Problem 1: The ratio of the value of any colligative property for KCl solution to that for sugar is nearly times. (a) 1 (b) 0.5 (c) 2 (d) 2.

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1 Solved Problems Chemistry: Liquid Solutions-1 Objective Problem 1: The ratio of the value of any colligative property for KCl solution to that for sugar is nearly times. (a) 1 (b) 0.5 (c) 2 (d) 2.5 Exp. colligative property i KCl = Normal colligative property = isugar = 1 (c) no.of particle furnished by KCl no. of particles before dissociation = 2 Problem 2: Osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose that will be isotonic with blood is wt/ vol. (a) 5.41% (b) 3.54% (c) 4.53% (d) 53.4% We know V = nrt for glucose and blood; If isotonic, glucose = blood W Thus, 7.65 V = V W = 54.1 g/ lit = 5.41 % (wt/vol) (a) Problem 3: The Van t Hoff factor for 0.1 M La (NO 3 ) 3 solution is found to be 2.74 the percentage dissociation of the salt is (a) 85% (b) 58% (c) 65.8% (d) 56.8% La (NO3)3 La NO i = = = 0.58 percentage dissociation = 58% (b) Problem 4: The vapour pressure of pure benzene at 50 C is 268 torr. How many mol 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) (b) (c) (d) 0.395

2 Chemistry: Liquid Solutions-2 o P P X B B 167 X B 268 Xsolute = 1 XB = (a) Problem 5: Maximum freezing point will be for 1 molal solution of, assuming equal ionization in each case (a) [Fe(H 2 O) 6 ]Cl 3 (b) [Fe(H 2 O) 5 Cl]Cl 2. H 2 O (c) [Fe(H 2 O) 4 Cl 2 ]Cl. 2H 2 O (d) [Fe(H 2 O) 3 Cl 3 ]. 3H 2 O Larger the value of i Smaller the f.p. (d) in a case of non electrolytes, i = 1 Problem 6: 1.0 molal aqueous solution of an electrolyte X 3 Y 2 is 25% ionized. The boiling point of the solution is (K b for H 2 O = 0.52 K kg/mol) (a) K (b) K (c) K (d) K XY 3X 2Y i = =1 + 4(0.25) = 2 Tb = i Kb M = = 1.04 Boiling point of the solution = K (b) Problem 7: Equal amounts of a solute are dissolved in equal amounts of two solvents A and B. The relative lowering of vapour pressure for the solution A has twice the relative lowering of vapour pressure for the solution B. If M A and M B are the molecular weights of solvents A and B respectively, then for dilute solution of A and B, with solute. M B (a) M A = M B (b) M A 2 (c) M A = 4M B (d) M A = 2M B o o P Ps n P Ps and o o A A n P N P N B B o P Ps o P A B B A A N w M M or o P P N M w M s A B A B o P B (d) MA 2 M B MA = 2MB

3 Chemistry: Liquid Solutions-3 Problem 8: The molal boiling point constant of water is C Kg mole 1. When 0.1 mole of glucose is dissolved in 1000 gm of water, the solution boils under atmospheric pressure at (a) C (b) C (c) C (d) C Tb = KbM = C B.P. of the solution = B.P. of pure solvent + Tb = = C (b) Problem 9: The relationship between osmotic pressure at 273 K when 10 gm glucose (P 1 ), 10 gm urea (P 2 ) and 10 g sucrose (P 3 ) are dissolved in 250 ml of water is (a) P 1 P 2 P 3 (b) P 3 P 1 P 2 (c) P 2 P 1 P 3 (d) P 2 P 3 P 1 n w = CRT RT or RT v M v 1 M M. wt. of glucose, sucrose and urea in the order urea glucose sucrose in the order urea glucose sucrose Problem 10: FeCl 3 on reaction with K 4 (Fe(CN) 6 ] in aqueous solution gives blue colour. These are separated by a semi permeable membrane AB. Due to osmosis there is 0.1 M K 4 [Fe(CN) 6 ] Side X 0.01 M FeCl 3 Side Y (a) blue colour formation in side x (b) blue colour formation in side y (c) blue colour formation in both sides (d) no blue colour formation Due to osmosis, solvent (i.e., water) from y side migrates towards x side. Since solute (i.e.) FeCl3 is not migrating, there is no blue colour formation. Problem 11: A weak electrolyte, AB, is 5% dissociated in aqueous solution. What is the freezing point of a molal aqueous solution of AB? K f for water is 1.86 deg/molal. (a) 3.8 C (b) C (c) 1.7 C (d) 0.78 C

4 Chemistry: Liquid Solutions-4 5 Degree of dissociation,, AB = AB A + + B No. of moles dissolved m 0 0 No. of moles after dissociation m(1 ) m m After dissociation 0.1(1 0.05) Total moles = Molality = 0.1 (1 0.05) = = 0.105m Tf = Kfm or Tf 0 Tf = 1.86 K/m 0.105m = deg Tf = 0 C = C (b) Problem 12: Calculate the boiling point of a one molar aqueous solution (density = 1.04 g ml 1 ) of potassium chloride, K b for water = 0.52 kg mol 1. Atomic masses of K = 39, Cl = 35.5 (a) C (b) C (c) C (d) None of these Volume of solution = 1000, Mass of the solution = V d = 1000 ml 1.04 g/ml = 1040g Amount of solute in 1000 ml solution = 1 M = 1 M Molecular Mass = 1M 74.5g/mol = 74.5g Mass of water = Mass of solution Mass of KCl = 1040g 74.5g = 965.5g Moles of solute 1 mol Molality of the solution = m Mass of solvent in kg (965.5/1000)kg Tb = i Kbm = K kg mol = C Boiling point of the solution = 100 C C = C (c) Problem 13: The vapour pressure of a solvent decreased by 10 mm Hg when a non volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg. (a) 0.8 (b) 0.6 (c) 0.4 (d) 0.2 P o Ps = P o mole fraction of solute; 10 = P o 0.2 Again, 20 = P o mole fraction solute Hence, n = 0.4, so, mole fraction of solvent = = 0.6 (b) Problem 14: The molal freezing point constant for water is 1.86 K. molarlity 1. If 34.2 g of cane sugar (C 12 H 22 O 11 ) are dissolved in 1000g of water, the solution will freeze at (a) 1.86 C (b) 1.86 C (c) 3.92 C (d) 2.42 C

