LIQUID SOLUTIONSSolution: A homogeneous mixture of two or more substances. (Solute + solvent = solution) Saturated Solution : A solution that is in equilibrium with pure solid solute. No more solute can be dissolved in it. 1. Methods of expressing the concentration of a solution The concentration of a solution can be expressed in a number of ways. The important methods are: (i) Mass percentage or per cent by mass: %(w/w) Mass percentage of solute = Mass of solute x 100 Mass of solution Mass of solute Mass of solute Mass of solvent = Mass of solute Volume of solution Density of solution .in = (ii) Percent mass by volume: te (iii) Parts per million (ppm) : Mass of solute x 100 Volume of soltuion ps %(w/v) = Mass of solute x 106 Mass of solution yS ppm = St ud (iv) Mole fraction: Let n moles of solute (A) and N moles of solvent (B) be presnet in a solution. n N Mole fraction of solute = = XA , Mole fraction of solvent = = XB Nn Nn In binary solution, XA + XB = 1 Mole fraction is independent of temperature of the solution. (v) Molality No. of moles of solute weight (in kg) of solvent Let wA grams of the solute of molecular mass mA be present in wB grams of the solvent, then Molality (m) = Molality (m) = wA x 1000 mA wB Relation between mole fraction and Molality : Page 1 of 33 X A 1000 wA 1000 = m = XB mB wB mB Note: (i) Molality is the most convenient method to express the concentration because it involves the mass of liquids rather than their volumes. It is also independent of the variation in temperature. www.StudySteps.in LIQUID SOLUTIONS (vi) Molarity (Molar concentration) No. of moles of solute Volume (in litre) of solution wA Molarity of the solution = x 1000 mA V Relation between molarity and % solute by mass : Let d = density of solution in g/mL and let it contains x% (w/w) solute by mass. Number of moles of solute in 1 litre x d 10 mass of solute in gram = = M grams molecular mass of solute solute x d 10 M = mA Molarity (M) = Molarity of dilution: Before dilution M1V 1 = After dilution M2V 2 Relationship between molality and molarity : 1000 M 1000 d M M solute molarity = d molarity msolute ps Molality (m) = MR = resultant molarity .in Molarity of mixing: M1V1 + M2V2 + M3V3 = MR(V1 + V2 + V3) yS te Illustration 1. The density of a solution containng 13% by mass of sulphric acid is 1.09 g/mL. Calculate the molarity of the solution. Solution: In solving such numericals, the following formula can be applied: % strength of soln. density of soln.10 Mol. mass 13 1.09 10 = 1.445 M 98 St M = ud Molarity = Illustration 2. The density of a 3 M sodium thiosulphate solution (Na 2S2O3) is 1.25 g/ mL. Calculate (i) the percentage by mass of sodium thiosulphate, (ii) the mole fraction of sodium thiosulphate and (iii) molalities of Na + and S2O32– ions. Solution: Page 2 of 33 M = x d 10 mA 3 = x 1.25 10 158 x = 37.92 (ii) No. of moles of Na2S2O3 = 474 = 3 158 Mass of water = (1250 – 474) = 776 g , Mole fraction of Na2S2O3 = No. of moles of water = 3 3 = = 0.065 43.1 3 46.1 (iii) No. of moles of Na+ ions = 2 × No. of moles of Na2S2O3 = 2 × 3 = 6 Molality of Na+ ion = No. of moles of Na ions Mass of water in kg www.StudySteps.in 776 = 43.1 18 LIQUID SOLUTIONS 6 × 1000 = 7.73 m 776 No. of moles of S2O32– ions = No. of moles of Na 2S2O3 3 Molality of S2O32– ions = × 1000 = 3.86 m 776 = Illustration 3. One litre of sea water weighs 1030 g and contains about 6 × 10 –3g of dissolved O2. Calculate the concentration of dissolved oxygen in ppm. Solution: Mass of O2 in mg = 6 × 10–3g × 103 mg/g = 6mg Mass of O in mg 6 2 ppm of O2 in 1030 g sea water = Mass of sea water in kg (1030 /1000)kg 6 1000 5.8ppm 1030 Thermostat Vapours P ps Liquid .in 2. Vapour Pressure te Manometer yS Physical equilibrium between liquid vapours ud The pressure exerted by the vapour (molecules in the vapour phase) over the surface of the liquid at the equilibrium at given temperature is called the vapour pressure of the liquid. Factors affecting vapour pressure Temperature : Vapour pressure Temperature St (i) The temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure is called its boiling point. (ii) Nature of liquid: Vapour pressure of liquid 1 The strength of intermolecular forces acting between molecules For example, ethyl alcohol has higher vapour pressure because of the weak intermolecular forces acting between its molecules than water which has stronger intermolecular forces acting between water molecules of volatile liquid has lower boiling point than a non-volatile liquid. 