – ) + x + y (C)  = (D) None x + y −1 1− x − y 12. In which of the following pairs of solutions will the values of the vant Hoff factor be the same? (A) 0.05 M K4 [Fe(CN)6] and 0.10 M FeSO4 (B) 0.10 M K4[Fe(CN)6] and 0.05 M FeSO4 (NH4)2SO4. 6H2O (C) 0.20 M NaCl and 0.10 M BaCl2 (D) 0.05 M FeSO4 (NH4)2SO4.6H2O and 0.02 M KCl.MgCl2.6H2O 13. 2 g of non-volatile hydrocarbon solute dissolved in 100 g of hypothetical organic solvent (molar mass = 50) was found to lower vapour pressure from 75.50 to 75 mm of Hg at 20°C. Given that hydrocarbon contains 96% of C. Then which of the following are true ? (A) molecular wt of solute = 150 (B) molecular formula = C12H6 (C) molecular wt. of solute = 132 (D) none of these 14. In the depression of freezing point experiment, it is found that the : (A) Vapour pressure of the solution is less than that of pure solvent (B) Vapour pressure of the solution is more than that of pure solvent (C) Only solute molecules solidify at the freezing point (D) Only solvent molecules solidify at the freezing point Consider following solutions : I : 1 M aqueous glucose solution II : 1M aqueous sodium chloride solution III : 1M aqueous ammonium phosphate solution IV : 1M benzoic acid in benzene Select correct statements for the above solutions : (A) All are isotonic solutions (B) III is hypertonic of I, II and IV (C) IV is hypotonic of I, II and III (D) II is hypotonic of III but hypertonic of I and IV 15. Solution & Colligative Properties 16. Which facts are true when we use van’t Hoff equation PV = nST for osmotic pressure P of dilute solutions? (A) The equation is identical to that of ideal gas equation (B) The solute particles in solution are analogous to the gas molecules and the solvent is analagous to the empty space between the gas molecules (C) Solute molecules are dispersed in the solvent the way the gas molecules are dispersed in empty space (D) The equation is not identical to that of ideal gas equation PART - IV : COMPREHENSIONS Read the following passage carefully and answer the questions. Comprehension # 1 PA = x APA & PB = xBPB PT = x APA + xBPB (Note : PA > PB ; A is more volatile than B) Vapour pressure of mixtures of Benzene (C6H6 ) & toluene (C7H8) at 50ºC are given by PM = 179 XB + 92, where XB is mole fraction of C6H6. 1. 2. What is vapour pressure of pure liquids ? (A) PB = 92 mm, PT = 179 mm (B) PB = 271 mm, PT = 92 mm (C) PB = 180 mm, PT = 91 mm (D) none of these Vapour pressure of liquid mixture obtained by mixing 936 g C6H6 & 736 g tolene is : (A) 300 mm Hg (B) 250 mm Hg (C) 199.4 mm Hg (D) 180.6 mm Hg Comprehension # 2 Answer the questions (given below) which are based on the following diagram. Consider some facts about the above phase diagram: Vapour pressure diagram for real solutions of two liquids A and B that exhibit a positive deviation from Raoult’s law. The vapour pressure of both A and B are greater than predicted by Raoult’s law. The dashed lines represented the plots for ideal solutions. 3. A : This is observed when A...B attractions are greater than average of A...A and B...B attraction: B : Hmix = +ve, Vmix = +ve C : Boiling point is smaller than expected such that vaporisation is increased D : Mixture can form azeotropic mixture Select correct facts (A) A, B, C (B) B, C, D (C) A, C, D (D) A, B, C, D Solution & Colligative Properties 4. Total vapour pressure of mixture of 1 mol of volatile component A (p° A = 100 mm Hg) and 3 mol of volatile component B (p°B = 60 mm Hg) is 75 mm. For such case : (A) There is positive deviation from Raoult’s law (B) Boiling point has been lowered (C) Force of attraction between A and B is smaller than that between A and A or between B and B. (D) All the above statements are correct. Comprehension # 3 Addition of non-volatile solute to a solvent always increases the colligative properties such as osmotic pressure, P, Tb and Tf. All these colligative properties are directly proportional to molality if solutions are dilute. The increases in colligative properties on addition of non-volatile solute is due to increase in number of solute particles. 5. For different aqueous solutions of 0.1 M NaCI, 0.1 M urea, 0.1 M Na 2SO4 and 0.1 M Na3PO4 solution at 27ºC, the correct statements are : (Consider the solutions to be dilute) 1. The order of osmotic pressure is, NaCl > Na2SO4 > Na3PO4 > urea T 2.  = b × ST for urea solution Kb 3. Addition of salt on ice increases its melting point 4. Addition of salt on ice brings in melting of ice earlier (A) 2, 3, 4 (B) 2, 4 (C) 1, 2, 3 (D) 3, 4 6. 1 g mixture of glucose and urea present in 250 mL aqueous solution shows the osmotic pressure of 0.74 atm at 27ºC. Assuming solution to be dilute, which are correct ? 1. Percentage of urea in mixture is 17.6. 2. Relative lowering in vapour pressure of this solution is 5.41 × 10 –4. 3. The solution will boil at 100.015ºC, if Kb of water is 0.5 K molality–1. 4. If glucose is replaced by same amount of sucrose, the solution will show higher osmotic pressure at 27ºC. 5. If glucose is replaced by same amount of NaCl, the solution will show lower osmotic pressure at 27ºC. (A) 1, 2, 3 (B) 1, 2, 3, 5 (C) 2, 4, 5 (D) 1, 4, 5 (i) Ideal solution (a) pT = pºA x A + pBº xB (P) Vapour Pressure Comprehension # 4 Let us consider a binary solution of two volatile liquids 'A' and 'B', when taken in a closed container. Both the components would evaporate and an equilibrium would be established between vapour phase and liquid phase. Let the total vapour pressure at this stage be p total and pA and pB are partial pressures of A and B. Mole fractions of these components in liquid solution are x A and xB, that of vapour phase are yA and yB respectively pºA & pBº are vapour pressure of pure A & pure B. Column-I Column-II Column-III O (ii) Non ideal solution (Positive deviation) A—A (b) interactions B—B interactions (iii) Non ideal solution (negative deviation) (c) Form azeotropic mixture  A—B interactions x Liquid Liquid + vapour y vapour Mole fraction of A (Q) PT = P°A + P°B (R) Ssurrounding = +ve V.P 1 (iv) Immiscible liquids (d) solution having vapours of fixed composition (S) Vapour Pressure of A Solution & Colligative Properties O 7. 8. 9. 10. º PA Mole fraction of A For Hexane + Heptane → solution correct set is : (A) (i) (a) (R) (B) (ii) (b) (P) (C) (ii) (d) (R) (D) (i) (a) (P) For water + ethanol → solution correct set is : (A) (ii) (b) (Q) (B) (ii) (c) (S) (C) (iv) (d) (S) (D) (iv) (c) (R) For water + H2SO4 → solution correct set is (A) (i) (a) (P) (B) (ii) (b) (Q) (C) (iii) (b) (R) (D) (iv) (c) (P) For a mixture of water and chlorobenzene correct set is (A) (i) (a) (P) (B) (ii) (b) (Q) (C) (iii) (b) (R) (D) (iv) (d) (Q) 1 * Marked Questions may have more than one correct option. PART - I : ADVANCED / PROBLEMS (PREVIOUS YEARS) 1. 