5 Chemistry: Liquid Solutions Kf w Tf = MW Tf = = 1.86 C (a) = 1.86 Problem 15: 20 g of a binary electrolyte (mol.wt. = 100) are dissolved in 500 g of water. The freezing point of the solution is 0.74 C, K f = 1.86 K.molality 1. The degree of ionisation of the electrolyte is (a) 50% (b) 75% (c) 100% (d) Kf w Tf = 0.74 = M W M 500 Normal mol.wt Now, = 1 + = 100 exp. mol.wt. 100 So, = 0 (d) M = 100 Problem 16: Y g of non - volatile organic substance of molecular mass M is dissolved in 250 g benzene. Molal elevation constant of benzene is K b. Elevation in its boiling point is given by M (a) (b) 4 KY b KY b M KY b KY b (c) (d) 4 M M 1000Kb Y T = 250 M (b) 4KbY = M Problem 17: The values of observed and calculated molecular weights of silver nitrate are and 170 respectively. The degree of dissociation of silver nitrate is (a) 60% (b) 83.5% (c) 46.7% (d) 60.23% Normal mol.wt i for AgNO3 = = 1 + Observed mol.wt. 170 = 1 = = 83.5 % (b) Problem 18: At 40 C, the vapour pressures in torr, of methyl alcohol - ethy alcohol solution is represented by the equation. P = 119X A where X A is mole fraction of PA methyl alcohol, then the value of Lim X A 1 X is A (a) 254 torr (b) 135 torr (c) 119 torr (d) 140 torr

6 Chemistry: Liquid Solutions-6 P = 119XA + 135; (a) Lim XA 1 P X A A = = 254 Problem 19: The amount of ice that will separate on cooling a solution containing 50g of ethylene glycol in 200g water to 9.3 C is : [K f = 1.86 K molality 1 ] (a) g (b) mg (c) 42 g (d) 42 mg 1000 Kf w T = W M or 9.3 = 62 W W = Ice separated = = g (a) Problem 20: The freezing point of aqueous solution contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is : ( K fh2o= 1.86K molality 1 ) (a) K (b) K (c) K (d) 250 K Subjective Tf = Tf for glucose + Tf for KCl + Tf for urea = freezing point = = K (c) = Problem 1: The freezing point depression of a M aq. solution of formic acid is 0.21 C. Calculate the equilibrium constant for the reaction, HCOOH (aq) H + (aq) + HCOO (aq) K f for water = 1.86 kg mol 1 K From question Tf = 0.21 C Molality of solution = m The reaction is HCOOH (aq) H + (aq) + HCOO (aq) Initially moles C 0 0 Moles after dissociation (1 )C C C (1 ) CCC i 1 C = 1 +, where = degree of dissociation As we know, Tf = Kf i Cm

7 Chemistry: Liquid Solutions = 1.86 (1+) ( in very dilute solution molarity and molality can be taken as same) = = or, = = [HCOOH] = (1 ) C = ( ) C = C [H + ] = C = C, [HCOO ] = C [H ][HCOO ] C0.036C Ka = = = C mol lit 1 [HCOOH] C = Ka = Problem 2: The freezing point of a solution of acetic acid (mole fraction is 0.02) in benzene is K. Acetic acid exists partly as a dimer 2A A 2. Calculate equilibrium constant for dimerisation. Freezing point of benzene is K and (K f for benzene is 5) Let acetic acid = A Benzene = B Assume, part of A forms dimer 2A A2 1 0 initially moles 1 /2 moles after dimer is formed i = = 1 /2 Mol. fraction of A = xa = 0.02 Mol. fraction of B = xb = 0.98 x A 1000 Molality of A in B = = m X Since, Tf = Kf i molality = 5 i or, 1 = 5 i i = = /2 = = 0.47 B B = mol kg 1 of Benzene Hence the molality of A after dimer is formed = (1 ) initial molality = (1 0.48) initial molality = Molality of A2 after dimer is formed 0.48 = molality = = = The equilibrium constant Keq = A 2 = = 3.39 kg mol A ( )

8 Chemistry: Liquid Solutions-8 Problem 3: A milimolar solution of pottassium ferricyanide is 70% dissociated at 27 C. Find out the osmotic pressure of the solution. 1 Given, concentration of solution = C = 1000 M = 10 3 M K3[Fe(CN)6] 3K + + [Fe(CN)6] 3 Initially moles Moles after dissociation Where = degree of dissociation = 70% = 100 = 0.7 Total moles of after dissociation = (1 ) = 1+ 3 i = i = 1+3 = 1+(3 0.7) = 3.1 Since osmotic pressure = icrt = = atm = atm Problem 4: If the solution of mercuric cyanide of strength 3 g / L has an osmotic pressure Nm 2 at 298 K, what is the apparent molecular weight and degree of dissociation of Hg(CN) 2? (Hg = , C = 12, N = 14) Let the apparent molecular weight = mo Osmotic pressure = = v n RT w 1 = RT m v o or, = mo mo = kg Since, number of particles after dissociation normal mol. w t. number of particles before dissociation observed mol. wt. Hg(CN)2 Hg CN 1 2 Number of particles after dissociation = (1 ) = i.e., = = or %

9 Chemistry: Liquid Solutions-9 Problem 5: If the apparent degree of ionization of KCl (KCl = 74.5 gm mol 1 ) in water at 290K is Calculate the mass of KCl which must be made up to 1 dm 3 of aqueous solution so as to produce the same osmotic pressure as the 4.0% solution of glucose at that temperature., Due to ionization of KCl KCl K + + Cl Initial moles Moles after dissociation 1 1 = 1+ i = 1 From question, degree of ionization = = 0.86 i = = 1.86 For Osmotic pressure of glucose, For 4% glucose solution, Weight of glucose = 40 gm Volume of solution = V = 1L = 1dm 3 Molecular weight of glucose C6H12O6= m = 180 glucose = n v 1 RT, where glucose = osmotic pressure of glucose = w m 1 v R T = RT 1 Similarly, KCl = in RT v = 1.86 n 1 v RT= 1.86 w1 m 1 v As both solutions are isotonic KCl = glucose w RT = RT W1 = = 8.9 gm m 1 w RT = RT 1 Problem 6: The freezing point of an aqueous solution of KCN containing mol kg 1 of solvent was found to be C. On adding mol of Hg(CN) 2, the freezing point of the solution was found to be C. If the complex formation takes place according to the following equation. Hg(CN) 2 + nkcn K n [Hg(CN) n+2 ]. What is the formula of the complex? K f (H 2 O) is 1.86 kg mol 1 K. Tf (KCN solution) = C molality of KCN solution = i = (1 + x) Tf = Kf m i