3. Vapour Pressure of solution (i) Vapour Pressure of a Solution Containing Non Volatile Solute - Raoult’s Law : Raoult’s Law : Acccording to this law, the partial pressure of any volatile constituent of a soltuion at a constant temperature is equal to the vapour pressure of pure constituent multiplied by the mole fraction of that constituent in the solution. Page 3 of 33 www.StudySteps.in 0. Therefore.. Illustration 4: The vapour pressure of pure water at 37°C is 47.9 0.3 torr 7 .StudySteps.. Also calculate vapour pressure lowering.78 mol . (RLVP) VP) .1 – 46.89 According to Raoult’s law..996 46..2 torr..78 27. then PAo P – P XB A A is called relative lowering of vapour pressure.. 500 20 n H2O 27.in If X B be the mole fraction of the solute B.78 X H2O 0.in . the partial vapour pressure of a solution containing a non-volatile solute is equal to the product of vapour pressure of the pure liquid (solvent PA ) and its mole fraction in the solution. the vapour pressure contribution by the non-volatile solute is negligible.2 torr Illustration 5.. What is the molality of the solution? Solution: Given: PAo = 59.. then PA PA X A XA XB 1 XA 1 – XB hence PA PA (1 – X B ) PA – PA X B .. (Here pB0 = negligible as solute is non-volatile) Let X A be the mole fraction of solvent A.3 torr.2 torr..11 mol Solution: n(glucose) 18 180 n H2O 27.(1) Vapour pressure PA0 PA 0 1 = o XA PA 1 0 XA XB Relative Lowering of vapour pressure: For a solution of a non-volatile solute in a liquid.1 0. Raoult’s law states that the relative lowering of vapour pressure P .9 By RLVP Xsolute = = .996 n H2O n (glu cos e) 27.1 torr..(2) ps PAo XB PAo PA .2 Page 4 of 33 www.11 27.LIQUID SOLUTIONS PA XA PA = PA0 XA .78 0. pA = 51. The vapour pressure of ethyl alcohol at 25°C is 59..1334 59..9 torr Lowring of vapour pressure PHo 2O – PH2O 47.(3) P A yS PAo PA te ( PAo PA is known as lowering of vapour pressure) St ud PA – PA Therefore. for a non-volatile solute A is equal to the mole fraction of the solute when the solvent alone is volatile.. Vapour pressure of solution PH2O PHo 2O X H2O 47. The vapour pressure of a solution of urea in ethyl alcohol is 51.. What is the vapour pressure of an aqueous solution at 37°C containing 20 g of glucose dissolved in 500 gm of water. 80 atm.00289 RLVP = XA = 0.30 (D) 56. Which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.3 A 6% (by weight) of non-volatile solute in isopentane at 300 K has a vapour pressure of 1426 mm Hg. wt.543 molal (D) 0.68 mm (B) 770.9 The vapour pressure of pure liquid solvent A is 0.9 mm of Hg. 32 (B) 92.2 g of a solute is dissolved in 60 g ether. When 7. vapour pressure is lowered by 32 units.StudySteps. Calculate relative lowering of vapour pressure of 0.41 mm Hg.75 (D) 0.32 mm Q. Solution : X B 1000 0. When a non-volatile substance B is added to the solvent.50 (B) 0. The vapour pressure of pure water is 23. Molality of solution is (A) 0.161 molal aqueous solution.518 (C) 1. If a non-volatile solute is added to it vapour pressure falls by 4 mm.77 (B) 150.23 (D) 190.in Q.346 molal 0.40 Page 5 of 33 St ud yS te ps .079 mol/kg of solvent (B) 0.8666 46 Illustration 6.8 Vapour pressure of pure water is 40 mm.00289 1.1 www. is (A) 20 gm (B) 10 gm (C) 7.158 mol/kg of solvent (D) 0.0864 molal (C) 1.76 mm Hg.25 (C) 0.4 Find out the weight of solute (M.1334 1000 = 3.316 mol/kg of solvent Q.60 atm .26 Q. 40).35 (C) 116.5 gm (D) 5 gm Q. Daily Practice Problem Sheet The vapour pressure of ether at 20ºC is 442 mm.7 Twenty gram of a solute are added to 100 g of water at 25ºC. mole fraction of the component B in the solution is (A) 0.6 The weight of a non-volatile solute (m.46 mm (D) 981. molality of solution is (A) 6.256 mol/kg of solvent (C) 0.LIQUID SOLUTIONS (molality) m = m= XB 1000 XB 1000 = 1 X B M solvent X A M solvent ( XA + XB = 1) 0.891 Q. If molecular weight of ether is 74 then molecular weight of solute is (A) 113. 