2. In the depression of freezing point experiment, it is found that : I. The vapour pressure of the solution is less than that of pure solvent. II. The vapour pressure of the solution is more than that of pure solvent. III. Only solute molecules solidify at the freezing point. IV. Only solvent molecules solidify at the freezing point. (A) I, II (B) II, III (C) I, IV (D) I, II, III. The van’t Hoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation is : (A) 91.3% cm3 (B) 87% (C) 100% (D) 74% 10–3 3. To 500 of water, 3.0 × kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression in freezing point? kf and density of water are 1.86 K kgmole–1 and 0.997 g cm–3 respectively (A) 0.186 K (B) 0.228 K (C) 0.372 K (D) 0.556 K 4. During depression of freezing point in a solution, the following are in equilibrium : (A) Liquid solvent-solid solvent (B) Liquid solvent-solid solute (B) Liquid solute-solid solute (D) Liquid solute-solid solvent 5. A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution of glucose at same temperature. The apparent percentage dissociation of Na2SO4 is : (A) 25% (B) 50% (C) 75% (D) 85% Solution & Colligative Properties 6. 1.22 g of benzoic acid is dissolved in 100 g of acetone and 100 g of benzene separately. Boiling point of the solution in acetone increases by 0.17ºC, while that of, in the benzene increases by 0.13ºC ; K b for acetone and benzene is 1.7 K kg mol–1 and 2.6 K kg mol–1. Find molecular weight of benzoic acid in two cases and justify your answer. 7. The elevation in boiling point of a solution of 13.44 g of CuCl 2 in 1kg of water using the following information, will be (Molecular weight of CuCl2 = 134.4 and Kb = 0.52 K molal–1) : (A) 0.16 (B) 0.05 (C) 0.1 (D) 0.2 8. When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol–1), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is : (A) 0.5 (B) 1 (C) 2 (D) 3 Comprehension # Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 Given : Freezing point depression constant of water (Kfwater) = 1.86 K kg mol–1 Freezing point depression constant of ethanol (Kfethanol) = 2.0 K kg mol–1 Boiling point elevation constant of water (Kbwater) = 0.52 K kg mol–1 Boiling point elevation constant of ethanol (Kbethanol) = 1.2 K kg mol–1 Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol–1 Molecular weight of ethanol = 46 g mol–1 In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. 9. 10. The freezing point of the solution M is (A) 268.7 K (B) 268.5 K (C) 234.2 K (D) 150.9 K The vapour pressure of the solution M is (A) 39.3 mm Hg (B) 36.0 mm Hg (C) 29.5 mm Hg (D) 28.8 mm Hg 11. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is (A) 380.4 K (B) 376.2 K (C) 375.5 K (D) 354.7 K 12. The Henry's law constant for the solubility of N 2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N 2 from air dissolved in 10 moles of water of 298 K and 5 atm pressure is : (A) 4 × 10–4 (B) 4.0 × 10–5 (C) 5.0 × 10–4 (D) 4.0 × 10–6 13. The freezing point (in ºC) of a solution containing 0.1 g of K 3[Fe(CN)6] (Mol. Wt. 329) in 100 g of water (Kf = 1.86 K kg mol–1) is : (A) – 2.3 × 10–2 (B) – 5.7 × 10–2 (C) – 5.7 × 10–3 (D) – 1.2 × 10–2 14. For a dilute solution containing 2.5 g of a non- volatile non- electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K b = 0.76 K kg mol–1) (A) 724 (B) 740 (C) 736 (D)718 Solution & Colligative Properties 15.* Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is (are) : (A) G is positive (B) Ssystem is positive (C) Ssurroundings = 0 (D) H = 0 16. MX2 dissociates into M2+ and X– ions in an aqueous solution, with a degree of dissociation () of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is 17. If the freezing point of a 0.01 molal aqueous solution of a cobalt(III) chloride-ammonia complex (which behaves as a strong electrolyte) is –0.0558ºC, the number of chloride(s) in the coordination sphere of the complex is [Kf of water = 1.86 K kg mol–1] 18. * Mixture(s) showing positive deviation from Raoult's law at 35ºC is(are) (A) carbon tetrachloride + methanol (B) carbon disulphide + acetone (C) benzene + toluene (D) phenol + aniline 19.