10 Chemistry: Liquid Solutions-10 Tf i = (1 + x) = = 2.0 K f m This gives x = 1, indicating 100% ionization of KCN Tf of the complex = Molality of Hg(CN)2 = mol kg 1 Kn [Hg(CN)n+2] nk + + [Hg(CN)n+2] initially moles 1 n.. moles after dissociation here i = (1 ) + n + = 1+n = 1 i = 1+ n Tf = Kf i molality = 1.86 I i = n = 3 or, n = 2 Hence, the complex is K 2 [Hg(CN) 4 ] Problem 7: Ethylene dibromide (C 2 H 4 Br 2 ) and 1,2 dibromopropane (C 3 H 6 Br 2 ) forms a series of ideal solution over the whole range of composition. At 85C, the vapour pressure of these pure liquids are 173 mmhg and 127mm Hg respectively. 10 gm of ethylene dibromide is dissolved in 80 gm of 1,2 dibromo propane. Calculate the partial pressures of each components and the total pressure of the solution at 85C. Calculate the composition of vapour in equilibrium with the above solution and express as mole fraction of ethylene dibromide What would be the mole fraction of 1,2 dibromopropane in solution with 50:50 mole mixture in the vapour. Let C2H4Br2 = A, ma = , where ma= molecular weight of A mb = molecular weight of B and B is C3H6Br2 = mb = = 202 w 10 Moles of A, na = A = ma 188 w 80 Moles of B, nb = B = mb 202 nb XB = = na nb xa + xb = 1 xa = = a) We know, total pressure PT = PA + PB Where PA, PB are partial pressures of A and B respectively.

11 Chemistry: Liquid Solutions-11 = P o o AXA PB XB, where P o o A,P B are partial pressures of pure component of A and B. PT = = = mm Hg Here partial pressure of A, PA = mm Hg Here partial pressure of B, PB = mm Hg. b) Mole fraction of ethylene dibromide in the vapour partial pressure of A YA = Total pressure c) In vapour phase na: nb = 50:50 YA = 0.5, YB = 0.5 Problem 8: Calculate the molecular weight of cellulose acetate if its 0.5% (wt./vol) solution in acetone (sp. gr. = 0.9) shows an osmotic rise of 23 mm against pure acetone at 27 C. 0.5% (wt. / vol) solution means 0.5 gm of cellulose acetate is dissolved in 100 ml solution. Osmotic pressure = 23 mm of pure acetone = 2.3 Cm of pure acetone = cm of Hg = cm of Hg 13.6 = atm = atm 76 Let the molecular weight of the cellulose acetate be M Here, number of mole of cellulose acetate (n) = 0.5 M Volume = v = 100 ml = 0.1 lit R = lit atm mol 1 K 1, T = ( ) = 300 K Osmotic pressure () = n v M = RT = M 0.1 Problem 9: 1kg of an aqueous solution of Sucrose is cooled and maintained at 4 C. How much ice will be separated out if the molality of the solution is 0.75? K f (H 2 O) = 1.86 Kg mol 1 K. Sucrose is a non electrolyte, Hence i = 1 Molecular weight of sucrose (C12H22O11) = m = 342 gm mol 1 Molality of the solution = 0.75 m = 0.75 mol kg 1 solvent = gm Sucrose per kg solvent = gm Sucrose per kg solvent hence, weight of 1 molal solution = = gm

12 Chemistry: Liquid Solutions-12 Sucrose present in 1 kg solution = = gm Weight of solvent (H2O) present in 1 kg solution = = gm Since depression in freezing point Tf = Kf i w m 1000, Where W = weight of solvent W w = weight of the solute or, 4 = W 342 W = gm i.e., weight of solvent required to maintain this solution at 4 C is W = Hence rest weight of H2O will convert into ice. Hence amount of ice formed = = gm Problem 10: River water is found to contain 11.7% NaCl, 9.5% MgCl 2, and 8.4%. NaHCO 3 by weight of solution. Calculate its normal boiling point assuming 90% ionization of NaCl, 70% ionization of MgCl 2 and 50% ionization of NaHCO 3 (K b for water = 0.52) nnacl = = 0.2 n MgCl 2 = = 0.1 n NaHCO 3 = = 0.1 inacl = 1+ = = 1.9 i MgCl 2 = = = 2.4 i NaHCO 3 = 1+ 2 = = 2.0 Weight of solvent = 100 ( ) = 70.4 g (inacl n NaCl imgcl n 2 MgCl i 2 NaHCO n 3 NaHCO ) K 3 b 1000 Tb = Weight of solvent ( ) = = 5.94 C 70.4 Boiling point of solution = = C Problem 11: An aqueous solution containing 288 gm of a non volalite compound having the stochiometric composition C x H 2x O x in 90 gm water boils at C at 1.00 atmospheric pressure. What is the molecular formula? K b (H 2 O) = K mol 1 kg T b (H 2 O) = 100 C Elevation in B.P. = = 1.24 C Since solute is non volatile Since the elevation in boiling point is Tb = Kb i molality