20 g of isopentane is now added to the solution. 60) that is required to dissolve in 180 g water to reduce the vapour pressure to 4/5th of pure water (A) 130 g (B) 150 g (C) 300 g (D) 75 g Q.2 The vapour pressure of pure benzene at 25ºC is 639.133 (D) 1.173 molal (B) 3.772 molal Q. The mass of this solute that is required in 100 g water to reduce the vapour pressure to one-half of the pure water is (A) 333 g (B) 666 g (C) 166 g (D) 256 g Q.in .89 mm (C) 1212. Then : (i) Molecular weight of solute is (A) 28.7 mm of Hg and the vapour pressure of a solution of a solute in C6H6 at the same temperature is 631.65 (ii) Vapour pressure of isopentane at 300 K is (A) 1541. will be (A) 2. Hence. The resulting solution has a vapour pressure of 1445 mm of Hg at the same temperature.161 = 1 X 18 B we know Molality = X B 1000 1 X B msolvent XA = 0. wt. its vapour pressure drops to 0.5 The molality of a solution containing a non-volatile solute if the vapour pressure is 2% below the vapour pressure of pure water.81 (C) 113.213 (B) 2. vapour pressure of solution is 22. 13 The mole fraction of the solvent in the solution of a non-volatile solute is 0.19 mmHg at a temperature at which pressure of pure water is 25 mm Hg ? (A) 342 g (B) 360 g (C) 375 g (D) 380 g (ii) Vapour Pressure of a Solution Containing Two Volatile Liquids Raoult’s law states that the partial vapour pressure of a component of a solution of two miscible liquids A and B at a given temperature is equal to the product of the vapour pressure of the pure component at that temperature and its mole fraction in the solution.10 Vapour pressure of CCl4 at 25ºC is 143 mm Hg.2 (B) 0.02 (D) 0.19 How many grams of sucrose must be added to 360 g of water to lower the vapour pressure by 1.11 Lowering of vapour pressure of 1.9 (C) 0.980. its vapour pressure is reduced to 9.99 mm Q.7 (C) 0.9 mm Hg (B) 24.42 mm Hg (D) 31.305 mol (B) 0.00 torr (D) 40. (A) 0.1.0 torr.34 mm (D) 143.5 gm of non-volatile solute (mol.00 torr Q.in Q.1 is equal to – (A) 23.15 The vapour pressure of water at room temperature is 23.49 Q. mass 65) is dissolved in 100 ml CCl4.8 mm Hg.StudySteps.00 m solution of a non-volatile solute in a hypothetical solvent of molar mass 40 g at its normal boiling point.partial vapour pressure of B in the solution) The total vapour pressure of an ideal solution containing components A and B is the sum of partial vapour pressures of all the components (Dalton’s law of partial pressures) Page 6 of 33 www.6 St ud yS te ps . The mole fraction of the solute in solution is 0. If the molecular mass of A is 200 amu. Mathematical Expression: Let us assume that a solution has nA moles of liquid A and n B moles of liquid B. the same solute being dissolved (A) 0. X B nA nB nA nB According to Raoult’s law. Let PA be the vapour pressure of the pure liquid A and PB is the vapour pressure of the pure liquid B.14 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. then the vapour pressure of the solution at 25ºC is [ Given : Density of CCl4 = 1.01 (B) 0. How many moles of nonvolatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure of 167. Mole fraction of B. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.58 g/cm3 ] (A) 141. If the molecular weight of the solvent is 0.44 Torr (B) 14. is (A) 29.LIQUID SOLUTIONS Q.16 The vapour pressure of pure A is 10 torr and at the same temperature when 1 g of B is dissolved in 20 g of A. PA PA X A (where PA is the partial vapour pressure of liquid A in the solution) Similarly.17 The vapour pressure of pure benzene C6H6 at 50ºC is 260 Torr.44 mm Hg Q.4 torr (C) 35.12 Torr (C) 312 Torr (D) 352 Torr Q.0 Torr at 50ºC ? (A) 0.in . X A . If 0.336 mol (D) 0.12 The vapour pressure of a liquid decreases by 10 torr when a non-volatile solute is dissolved. PB PB X B ( PB .2 (D) none of these Q. then the molecular mass of B is – (A) 100 amu (B) 90 amu (C) 75 amu (D) 120 amu Q.8 (D) 0.15 (B) 5.39 mm (C) 199. what is the weight ratio of solvent to solute. What would be the mole fraction of the liquid if the decrease in vapour pressure is 20 torr.663 mol Q.605 mol (C) 0.2 mm Hg (C) 21.980 (C) 0.18 Lowering of vapour pressure due to a solute in 1 molal aqueous solution at 100ºC is (A) 13. nB nA Mole fraction of A. The relative lowering of vapour pressure is (A) 0.93 mm (B) 94.23 torr (B) 30.3 times the molecular weight of the solute. Find the vapour pressure of the pure liquid A..