* For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here xL and xM represent mole fractions of L and M, Z respectively, in the solution. The correct statement(s) applicable to this system is (are) [JEE(Advanced) 2017, 4/122] pL (A) The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed from xL = 0 to xL = 1. (B) Attractive intermolecular interactions between L-L in pure liquid L and M-M in pure liquid M are stronger than those between L-M when mixed in solution 0 1 XM (C) The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed when xL → 0 (D) The point Z represents vapour pressure of pure liquid L and Raoult's law is obeyed when xL → 1 20. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol–1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [Molecular weight of ethanol is 46 g mol–1] Among the following, the option representing change in the freezing point is (B) Ice Water + Ethanol (C) V.P./bar 270 273 1 21. Ice Water + Ethanol 271 273 (D) Water + Ethanol 270 273 1 T/K Water Ice V.P./bar 1 Water T/K V.P./bar (A) V.P./bar Water 1 T/K Water Ice Water + Ethanol 271 273 T/K Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions xA and xB, respectively, has vapour pressure of 22.5 Torr. The value of xA / xB in the new solution is ____. (given that the vapour pressure of pure liquid  is 20 Torr at temperature T) Solution & Colligative Properties The plot given below shows P–T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal 760 solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents. On addition of equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is ____. [JEE(Advanced) 2018, 3/120] 1 2 3 4 1. Solvent X Pressure(mmHg) 22. 2. Solution of NaCl in solvent X 3. Solvent Y 367 368 362 360 4. Solution of NaCl in solvent Y Temperature (K) PART - II : MAIN / AIEEE PROBLEMS (PREVIOUS YEARS) (MAIN) OFFLINE PROBLEMS 1. 6.02 × 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (1) 0.001 M (2) 0.01 M (3) 0.02 M (4) 0.1 M. 2. Which one of the following aqueous solutions will exhibit highest boiling point ? (1) 0.01 M Na2SO4 (2) 0.01 M KNO3 (3) 0.015 M urea (4) 0.015 M glucose 3. If  is the degree of dissociation of Na2SO4, the vant Hoff’s factor (i) used for calculating the molecular mass is : (1) 1 +  (2) 1 –  (3) 1 + 2 (4) 1 – 2. 4. Equimolar solutions in the same solvent have : (1) same boiling point but different freezing point (2) same freezing point but different boiling point (3) same boiling and same freezing points (4) differnet boiling and freezing points 5. Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 mL of 1.2 M second solution. What is the molarity of the final mixture? (1) 1.20 M (2) 1.50 M (3) 1.344 M (4) 2.70 M 6. Benzene and toluene form nearly ideal solutions. At 20ºC, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20ºC for a solution containing 78 g of benzene and 46 g of toluene in torr is : (1) 50 (2) 25 (3) 37.5 (4) 53.5 7. Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is (1) 3.28 mol Kg–1 (2) 2.28 mol Kg–1 (3) 0.44 mol Kg–1 (4) 1.14 mol Kg–1 8. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be : (1) 700 (2) 360 (3) 350 (4) 300 9. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60g mol –1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm –3, molar mass of the substance will be (1) 105.0 g mol–1 (2) 210.0 g mol–1 (3) 90.0 g mol–1 (4) 15.0 g mol–1 Solution & Colligative Properties 10. The vapour pressure of water at 20ºC is 17.5 mm Hg. If 18 g of glucose (C 6H12O6) is added to 178.2 g of water at 20°C, the vapour pressure of the resulting solution will be : (1) 15.750 mm Hg (2) 16.500 mm Hg (3) 17.325 mm Hg (4) 17.675 mm Hg 11. At 80ºC, the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a mixture solution of 'A' and 'B' boils at 80º C and 1 atm pressure, the amount of 'A' in the mixture is (1 atm = 760 mm Hg) (1) 34 mol percent (2) 48 mol percent (3) 50 mol percent (4) 52 mol percent 12. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statement is correct regarding the behaviour of the solution ? (1) The solution is non-ideal, showing +ve deviation from Raoult’s Law. (2) The solution in non-ideal, showing –ve deviation from Raoult’s Law. (3) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s Law. (4) The solution formed is an ideal solution. 13. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively : (1) 300 and 400 (2) 400 and 600 (3) 500 and 600 (4) 200 and 300 14. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (Tf), when 0.01 mole of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1) (1) 0.0372 K (2) 0.0558 K (3) 0.0744 K (4) 0.0186 K 15. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol–1 and of octane = 114 g mol–1) (1) 72.0 kPa (2) 36.1 kPa (3) 96.2 kPa (4) 144.5 kPa 16. Kf for water is 1.86 K kg mol–1. If your automobile radiator holds 1.0 kg of water, how may grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8ºC ? (1) 72 g (2) 93 g (3) 39 g (4) 27 g 17. Consider separate solution of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125 M Na3PO4(aq) at 25C. Which statement is true about these solution, assuming all salts to be strong electrolytes ? (1) They all have the same osmotic pressure. (2) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure. (3) 0.125 M Na3PO4(aq) has the highest osmotic pressure. (4) 0.500 M C2H5OH(aq) has the highest osmotic pressure. 18. The vapour pressure of acetone at 20ºC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20ºC, its vapour pressure was 183 torr. The molar mass (g mol –1) of the substance is: (1) 32 (2) 64 (3) 128 (4) 488 19. 18 g glucosse (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is. (1) 76.0 (2) 752.4 (3) 759.0 (4) 7.6 20. The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (Kf for benzene = 5.12 K kg mol–1) (1) 80.4% (2) 74.6% (3) 94.6% (4) 64.6% 21. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point ? (1) [Co(H2O)4Cl2]Cl.2H2O (2) [Co(H2O)3Cl3].3H2O (3) [Co(H2O)6]Cl3 (4) [Co(H2O)5Cl]Cl2.H2O Solution & Colligative Properties MAIN ONLINE PROBLEMS 1. Choose the correct statement with respect to the vapour pressure of a liquid among the following : (1) Increases linearly with increasing temperature (2) Increase non-linearly with increasing temperature (3) Decreases linearly with increasing (4) Decreases non-linearly with increasing temperature 2. The observed osmotic pressure for a 0.10 M solution of Fe(NH 4)2(SO4)2 at 25ºC is 10.8 atm experimental (observed) and theoritical values of Van't Hoff factor (i) will be respectively : (R = 0.082 L atm k–1 mol–1) (1) 5 and 4.42 (2) 4 and 4.00 (3) 5 and 3.42 (4) 3 and 5.42 3. For an ideal solution of two components A and B, which of the following is true ? (1) Hmixing < 0 (zero) (2) Hmixing > 0 (zero) (3) A–B interaction is stronger than A–A and B–B interactions (4) A–A, B–B and A–B interactions are identical 4. A solution at 20ºC is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively : (1) 35.0 torr and 0.480 (2) 30.5 torr and 0.389 (3) 38.0 torr and 0.589 (4) 35.8 torr and 0.280 5. Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by : (1) partial ionization (2) dissociation (3) complex formation (4) association 6. The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L –1. The solubility (in g L–1) at 750 torr partial pressure is : (1) 0.02 (2) 0.015 (3) 0.0075 (4) 0.005 7. An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is : (1) 0.67 (2) 0.33 (3) 0.80 (4) 0.50 8. 5 g of Na2SO4 are was dissolved in x g of H2O. The change in freezing point was found to be 3.82ºC. If Na2SO4 is 81.5% ionised, the value of x (Kf for water = 1.86ºC kg mol–1) is approximately : (molar mass of S = 32 g mol–1 and that of Na = 23 g mol–1) (1) 45 g (2) 65 g (3) 25 g (4) 15 g 9. A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol–1) (1) 0.675 (2) 0.162 (3) 0.486 (4) 0.325 10. Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volotatile solute separately in the solvents X and Y. The molecular weights of the solvents are M X and MY, respectively where MX = 3 MY. The relative lowering of vapour pressure of the solution in X is "m" times that of the solution in Y. 4 Given that the number of moles of solute is very small in comparison to that of solvent, the value of "m" is : 3 1 1 4 (1) (2) (3) (4) 4 2 4 3 11. The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol–1) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75 %, is : (1) 37.5 g (2) 75 g (3) 150 g (4) 50 g Solution & Colligative Properties 12. Which one of the following statements regarding Henry's law is not correct ? (1) Different gases have different KH (Henry' law constant) values at the same temperature. (2) The value of KH increases with increase of temperature and KH is function of the nature of the gas (3) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution. (4) Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids. 13. A solution containing 62 g ethylene glycol in 250 g water is cooled to –10ºC. If Kf for water is 1.86 K kg mol–1, the amount of water (in g) separated as ice is: (1) 16 (2) 32 (3) 48 (4) 64 14. Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapor pressures of pure A and pure B are 7 × 103 Pa and 12 × 103 Pa, respectively. The composition of the vapor in equilibrium with a solution containing 40 mole percent of A at this temperature is : (1) xA = 0.28 ; xB = 0.72 (3) xA = 0.37 ; xB = 0.63 15. Elevation in the boiling point for 1 molal solution of glucose is 2 k. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 k. the relation between kb and Kf is : (1) Kb= 0.5Kf 16. (2) xA = 0.76 ; xB = 0.24 (4) xA = 0.4 ; xB = 0.6 (2) Kb =Kf (4) kb = 2 kf The freezing point of a diluted milk sample is found to be –0.2°C, while it should have been –0.5°C for pure milk. How much water has been added to pure milk to make the diluted sample ? (1) 1 cup of water to 3 cups of pure milk (3) 2 cups of water to 3 cups of pure milk 17. (3) Kb = 1.5 Kf (2) 1 cup of water of 2 cups of pure milk (4) 3 cups of water to 2 cups of pure milk K2HgI4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is : (1) 1.8 (2) 1.6 (3) 2.0 (4) 2.2 18. Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is: (1) 4A (2) 2A (3) 3A (4) A 19. Molecules of benzoic acid (C6H5COOH) dimerise in benzene. ‘w’ g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is : (Given that Kf = 5 kg mol–1, Molar mass of benzoic acid = 122 g mol–1) (1) 2.4 g (2) 1.8 g (3) 1.0 g (4) 1.5 g Solution & Colligative Properties EXERCISE - 1 PART - I A-1. Some of the characteristics of supersaturated solution are given below (i) If a crystal of solute is added to supersaturated solution, crystallisation occurs rapidly. (ii) Supersaturated solutions contain more solute than they should have at a particular temperature. A-2. Certain compounds combine with the moisture of atmosphere and are converted into hydroxides or hydrates. Such substances are called hygroscopic. e.g., anhydrous CuSO 4, quick lime (CaO), anhydrous Na2CO3 etc. A-3. The overall energy change associated with dissolution depends on the relative magnitude of the solute– solute, solvent–solvent and solute–solvent interactions. The process is exothermic if the new interaction release more energy than disrupting the old interactions requires, it is endothermic if opposite is true. B-1. 0.04 g/mL B-3. Molality = 11.44 m, Molarity = 7.55 M C-1. PO2 = 810 mm Hg, PH2O = 355 mm Hg , Ptotal = 1165 mm Hg C-2. 2.4 atm D-1. 7.62 bar E-2. WW = 3.973 WN F-1. pA0 = 400 mm of Hg, pB0 = 600 mm of Hg F-2. 27.3 mole % F-3. PB0 F-5. Y’B = 0.932. B-2. D-2. (i) 30%, (ii) 0.046 (iii) Na+ = 5.42 m, S2O32– = 2.71 m 4 mmol. 0 = 0.7 atm & PA = 1.9 atm G-1. V < 80ml. E-1. 5.15 g F-4. 14.16 mole percent benzene G-2. There is positive deviation from Raoult’s law, S > 0. H-1. Solute KCl H2SO4 CH3COOH (in water) CH3COOH (in benzene) Urea NaBr A Dissociation/association reaction KCl → K+ + Cl– H2SO4 → 2H+ + SO42– CH3COOH → H+ + CH3COO– 2CH3COOH → (CH3COOH)2 No association or dissociation NaBr → Na+ + Br– 3A → A3 I-1. Degree of dissociation /association n i 1 1 0.2 0.5 No association or dissociation 0.8 1 2 3 2 0.5 Not defined 2 0.33 2 3 1.2 0.75 1 1.8 0.33 H-2. 75% (a) 60 g/mol, (b) 333.3 g J-1. (a) M = 94.52, (b) m = 135 J-4. 1.075, 7.5. K-1. (a) Urea < NaCl < Na2SO4 < Na3PO4 K-2.  = 0.81 K-3. 50% K2SO4 K-4. A-1. (B) A-2. (B) B-1. (B) C-1. (B) C-2. (C) C-3. (B) C-4. (B) D-1. (C) D-2. (A) D-3. (B) D-4. (D) D-5. (B) E-1. (C) F-1. (B) F-2. (C) F-3. (C) F-4. (A) F-5. (A) G-1. (C) G-2. (A) G-3. (A) G-4. (C) G-5. (A) G-6. (C) H-1. (C) H-2. (A) H-3. (B) H-4. (B) I-1. (B) I-2. (B) I-3. (B) I-4. (C) I-5. (B) I-6. (A) J-1. (D) J-2. (B) J-3. (A) J-4. (A) J-5. (C) J-6. (D) J-7. (A) J-8. (B) J-9. (B) K-1. (C) K-2. (A) K-3. (A) K-4. (A) J-2. 232 l-2. 746.2 mm of Hg J-3. S8 (b) 6.15 atm Volume must have been made 5 times PART - II Solution & Colligative Properties PART - III 1. (A) – (r); (B) – (s); (C) – (q); (D) – (p) 2. (A) – (p,s,r); (B) – (p,q,t); (C) – (p); (D) – (p,q,t) EXERCISE - 2 PART - I 1. (C) 2. (D) 3. (B) 4. (C) 5. (D) 6. (D) 7. (B) 8. (B) 9. (B) 10. (A) 11. (B) 12. (A) 13. (D) 14. (B) 1. 18 2. 5 3. 21 4. 20 5. 27 6. 4 (A,B,C,D) 7. 2 8. 48 9. 43 10. 2 11. 60 12. 60 13. 4 1. (ABD) 2. (ABCD) 3. (ABD) 4. (ABCD) 5. (AC) 6. (A) 7. (AB) 8. (B) 9. (AB) 10. (B) 11. (ABC) 12. (BD) 13. (AB) 14. (AD) 15. (BCD) 16. (ABC) 1. (B) 2. (C) 3. (B) 4. (D) 5. (B) 6. (A) 7. (D) 8. (B) 9. (C) 10. (D) PART - II PART - III PART - IV EXERCISE - 3 PART - I 1. (C) 2. (B) 3. (B) 4. (A) 5. (C) 6. 122, 224 7. (A) 8. (A) 9. (D) 10. (B) 11. (B) 12. (A) 13. (A) 14. (A) 15. (BCD) 16. 2 17. (1) 18. (AB) 19. (BD) 20. (A) 21. 19 22. 0.05 (3) 5. (3) PART - II JEE(MAIN) OFFLINE PROBLEMS 3. (3) 4. 1. (2) 2. (1) 6. (1) 7. (2) 8. (3) 9. (2) 10. (3) 11. (3) 12. (1) 13. (2) 14. (2) 15. (1) 16. (2) 17. (1) 18. (2) 19. (2) 20. (3) 21. (2) JEE(MAIN) ONLINE PROBLEMS 1. (2) 2. (1) 3. (4) 4. (3) 5. (4) 6. (2) 7. (4) 8. (1) 9. (4) 10. (2) 11. (Bonus) 12. (4) 13. (4) 14. (1) 15. (4) 16. (4) 17. (1) 18. (3) 19. (1)