13 Chemistry: Liquid Solutions-13 = Kb 1 w m 1000 W 1.24 = m m = gm mol 1 m = mol.wt. of solute molar mass of CxH2xOx = 12x + 1 2x + 16x = 30x 30x = x = 44 Hence the molecular formula is = C44H88O44 Problem 12: 30 ml of CH 3 OH ( d = gm Cm 3 ) and 70 ml of H 2 O (d = gm cm 3 ) are mixed at 25 C to form a solution of density gm cm 3. Calculate the freezing point of the solution. K f (H 2 O) is 1.86 Kg mol 1 K. Also calculate its molarity Weight of CH3OH (w1) = 30cm gm/cm 3 = gm Weight of solvent (H2O) (w2) = 70 cm gm cm 3 = gm w molality of solution = m1 w Where m1 = molecular weight of CH3OH = 32 = = m As we know that Depression in freezing point, Tf = Kf i molality, for CH3OH, i = 1 Tf = C = C Freezing point of the solution = C = C Weight of solution = weight of solute + weight of solvent = = gm volume of the solution = wt. of the solution density of the solution = = ml w1 molarity of solution = m 1000 V(in ml) = mol lit 1 = 7.63 M 1 Problem 13: A complex is represented as CoCl 3.xNH 3. Its 0.1 m solution in aqueous solution shows T f = K f (H 2 O) = 1.86 mol 1 K and assume 100% ionization and co ordination number of Co(III) is six. What is the complex? From question, molality of solution = Cm = 0.1 Depression in freezing point = Tf = C Kf (H2O) = 1.86 kg mol 1 K 1 As we know that Tf = Kf i m, Where i = Vant Hoff s factor = i

14 Chemistry: Liquid Solutions-14 i = = 3 or, i 3 indicates that complex ionize to form three ions since coordination number is 6 hence x = 5 i.e., CoCl3.5NH3 [Co(NH3)5Cl] Cl 1 Cation 2 anions So, the complex is [Co(NH3)5Cl]Cl2 Problem 14: A solution comprising 0.1 mol of naphthalene and 0.9 mol of benzene is cooled until some benzene freezes out. The solution is then decanted off from the solid and warmed to 353 K, where its vapour pressure is found to be 670 torr. The freezing and normal boiling point of benzene are K and 353 K, respectively. Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume conditions of ideal solution. K f for benzene = 5K kg mole 1. Molality of the resulting solution (i.e. after the benzene freezes out) can be determined on the basis Raoult s law. ( ) Xnaph 1000 m 760 = 1.72 mol kg 1. X benzene Weight of benzene in the resulting solution = = 58.1 g Amount of benzene frozen = = 12.1 g Tf = Kfm = = 8.60 K Hence the temp to which the solution was cooled= = K Problem 15: Find Ka, the ionization constant of tartaric acid if a molal aqueous solution of tartaric acid freezes at C. Assume that only the first ionization is of importance and that 0.1 m = 0.1M. Kf = 1.86 kg mol 1 K. Assuming that the tartaric acid be a monobasic as AH. It ionizes as AH A + H + Initially conc. C 0 0 Conc. after dissociation C(1 ) C C, Here, i = C(1 ) C C C(1 ), where = degree of dissociation = = 1+ C C Molal concentration = 0.1 Tf = Kf Cm i = (1+) = [A ][H ] CC C Ka = = [AH] C(1 ) 1 Ka = = Ka =

15 Assignments (New Pattern) SECTION I Chemistry: Liquid Solutions-15 Single Choice Questions 1. For a dilute solution, Raoult s law states that (a) the lowering of vapour pressure is equal to the mole fraction of the solute (b) the relative lowering of vapour pressure is equal to the mole fraction of the solute (c) the relative lowering of vapour pressure is proportional to the amount of the solute in solution. (d) the vapour pressure of the solution is equal to the mole fraction of the solvent 2. The molal elevation constant is the ratio of the elevation in B.P. to (a) molarity (b) molality (c) mole fraction of solute (d) mole fraction of solvent. 3. The osmotic pressure of a 5% solution of cane sugar at 150 C is (Mol. Wt. of cane sugar = 342) (a) 4 atm. (b) 3.4 atm. (c) 3.55 atm. (d) 2.45 atm. 4. Which of the following compounds corresponds Van t Hoff factor (i) to be equal to 2 for dilute solution? (a) KSO 2 4 (b) NaHSO 4 (c) Sugar (d) MgSO ml of liquid A was mixed with 25 ml of liquid B to give a non-ideal solution of A-B mixture. The volume of this mixture would be (a) 75 ml (b) 125 ml (c) close to 125 ml but not exceeding 125 ml (d) just more than 125 ml. 6. Which of the following solutions will have the highest boiling point? (a) 1% glucose (b) 1% sucrose (c) 1% NaCl (d) 1% CaCl The osmotic pressure of 1 m solution at 27 C is (a) 2.46 atm (b) 24.6 atm (c) 1.21 atm (d) 12.1 atm. 8. Very dilute solutions which show deviations (positive or negative) from Raoult s law are called (a) ideal solutions (b) true solutions (c) non-ideal solutions (d) colloidal solutions M solution each of urea, common salt and Na2SO4 are taken. The ratio of depression of freezing point is (a) 1:1:1 (b) 1:2:1 (c) 1:2:3 (d) 2:2:3

16 Chemistry: Liquid Solutions The relative lowering of vapour pressure is equal to the mole fraction of the solution. This law is called (a) Henry s law (b) Raoult s law (c) Ostwald s law (d) Arrhenius law. 11. A X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. The value of X is (a) 14 (b) 3.2 (c) 1.4 (d) Molal depression constant for water is 1.86 C. The freezing point of a 0.05 molal solution of a non-electrolyte in water is (a) 1.86 C (b) 0.93 C (c) 0.093C (d) 0.93 C. 13. When mercuric iodide is added to the aqueous solution of potassium iodide, the (a) freezing point is raised (b) freezing point does not change (c) freezing point is lowered (d) boiling point does not change. 14. Nitrobenzene freezes at C molal solution of a solute in nitrobenzene causes freezing point depression of 2 C. K f for nitrobenzene is 1 (a) 2K m 1 (b) 4K m 1 (c) 8K m 1 (d) 12K m. 15. The freezing point of a 0.05 molal solution of a non-electrolyte in water is (a) 1.86 C (b) 0.93 C (c) C (d) 0.93 C. 16. The freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is (a) 1.86 C (b) 3.72 C (c) C (d) C. 17. The freezing point of equimolal aqueous solution will be highest for (a) C6H5NH3Cl (aniline hydrochloride) (b) Ca(NO 3) 2 (c) La(NO 3) 3 (d) C6H12O 6 (glucose). 18. Which one of the following pairs of solution can we expect to be isotonic at the same temperature? (a) 0.1 M urea and 0.1 M NaCl (b) 0.1 M urea and 0.2 M MgCl 2 (c) 0.1 M NaCl and 0.1 M Na 2SO 4 (d) 0.1 M Ca(NO 3) 2and 0.1 M Na 2SO 4