(2) About relation shows that graph between P and XB is a straight line having slope PBo – PAo and intercept PAo Similarly P = PBo + ( PAo – PBo)XA .. 3..in .. of the solution for any composition and is given by the sum of the partial vapour pressure of liquids of A and B.LIQUID SOLUTIONS P PA PB PAo X A PBo X B .. Page 7 of 33 o o Ptotal = PA PB PA X A PB X B nA = 28 = 0. mass = 140) has a vapour pressure of 160 mm at 37ºC. 140 Liquid B is water.. According to Raoult’s law. Straight line II represents the plot of partial vapour pressure of liquid (B) PB and its mole fraction St 2.in P = PAo (1–XB) + PBo XB PAo – PAo XB + PBo XB P = PAo + ( PBo – PAo)XB . ud When mole fraction of liquid A is X A 1 . P PA PB Illustration 7..e. when XB 0 PB 0 . P= A PBo A I.. i.2.0 18 www. Straight line (III) represents the total vapour pressure.. nB = 72 = 4...(3) XA 0 PA 0 .. P.) Solution : For two miscible liquids. the liquid A is pure and its vapour pressure is equal to PAo as shown by line (I) (X B ) .. the liquid B is pure and its vapour pressure is equal to PBo as shown by line (II). P o XA A =P II. 72. According to Raoult’s law this should be a straight line when X A 1.. X B 1. PA PAo yS 1. (The vapour pressure of water at 37ºC is 150 mm.StudySteps. P B =Po B X B XA + XB = 1 0 XA 1 XB 1 0 XA = 1– XB . PB PBo When mole fraction of liquid B is X B 1 .(1) PAo Vapour pressure o P XB o P X A+ B III.. An aqueous solution containing 28% by mass of a liquid A (mol. this should be a straight line. ps Above figure shows the relationship between partial vapour pressure and mole fraction of an ideal solution at constant temperature.. te Straight line I represents the plot of vapour pressure of liquid A (PA ) and its mole fraction (X A ) . Its mass is (100–28). Determination of compostion in vapour phase Dalton’s Law v/s Raoult’s Law: The composition of the vapour in equilibrium with the solution can be calculated applying Daltons’ law of partial pressures.in .0 = 4. Page 8 of 33 A B B A B A B A B vapours (Dalton's law) y A = more fraction of A in vapour phase y B = more fraction of B in vapour phase A ..(2) P yB PBo =1 yB 1 y = Ao + o P PB PA .(1) PBo X B P Above formula is used for calculation of mole fraction of B in vapour phase yB = PBo XB = yB × P From (1) .. XA = on adding P yA From (2) PAo XA + XB = P yA PAo + P yB PBo XB = . PB = yB × P o A .2 4.. Let pA and pB be the partial pressure of vaours A and B respectively and total pressure P. PB0 = 150 mm So 0 .2 4 ... Partial pressure = Mole fraction × Total pressure For A PA = yA × P PAo X A P XA = yA × P yA = P Above formula is used for calculation of mole fraction of A in vapour phase For B.2 Given Ptotal = 160 mm..2 + 4..15mm 0 ..2 0 4 .15 4.0 PA 5 ..in A A B B A B A B A B A solution (Raoult's law) X A = more fraction of A in liquid phase X B = more fraction of B in liquid phase te A ps (Y A + Y B = 1) (X A + X B = 1) yS Solution of (A & B) St ud From Raoult's law P = PAoXA + PBo XB o PA= PAXA and PB = PBo XB From Dalton's law.2 160 = p 0A 17.StudySteps..(3) www.2 360.LIQUID SOLUTIONS Total number of moles = 0.. Let the mole fractions of vapours A and B be YA and YB respectively.0 4 .. in ‘XA’ of ethyl alcohol = 60 40 = 1.03 mm (D) 129.5107 × 44.e. Daily Practice Problem Sheet Q.40 = 0.31 mm (C) 52.7= 43.4 mm.. 24.13 2.63 mm (C) 104.5 =22. No.5 mm Q.in . P. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Illustration: 8.53 mm (B) 76. 