17 Chemistry: Liquid Solutions Vapour pressure of a solution of 5 g of a non-electrolyte in 100 g of water at a particular 2 2 temperature is 2985 N/m. The vapour pressure of pure water is 3000 N/m, the molecular weight of the solute is (a) 60 (b) 120 (c) 180 (d) g of a substance dissolved in 15 g of a solvent boiled at a temperature higher by C than that of the pure solvent. Find out the molecular weight of the substance. ( Kf for solvent is 2.16 C). (a) 1.01 (b) 10.1 (c) 100 (d) A 5% solution of cane sugar (Mol. Wt. = 342) is isotonic with 1% solution of substance X. The molecular weight of X is (a) (b) 68.4 (c) 34.2 (d) Which of the following statements is correct, if the intermolecular forces in liquids A, B and C are in the order A < B < C? (a) B evaporates more readily than A (b) B evaporates less readily than C (c) A and B evaporate at the same rate (d) A evaporates more readily than C 23. The vapor pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole-fraction of solute in solution is 0.2. What would be the mole-fraction of solvent if decrease in vapour pressure is 20 mm of Hg? (a) 0.8 (b) 0.6 (c) 0.4 (d) An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanol are kpa and kpa respectively, the composition of vapour (in terms of mole fraction) will be (a) MeOH, EtOH (b) MeOH, EtOH (c) MeOH, EtOH (d) MeOH, EtOH. 25. Vapour pressure of CCl 4 at 25 C is 143 mm Hg. 0.5 gm of a non-volatile solute (Mol. Wt. 65) is dissolved in 100 ml CCl 4. Find the vapour pressure of the solution. (Density of 3 CCl g / cm ). (a) mm (b) mm (c) mm (d) mm.

18 Chemistry: Liquid Solutions-18 SECTION II May be more than one choice 1. Which is/are correct statements? (a) 1 L of 1 N H3PO4 solution contains one-third of molecular weight when it is completely neutralized by NaOH (b) 1 L of 1 N H3PO4 solution contains 1 mol of H3PO4 when it is reduced to (c) 1 L of 1 NH3PO4 solution contains 0.5 mol of H3PO4 when it is reduced to (d) H3PO4 is a monobasic acid 2 HPO 3 HPO For the following disproportionation reaction 5Br2 + 6OH 5Br + BrO 3 Correct statements are: (a) equivalent weight of Br2 when it is reduced to Br is 80 (b) equivalent weight of Br2 when it is oxidized to BrO is 16 3 (c) equivalent weight of Br2 in the net reaction is 96 (d) equivalent weight of Br is H2O 3. 1 mol benzene 0 (P 42 mm) and 2 mol toluene benzene 0 (P 36mm) will have: toluene (a) total vapour pressure 38 mm (b) mol fraction of vapours of benzene above liquid mixture is 7/19 (c) positive deviation from Raoult s law (d) negative deviation from Raoult s law 4. At 40 C, the vapour pressure in torr of methanol and ethanol solution is P = 119x Where x is the mol. 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 mm (d) mixture is completely immiscible 5. Consider following cases: I : 2 M 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 ionizes 100%; which is/are true statement(s)? (a) both are isotonic (b) I is hypertonic (c) II is hypertonic (d) none is correct 6. Consider following solutions: I : 1 M aq. Glucose II : 1 M aq. Sodium chloride III : 1 M benzoic acid in benzene IV : 1 M ammonium phosphate Select correct statement(s): (a) All are isotonic solutions (c) I, II, IV are hypertonic of III (b) III is hypotonic of I, II, IV (d) IV is hypertonic of I, II, III

19 Chemistry: Liquid Solutions If PA is the vapour pressure of a pure liquid A and the mol fraction of A in the mixture of two liquids A and B is x, the partial vapour pressure of A is: (a) (1 x)pa (b) xpa x (1 x) (c) P (d) A P A (1 x) X 8. When mercuric iodide is added to the aqueous solution of potassium iodide, the: (a) freezing point is raised (b) freezing point is lowered (c) freezing point does not change (d) boiling point does not change 9. Which is/are correct statement(s)? (a) when mixture is less volatile, there is positive deviation from Raoult s law (b) when mixture is more volatile, there is negative deviation from Raoult s law (c) when mixture is less volatile, there is negative deviation from Raoult s law (d) when mixture is more volatile, there is positive devation from Raoult s law 10. When a solute is added to a pure solvent, the (a) vapour pressure of the solution becomes lower than that of the pure solvent (b) rate of evaporation of the pure solvent is reduced (c) solute does not affect the rate of condensation (d) rate of evaporation of the solution is equal to the rate of condensation of the solution at a lower vapour pressure than that in the case of the pure solvent 11. According to Raoult s law the relative decrease in the solvent vapour pressure over the solution is equal to (a) the mole fraction of the solvent (b) the mole fraction of the solute (c) the number of moles of the solute (d) i times the mole fraction of the solute which undergoes dissociation or association in the solvent (i = van t Hoff factor) 12. Which of the following combinations are correct for a binary solution, in which the solute as well as the solvent are liquid? (a) C6H6 and C6H5CH3; Hsoln 0; Vsol. = 0 (b) CH3COCH3 and CHCl3; Hsoln 0; Vsol 0 (c) H2O and HCl; Hsoln 0; Vsol 0 (d) H2O and C2H5OH; Hsoln 0; Vsol Which of the following statements are correct for a binary solution which shows negative deviation from Raoult s law? (a) The negative deviation from linearity diminishes and tends to zero as the concentration of the solution component approaches unity (b) When solutions form, their volumes are smaller than the sum of the volumes of their components (c) Heat is released during the formation of the solution (d) Heat is evolved during the formation of the solution