48. of 145 mm. 1. If the vapour pressure of water is 155 mm then vapour pressure of A at the same temperature will be (A) 205 mm (B) 105 mm (C) 185 mm (D) 52. the one having relatively greater vapour pressure. Find out : (i) The vapour pressure of ethylene bromide & propyline bromide in a 60% by weight solution of ethylene bromide in propylene bromide at same temperature.2 The vapour pressure of pure ethylene bromide and propylene bromide are 170 and 127 mm of Hg at a temperature.LIQUID SOLUTIONS Above formula is used to calculate total vapour pressure when mole fractions are given in vapoure phase Avoid confusion P = PAoXA + PBoXB This formula is used to calculate total pressure when mole fraction are given in liquid phase 1 yA y o Bo This formula is used to calculate total pressure when mole fraction P PA PB are given in vapour phase Note: Thus. 48. is (A) 52. 24.31 mm (ii) Total vapour pressure of solution.304 1. The vapour pressures of ethanol and methanol are 44.73 + 43.73 mm Hg yS te ‘XB’ of methyl alcohol = Partial pressure of methyl alcohol = XB.21 mm Page 9 of 33 www.4893 1. of moles of C2H5OH = 1.4893 × 88.40 mm Hg ud Total vapour pressure of solution = 22. Calculate the total vapour pressure of the solution and the mole fracion of methanol in the vapour.25 Partial pressure of ethyl alcohol = XA.304 = 0.304 1.71 mm (C) 101.PA0 = 0.13 mm Hg St Mole fraction of methyl alcohol in the vapour = Partial pressure of CH 3OH Total vapour pressure = 43.5107 1.9 mm. in case of ideal solution the vapour phase is phase is richer with more volatile component i.304. = 128) 64% by weight has a V.40 = 66.25 = 0.25 46 32 ps Solution: No.5 mm and 88.4 mm. of moles of CH3OH = = 1.25 .63 mm (D) 104. wt.7 mm Hg respectively.StudySteps. is (A) 153.6563 66.PB0 = 0.9 mm.1 An aqueous solution containing liquid A (M. hence. the mole fraction of benzene in vapour phase when vapours are in equilibrium with liquid mixture.3 and 2.9 Solution of two volatile liquids A and B obey Raoult’s law. 1200 Q.3 and 2.3 cm of Hg (C) 5. 314.8.9 mm.57 mm (C) 276.6 At 40ºC the vapour pressure in torr of methanol and ethanol solution is P = 119 x + 135 where x is the mole fraction of methanol. The vapour pressure (in torr) of equimolar mixture of the two liquids will be (A) 166 (B) 83 (C) 140 (D) 280 Q.in . 0. 300 (C) 400. 2 2 (D) 1 3 . 0. 600 (D) 800.3 and 3. is (A) 0.11 The vapour pressure of two liquids are 15000 and 30000 in a unit. At a certain temperature. 0.80 (B) 0.LIQUID SOLUTIONS Two liquids A and B form ideal solution at 300 K. Assume ideal behaviour.305 (D) 0. the vapour pressure of pure B at 25ºC is (A) 1 atm (B) 14 mm of Hg (C) 140 mm of Hg (D) 56 cm of Hg Q. the vapour pressure of solution decreases by 10 mm of Hg.3 – 2XB) in cm of Hg .41 (B) 0.1 cm of Hg (B) 2. If two forms ideal solution.in Q. When equimolar solution of liquids is made. the mole fraction of A in vapour phase is 0. 314.4g of toluene are mixed. 550 (B) 200.465 Page 10 of 33 www.28 mm (D) 138. 0.3 Q.3 and 3.8 Vapour pressure (in torr) of an ideal solution of two liquids A and B is given by : P = 52 XA + 114 where XA is the mole fraction of A in the mixture. 0. it is found that when the total vapour pressure above solution is 400 mm of Hg.57 mm (B) 276.5 (D) 0.4 g of benzene and 64.695.3 cm of Hg (D) 5. Vapour pressures of A and B in pure state will be(A) 330.50.65 Q. 23. (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 Q.3 cm of Hg Q. Find out the mole fraction of each component in vapour state for the solution of two having mole fraction of ether 0.35.4 The vapour pressure of two volatile liquid mixture is PT = (5. (A) 0.201 (C) 0.5 A mixture contains 1 mole volatile liquid A (PºA = 100 mg Hg) and 3 moles volatile liquid B(PºB = 80 mm). 3 3 (C) 1 1 .StudySteps. 3 3 (B) 1 2 . 0. 4 4 Q.28 mm St ud yS te ps .20. If solution be haves ideally. The vapour pressure of a solution containing one mole of A and four mole of B is 560 mm of Hg.799.65 then the vapour pressures of two pure liquids at the same temperature will be (A) 138.59. 0. What are vapour pressure of pure liquids A and B ? (A) 3. where XB is mole fraction of B in mixture.9 mm.4 mm. 628.535.4 mm.75 (C) 0. Then the mole fraction of A and B in vapour phase will be (A) 2 1 . 0.25. 