20 Chemistry: Liquid Solutions A binary liquid (AB) shows positive deviation from Raoult s law when 0 liq 0 liq (a) p p X andp p X A A A A B B (b) Intermolecular forces: A-A, B-B A-B (c) Vmix 0 (d) Hmix The azeotropic solutions of two miscible liquids (a) can be separated by simple distillation (b) may show positive or negative deviation from Raoult law (c) are supersaturated solutions (d) behave like a single component and boil at a constant temperature 16. Which of the following is/are characteristics of hydrophilic solutions? (a) High concentration of dispersed phase can be easily attained. (b) Coagulation is reversible (c) Viscosity and surface tension are nearly same as that of water (d) The charge of the particle depends on the ph value of the medium; it may be positive, negative or neutral. 17. A substance effloresces (a) due to the formation fo a crust on its crystal surface (b) when its vapour pressure is greater than that of the water vapour in air (c) till it melts (d) when all of the above happen 18. In which of the following pairs of solutions will the values of the van t 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 19. When solute is added to a pure solvent, the (a) vapuor pressure of the solution becomes lower than the vapors pressure of the pure solvent (b) rate of evaporation of the pure solvent is reduced (c) solute does not affect the rate of condensation (d) none of these 20. If P 0 and Ps are the vapour pressures of the solvent and its solution respectively and N1 and N2 are the mole fractions of the solvent and solute respectively, then (a) PS = P 0 N2 (b) P 0 PS = P 0 N2 (c) PS = P 0 N1 (d) (P 0 PS)/PS = N1/(N1 + N2) 21. Which of the following form ideal solution? (a) C6H5Cl C6H5Br (b) C6H6 C6H5CH3 (c) Hexane Heptane (d) Ethanol + Cyclohexane

21 22. At constant temperature, the osmotic pressure of a solution is (a) directly proportional to the concentration (b) inversely proportional to the molecular weight of the solute (c) directly proportional to the square of the concentration (d) directly proportional to the square root of the concentration 23. Ideal solution is formed when its components (a) have zero heat of mixing (b) have zero volume change (c) obey Raoult s law (d) can be converted into gases 24. In the depression of freezing point experiment, it is found that (a) The vapour pressure of the solution is less than that of pure solvent (b) The 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 Chemistry: Liquid Solutions The colligative properties of a solution are (a) molality 1 (b) = molecular mass of the solute (c) proportional to each other (d) independent of the nature of the solute, i.e., electrolyte or non-electrolyte SECTION III Comprehension Type Questions Write-up I According to Raoult s law (which is applicable for a mixture of volatile liquids) the partial vapour pressure of a liquid is directly proportional to mol fraction of that component. Further assuming ideal behaviour for vapours and applying Dalton s law we can write different equations as follows: 0 0 PA PAx A, PB PBxB PTotal PA PB PAxA PBxB P B (PA P B)xA Further mol fraction of a component in vapour phase is equal to the ratio of partial vapour pressure to total vapour pressure of mixture. The vapour pressure of two pure liquids A and B which form an ideal solution are 300 and 500 torr respectively at temperature T. A mixture of the vapours of A and B for which the mol fraction of A is 0.25 is slowly compressed at temperature T. 1. The total pressure when first drop of condensate is formed will be (a) 428 torr (b) 400 torr (c) 388 torr (d) 358 torr 2. The pressure when only the last bubble of vapour remains will be (a) 350 torr (b) 375 torr (c) 525 torr (d) 450 torr

22 Chemistry: Liquid Solutions The mol fraction of B in the last bubble of vapour will be (a) 0.16 (b) 0.84 (c) 0.20 (d) 0.80 Write-up II Addition of a non-volatile solute always lower the vapour pressure of solvent, therefore, it will be in equilibrium with solid phase at a lower pressure and hence at a lower temperature. The difference between the freezing points of the pure solvent and its solution is called depression of freezing point and depression in freezing point is when 1 mol of solute is dissolved in 1000 gm of solvent is called molal depression constant (K f ) & K f for H2O = 1.86 K mol 1 kg. To understand this, a solution of 25 % (w/w) MgCl2 containing impurity of the chloride of a trivalent metal M having atomic mass 43.5 gm (24% by weight of MgCl2) has been prepared. In this solution MgCl2 is completely dissociated, on the other hand only 50 % dissociation of MCl 3 takes place and molality is approximately equal to molarity. 4. Total moles of solute in the solution is - (a) 0.6 (b) 0.7 (c) 0.5 (d) Depression in freezing point of the solution is - (a) (b) (c) (d) If the solution is cooled to 21.7 ºC, will it freeze out. If so what weight of ice will formed (a) 15 gm (b) 60 gm (c) 40 gm (d) 51 gm Write-up III VP III P = P + P A B II I XA = 1 XB = 0 Mole fraction XA = 0 XB = 1 The above graph represents the deviation in the vapour pressure as compared to the expected value according to Raoult s law. 7. In the above graph deviation is (a) Positive (b) Negative (c) No deviation (d) can not be predicted

23 8. For the negative deviation conditions required are (a) PA PA X A;PB PB XB (b) P P X ;P P X (c) Hmixing = +ve (d) Vmixing = +ve 9. For the non ideal solutions the conditions are (a) Hmixing 0 or + Ve (b) PA = (c) 0 B B B Chemistry: Liquid Solutions-23 A A A B B B 0 PA XA P P X (d) Vmixing = 0 Write-up IV Colligative properties of a solution depend upon the number of moles of the solute dissolved and do not depend upon the nature of the solute. However, they are applicable only to dilute solutions in which the solutes do not undergo any association or dissociation. For solutes undergoing such changes, van t Hoff introduced a factor caused van t Hoff factor (i). This has helped not only to explain the abnormal molecular masses of such solutes in the solution but has also helped to calculate the degree of association or dissociation. 10. Which of the following is a colligative property? (a) Boiling point (b) Freezing point (c) Osmotic pressure (d) Vapour pressure 11. During depression of freezing point in a solution, the following are in equilibrium: (a) liquid solvent, solid solvent (b) liquid solvent, solid solute (c) liquid solute, solid solute (d) liquid solute, solid solvent Write-up V A solution of sucrose (molar mass = 342) has been prepared by dissolving 68.4g of sucrose in one kg of water. Kf for water is 1.86 K kg mol 1 and vapour pressure of water at 298 K is atm. 12. The vapour pressure of the solution at 298 K will be (a) atm (b) atm (c) atm (d) atm 13. The osmotic pressure of the solution at 298 K will be (a) 4.29 atm (b) 4.49 atm (c) 4.69 atm (d) 4.89 atm 14. The freezing point of the solution will be (a) C (c) C (b) C (d) C