628.45 and in liquid phase 0.12 At 30ºC. At the same temperature if one mole of B is taken out from the solution. total vapour pressure of the solution is approximately (A) 80 mmHg (B) 85 mmHg (C) 90 mmHg (D) 92 mmHg Q.10 The vapour pressure of benzene and toluene at 20ºC are 75 mm of Hg and 22 mm of Hg respectively. If vapour pressure of solution is 84 mm of Hg at 25ºC. the vapour pressure of pure ether is 646 mm and of pure acetone is 283 mm.7 Vapour pressure of pure A is 70 mm of Hg at 25ºC it forms an ideal solution with B in which mole fraction of A is 0.5. the vapour pressure of pure A and B is 108 and 36 torr respectively.92 .8 (B) 0. What is the mole fraction of A in the solution ? (A) 0.50 (D) 0.1 (B) 0. 0. approximately.25 (C) 0.33 (C) 0.23 Mole fraction of the toluene in the vapour phase which is in equilibrium with a solution of benzene (pº = 120) and toluene (pº = 80 ) having 2. 0. If the mole fraction of benzene and toluene in vapour phase are 0.2 (B) 0.StudySteps.16 1 mole benzene (Pº = 42 mm) and 2 moles toluene (Pº = 36 mm) will have (A) total vapour pressure 38 mm (B) mole fraction of vapours of benzene above liquid mixture is 7/19 (C) ideal behaviour (D) all of the above te ps Q. of pure B pºB = 150 mmHg Distillate of vapours of a solution containing 2 moles A and 3 moles B will have total vapour pressure.37 (C) 0.68 mm (B) 79.18 A solution of benzene (pº = 120 torr) and toluene (pº = 80 torr) has 2.75 .40 Page 11 of 33 www.25 (C) 0. P. The vapour pressure of benzene and toluene are respectively 75 mm and 22 mm at 20ºC.LIQUID SOLUTIONS Q.33.2 (C) 0. of pure A pºA = 100 mmHg V.14 At 90ºC. then X 'A (B) X > 1 A X 'A (C) X <1 A (D) X'A + X A = 1 Q.34 Q. 0.25 (C) 0. If XA and XA are the mole fractions of A in the solution and vapour in equilibrium.0 Q.08 (D) 0.5 (D) 1.17 At a given temperature. is (A) 0.22 At a certain temperature pure liquid A and liquid B have vapour pressures 10 torr and 37 torr respectively. 0.60 ud yS Q. the vapour pressure of toluene is 400 mm and that of xylene is 150 mm.15 V.50 (B) 0. P.56 mm (D) 29.5 atm.32 mm (C) 58. the mole fraction of A in vapour in equilibrium with the solution at a given temperature is(A) 0.13 Benzene and toluene form an ideal solution. If XA is the mole fraction of component A in the vapour in equilibrium.46 (D) 0.in Q.60 (D) 0. If the mole fraction ratio of xA : xB = 1 : 3.654 (C) 0. under ideal behaviour conditions ? (A) 0.67 Q.20 Mole fraction of a liquid A in an ideal mixture with another liquid b is XA. the vapour in equilibrium with the liquid has the components A and B in the partial pressure ratio PA : PB = 1 : 7 .0 mol of each component. What will be the mole fraction of B in the vapour phase which is in equilibrium with a solution containing equimole fraction of A and B.8.40 X 'A (A) X = 1 A St Q.0 moles of each. At equilibrium the mol fraction of toluene in vapour phase is (A) 0.63 and 0.63. 0. PºB = vapour pressure of pure B ) X 'A (A) P º A X A (B) Pº A X A X 'A (C) P º B X 'A XA (D) Pº B X A X 'A Q.346 (B) 0.in . is – (A) 0. For a certain ideal solution of A and B.66 . 0.60 (D) 0.54. The composition of liquid mixture that will boil at 90ºC when the pressure of mixture is 0.25 (B) 0. the total pressure of the liquid mixture is (PºA = vapour pressure of pure A . then : (i) The vapour pressure of mixture is (A) 39.37 respectively.50 (B) 0. 0.21 Two liquids A and B have PºA : PºB = 1 : 3 at a certain temperature.2. on condensation (A) 135 mm (B) 130 mm (C) 140 mm (D) 145 mm .5 (D) none of these Q. 0.19 Liquids A and B form an ideal solution and the former has stronger intermolecular forces.24 mm (ii) Mole fraction of benzene & toluene in liquid phase is (A) 0. The total vapour pressure of an ideal solution containing liquids A and B is given by the following equation. 960.01 mm respectively.5 (C) 3. These two compounds form ideal liquid and gaseous mixtures.StudySteps.07 (B) 0. 865. B = 43 % (C) A = 33% . is (A) 60. The vapour pressure of pure A and B respectively are (A) 165.44 mm (B) 30.15 mm (D) 276.65 (D) 0.60 mole fraction of toluene.93 (C) 0.29 An aqueous solution containing 28% by mass of a liquid A (mol.22 mm (C) 120.75. Total vapour pressure of the solution is approximately (A) 70 mm (B) 35 mm (C) 105 mm (D) 140 mm 5.32 mm (C) 360.in .52 mm Q.35.28 Benzene and toluene from nearly ideal solutions.30 The vapour pressure of ethyl alcohol and methyl alcohol are 45 mm and 90 mm.78 Q.12 mm (D) 76. If the vapour pressure of methanol and ethanol at 350 K are 8.5 (iii) This condensed liquid again brought to the same temperature then what will be the mole fraction of benzene in vapour phase (A) 0. B = 73% (B) A = 57 %. the vapour pressure of a solution containing 0.22 mm (B) 300. The vapour pressure of water at 37ºC is 150 mm.24 torr .00 torr (C) 213. 840. 772.15 torr . then mole fraction of methanol in vapour phase is (A) 0.25 64 gm of methanol & 46 gm of ethanol forms in an ideal solution at 350 K.00 torr (D) 312. When the total vapour pressure above the solution is 400 torr.40 and in the liquid phase 0.62 (B) 0.LIQUID SOLUTIONS Q. B = 70% Q.5 × 104 N m-2 respectively.1 × 104 and 4.8 (B) 1. then the pressure of pure liquid A.26 Two liquids A and B forms an ideal solution at temperature T.80 (D) 0.33 torr. is (A) 180.22 (C) 0.06 mm and 103.00 torr (B) 240. Determine : (i) vapour pressure of the solution obtained by mixing 936 gm of benzene and 736 gm of toluene (A) 199. Then : (i) The total pressure of the mixture is (A) 526 torr (B) 965 torr (C) 440 torr (D) 715 torr (ii) composition of the liquid mixture is (A) A = 27%.5 (D) 4.27 Vapour pressure of a mixture of benzene and toluene is given by P = 179 xB + 92 where xB is mole fraction of benzene.45 St Q.55 mm Q. the mole fraction of A in the vapour phase is 0.54 torr .00 torr ud yS te ps . The mole fraction of A in the vapour phase is 0. B = 67% (D) A = 30%.in Q. An ideal solution is formed at the same temperature by mixing 60 g of C2H5OH with 40 g of CH3OH. If at 27ºC the vapour pressures of pure toluene and pure benzene are 32.24 The vapour pressure of pure liquid A at 300 K is 575 torr and that of pure liquid B is 390 torr. P PA PB PAo X A PBo X B www. mass = 140) has a vapour pressure of 160 mm at 37ºC.4 mm (B) 271 mm (C) 280 mm (D) 289 mm (ii) If vapours are removed and condensed into liquid then what will be the ratio of mole fraction of benzene and toluene in first condensate (A) 2. Type of solutions (i) Ideal Solution Page 12 of 33 An ideal solution may be defined as the one which obeys Raoult’s law over all concentration ranges at a given temperature. both have identical intermolecular forces. ethyl bromde + ethyl iodide. A perfect ideal solution is rare but many liquids form nearly ideal solution. P PAo XA + PBo XB Vmix 0 . Liquids form ideal solution only when they have nearly same molecular size and related constitution so that they have similar molecular environment in the pure state as well as in solution.in Characteristics Of Ideal Solution: ps Mixing of two substances results in an ideal solution only when: 1. A–A and B–B) must be identical so that the escaping tendency of an A or B molecule is independent of whether it is surrounded by A molecules.StudySteps. n-butyl chloride + n-butyl bromide ethyl alcohol + methyl alcohol Conditions for Forming Ideal Solution: Two liquids on mixing form an ideal solution only when 1. Similarly the escaping tendency of B remains the same. all attractive forces between A and B molecules or between A and A molecules or between B and B molecules (A–B. The escaping tendency of pure liquid A in solution remains the same. Vmixing 0 that is the total volume of the solution is equal to the sum of the volume of the pure yS liquids mixed to form the solution. 3. B. chlorobenzene + bromobenzene.in . Characteristic of nonideal solution Page 13 of 33 (i) (ii) (iii) PA PAo X A + PB PBo X B . P o XA A =P II. both have similar structures and polarity so that they have similar molecular environment. They obey Raoult’s Law H mixing 0 that is no heat is absorbed or released during dissolution 3. both have similar molecular sizes. 