24 Chemistry: Liquid Solutions-24 SECTION IV Subjective Questions LEVEL I 1. A solution of A and B with 30 per cent moles of A is in equilibrium with its vapour which contains 60 mole per cent of A. Assuming ideality of the solution and the vapour, calculate the ratio of the vapour pressure of pure A to that of pure B. 2. At 37 C, the osmotic pressure of blood is 7.65 atm. How much glucose should be used per litre for an intraveninous injection that is to have the same osmotic pressure as blood. [M2 = 180g mol 1 ) 3. 1kg of an aqueous solution of Sucrose is cooled and maintained at 4 C. How much ice will be separated out if the molality of the solution is 0.75? Kf (H2O) = 1.86 Kg mol 1 K. 4. The density of 2.0 M solution of acetic acid (Mm = 60 g mol 1 ) in water is 1.02 Kg/ml. Calculate the mole fraction of acetic acid. 5. An aqueous solution of sugar containing 9 gm per 50 ml of the solution is isotonic with an aqueous solution of gm. NaCl per 100 ml. Calculate the molecular weight of sugar [Temperature = 300 K] g of monobasic acid when dissolved in 100 g of water lowers the freezing point by C. 0.2 gm of the same acid when dissolved and titrated required 15.1 ml of N/10 alkali. Calculate degree of dissociation of the acid. ( Kf for water is 1.86) 7. The Boiling Point of 5% solution of a non-volatile solute in water is C. The boiling point of pure water is 100 C. Calculate molecular mass of the solute. Kb for water is 0.52 Km Find the boiling point and freezing point of a solution containing 0.520g glucose (C6H12O6) dissolved in 80.2g water. For water Kf = 1.86 K/m, Kb = 0.52 K/m. 9. In a cold climate, water gets frozen causing damage to the radiator of a car. Ethylene glycol is used as an anti-freezing agent. Calculate the amount of ethylene glycol to be added to 4 kg of water to prevent it from freezing at 6 C. Kf for water = 1.86 kg mol 1 K gm of CoCl3.6NH3 (mol weight = 267) was dissolved in 100 gm of water. The freezing point of the solution was 0.29 C. How many moles of the solute particles exist in solution for each mole of solute particles introduced? LEVEL II 1. When a compound is dissolved in 1000gm of benzene, the freezing point of benzene is depressed by 1.28 K. If the same amount of compound is dissolved in 1 kg of water, the freezing point of water was depressed by 1.4 K. If the compound completely dissociates in water but remains undissociated in benzene, calculate into how many ions does a molecule dissociate in water. Kf for benzene = 5.12 Kf for water = 1.86

25 Chemistry: Liquid Solutions m solution of CH3COOH is 3% dissociated at 25 C. Calculate freezing point and osmotic pressure of the solution Kf for water = 1.86 C m When 36.0 g of a non-volatile solute having the empirical formula CH2O is dissolved in 1.20 kg of water, the solution freezes at 0.93 C. What is the molecular formula of the compound? 4. A complex is represented as CoCl3.xNH3. Its 0.1 m solution in aqueous solution shows Tf = Kf(H2O) = 1.86 mol 1 K and assume 100% ionization and co ordination number of Co(III) is six. What is the complex? 5. A solution containing 30g of a non-volatile solute in exactly 90g water has a vapour pressure of 21.85mm of Hg at 25 C. Further 18g of water is then added to solution, the new vapour pressure becomes mm of Hg at 25 C. Calculate, a) Molecular weight of solute b) Vapour pressure of water at 25 C 6. Ethylene dibromide (C2H4Br2) and 1,2-dibromopropane (C3H6Br2) form a series of ideal solution over the whole range of composition. At 85 C, the vapour pressures of these pure liquids are 173 mm Hg and 127 mm Hg respectively. a) If 10g of ethylene dibromide is dissolved in 80g of 1, 2-dibromopropane. Calculate the partial pressures of each components and the total pressure of the solution at 85 C. b) Calculate the composition of the vapour in equilibrium with the above solution and express as mole fraction of ethylenedibromide. c) What would be the mole fraction of 1,2-dirbomopropane in solution at 85 C equilibriated with 50:50 mole mixture in the vapour? 7. Calculate the amount of ice that will separate out on cooling a solution containing 50g of ethylene glycol in 200g water to 9.3 C (Kf for water = 1.86 K mol 1 kg). 8. The formula weight of an acid is In a titration, 100 cm 3 of solution of this acid containing 39.0g of this acid per litre were completely neutralized by 95.0 cm 3 of aqueous NaOH containing 40g of NaOH per litre. What is the basicity of this acid? 9. A certain mass of a substance, when dissolved in 100g C6H6, lowers the freezing point by 1.28 C. The same mass of solute dissolved in 100g water lowers the freezing point by 1.40 C. If the substance has normal molecular weight in benzene and is complexly ionized in water. Into how many ions does it dissociate in water? Kf for H2O and C6H6 are 1.86 and 5.12 Kg mol At 25 C, 1 mole of A having a vapour pressure of 100 torr and 1 mol of B having a vapour pressure of 300 torr were mixed. The vapour at equilibrium is removed, condensed and the condensate is heated back to 25 C. The vapours now formed are again removed, recondensed and analysed. What is the mole fraction of A in this condensate?

26 Chemistry: Liquid Solutions-26 LEVEL III (Judge yourself at JEE level) 1. The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution, containing 7.0g of the salt per 100g water at 100 C is 70 per cent. If the vapour pressure of water at 100 C is 760 mm, calculate vapour pressure of the solution. 2. A very small amount of a non-volatile solute (that does not dissociate) is dissolved in 56.8 cm 3 of benzene (density 0.89g cm 3 ). At room temperature, vapour pressure of this solution is mm Hg while that of benzene is 100 mm Hg. Find the molality of this solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene, what is the value of molal freezing depression constant of benzene? 3. A solution of non-volatile solute in water freezes at 0.30 C. The vapour pressure of pure water at 298K is mm Hg and Kf for water is 1.86 degree/molal. Calculate the vapour pressure of this solution at 298K. 4. The molar volume of liquid benzene (density = 0.877g ml 1 ) increases by a factor of 2750 as it vaporizes at 20 C and that of liquid toluene (density = g ml 1 ) increases by a factor of 7720 at 20 C. A solution of benzene and toluene at 20 C has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the vapour above the solution. 5. What weight of the non-volatile solute, urea (NH2 CO NH2) need to be dissolved in 100g of water, in order to decrease the vapour pressure of water by 25%? What will be the molarity of the solution? 6. Addition of 0.643g of a compound to 50 ml of benzene (density g/ml) lowers the freezing point from 5.51 C to 5.03 C. If Kf for benzene is Calculate the molecular weight of the compound 7. The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A nonvolatile non-electrolyte solid weighing 2.175g is added to 39.0g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance? 8. The vapour pressure of a dilute aqueous solution of glucose (C6H12O6) is 750 mm of mercury at 373 K. Calculate (i) molality and (ii) mole fraction of the solute. 9. For the reaction, N2O5(g) 2NO2(g) + 0.5O2(g), calculate the mole fraction of N2O5(g) decomposed at a constant volume and temperature, if the initial pressure is 600 mm Hg and the pressure at any time is 960 mm. Assume ideal gas behaviour. 10. To 500 cm 3 of water, 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 kg 1 mol 1 and g cm 3 respectively.

27 Chemistry: Liquid Solutions gm of NH4Cl (mol. weight = 53.5) when dissolved in 1000 gm of water lowered the freezing point by C. Calculate the degree of hydrolysis of the salt if its degree of dissociation is The molal depression constant of water is 1.86 kg mol 1 K. 12. The vapour pressure of two miscible liquids (A) and (B) are 300 and 500mm of Hg respectively. In a flask 10 mole of (A) is mixed with 12 mole of (B). However, as soon as (B) is added, (A) starts polymerising into a completely insoluble solid. The polymerization follows first-order kinetics. After 100 minute, mole of a solute is dissolved which arrest the polymerization completely. The final vapour pressure of the solution is 400mm of Hg. Estimate the rate constant of the polymeristaion reaction. Assume negligible volume change on mixing and polymerization and ideal behaviour for the final solution. SECTION V Miscellaneous Questions 1. Match List I, List II & List - III and select the correct answer using the codes given below the lists. List-I List II List III X1 B.P Liq. Vap. Composition B.P Y1 Hmix 0 Vmix 0 Z1 Solvent solvent and solute solute interaction is equal to solute solvent interaction X2 B.P Vap. Liq. Composition B.P Y2 Hmix 0 Vmix 0 Z2 Solvent solvent and solute solute interaction is less than that of solute solvent interaction X3 B.P Vap. Liq. Composition B.P Y3 H = 0, V = 0 Z3 Solvent solvent and solute solute interaction is greater than that of solute solvent interaction Codes X1 X2 X3 (a) Y3, Z2 Y2Z3 Y1Z1 (b) Y1, Z1 Y3Z2 Y1Z3 (c) Y1, Z2 Y2Z3 Y3Z2 (d) Y3, Z1 Y2Z3 Y1Z2

28 Chemistry: Liquid Solutions-28 The following questions (2 to 6) consists of two statements, one labelled as ASSERTION (A) and REASON (R). Use the following key to choose the correct appropriate answer. (a) If both (A) and (R) are correct, and (R) is the correct explanation of (A). (b) If both (A) and (R) are correct, but (R) is not the correct explanation of (A). (c) If (A) is correct, but (R) is incorrect. (d) If (A) is incorrect, but (R) is correct. ASSERTION (A) REASON (R) 2. A solution which contains one gram equivalent of solute per litre of the solution is called normal solution. 3. Van t Hoff factor for benzoic acid in benzene is less than one. 4. If on mixing the two liquids, the solution becomes hot, it implies that it shows negative deviation from Raoult s law. 5. If a liquid more volatile than the solvent is added to the solvent, the vapour pressure of the solution may increase, i.e., ps p o M solution of glucose has higher increment in the freezing point than 0.1 M solution of urea. A normal solution means a solution in which the solute does not dissociate or associate. Benzoic acid behaves as a weak electrolyte in benzene Solution which show negative deviation are accompanied by decrease, in volume. In the presence of more volatile liquid solute, Raoult s law does not hold good. Kf for both has different values.

29 Answers to Assignments Chemistry: Liquid Solutions (b) 2. (b) 3. (c) 4. (d) 5. (c) 6. (d) 7. (b) 8. (c) 9. (c) 10. (b) SECTION - I 11. (b) 12. (c) 13. (c) 14. (c) 15. (c) 16. (b) 17. (d) 18. (d) 19. (c) 20. (c) 21. (b) 22. (d) 23. (c) 24. (b) 25. (a) 1. (a, c) 2. (a, b, c, d) 3. (a, b) 4. (b) 5. (a) 6. (b, c, d) 7. (b) 8. (b) 9. (c, d) 10. (a, b, c, d) 11. (b, d) 12. (b, d) 13. (a, b, c) 14. (a, b, c) SECTION - II 15. (b, d) 16. (a), (c) 17. (a, b) 18. (b, d) 19. (a, b, c) 20. (b, c) 21. (a, b, c) 22. (a, b) 23. (a, b, c) 24. (a, d) 25. (a, b, c) 1. (a) 2. (d) 3. (b) 4. (b) 5. (a) 6. (a) LEVEL III 7. (b) 8. (a) 9. (a) 10. (c) 11. (a) 12. (d) 13. (d) 14. (c) SECTION - IV LEVEL I g gm g/mol K g LEVEL II atm 3. C 2 H 4 O 2 4. [Co(NH 3 ) 5 Cl]Cl and mm 6. a) P A = 20.41; b) P B = c) 0.154; d) g ions 10. X a = 0.1 LEVEL III mm K m mm m g/mol g mol , 0.73 mol kg (initial time), 0.25 (any time) K 11. h = SECTION - V 1. (d) 2. (b) 3. (c) 4. (b) 5. (c) 6. (d)