2. te 2. P= A PBo A I. Liquid heptane and hexane form an ideal solution because the interaction between a hexane molecule and another hexane molecule is the same as the interaction between two heptane molecules other examples of ideal solutions are: benzene + toluene. P B =Po B X B 0 XA 1 XB 1 0 Note : Components of ideal solution can be saperated in pure form by fractional distillation (ii) Non-Ideal Solutions Solutions which do not obey Raoult’s law over all concentration ranges at constant temperature are called non-ideal solutions. Vapour pressure St ud Graphical Representation of Vapour Pressure of Ideal Solutions: Figure shows the relationship between partial vapour pressure and mole fraction of an ideal solution at constant temperature.LIQUID SOLUTIONS In an ideal solution of two components A and B. . Hmix 0 www. PAo o P XB o P X A+ B III. molecules or varying proportions of A and B molecules. P. 2. Two liquids A and B on mixing form this type of solution when 1. pB < pB0XB. heat is evolved. te ps Graphical representation of vapour of non-ideal solution showing +ve deviation yS P> PA + PB St ud Vapour pressure V. Escaping tendency of both components ‘A’ and ‘B’ is lowered showing lower vapour pressure than expected ideally. CCl4 + CH3OH. size and character 3. pA < PA0 XA. Characteristic of non-ideal solution showing +ve deviation 1.in . 2. CCl4 + CHCl3. Non ideal solutions showing +ve deviation Non ideal solutions showing –ve deviation.LIQUID SOLUTIONS Types of Non-Ideal Solutions 1. pB > pBoXB . 1. XA = 0 XB = 1 P A > P o B X B Mole fraction XA = 1 XB = 0 2. water + methanol. Tow liquids A and B on mixing form this type of solution when 1. 3. A—B attractive force should be weaker than A—A and B—B attractive forces. V mix > 0. (Exothermic dissolution.in acetone + CS2 CCl4 + toluene. Non ideal solutions showing +ve deviation Condition for forming non-ideal solution showing +ve deviation from Raoult’s law. Non ideal solutions showing –ve deviation.) 3. = min. Condition for forming non-ideal solution showing +ve deviation from Raoult’s law. Characteristic of non-ideal solution showing +ve deviation 1.) 3. Hmix < 0. Do not obey Raoult’s law 2. pA > pA0 XA. (Volume is decreased during dissoluton) 4. ‘A’ and ‘B’ have different shape. cyclohexane + ethanol ac- . pA + pB > poAXA + pBoXB Example : acetone + ethanol water + ethanol. ‘A’ and ‘B’ have different shape. Hmix > 0. Do not obey Raoult’s law.P. size and charater. (endothermic dissolution heat is absorbed. Vmix < 0. A—B attractive force should be greater than A—A and B—B attractive forces. (Volume is increased after dissolution) 4.= max. 2. pA + pB < poAXA + pBoXB Page 14 of 33 www. etone + benzene.StudySteps. ‘A’ and ‘B’ escape easily showing higher vapour pressure than the expected value. B. 2. the maximum in the temperature composition curve is obtained. vapour pressure and Max. The point of maximum vapour pressure means that the bp at this composition will be minimum and constant. chloroform + benzene water + HCl CH3OH + CH3COOH. their escape tendencies and hence the vapour pressure decreases than expected on the basis of Raoult’s law.in .57 373 351. B. Page 15 of 33 Boiling point (K) B Mass% of B A B Azeoterope H 2O C2 H 5OH 95.= min.10 H 2O C3 H 7 OH 71. the deviations are so extreme as to lead to a maximum in the curve. yS te The non-ideal solution showing large deviations from Raoult’s law can not be purified by distillation.P. acetic acid + pyridine. vapour pressure and Min. chlorofrom + diethyl ether. Cl C CH 3 OC H CH 3 Cl V. +P B P < PA P A >P B o XA <PA PA o X B XA = 0 XB = 1 Mole fraction XA = 1 XB = 0 ps (iii) Azeotropes .00 334 351.69 373 370.P. and constant. www. St ud 1. BP : When liquid in a solution do not have great chemical affinity for each other (+ve deviation from ideality) their higher escape tendencies increase the vapour pressure much more than expected on the basis of Raoult’s law. In many cases.in Vapour pressure Graphical representation of vapour of non-ideal solution showing +ve deviation Azeotropes with Max. Cl acetone + chloroform. bp : When liquids in solution form chemical bonds (–ve deviation from ideality).StudySteps. = max. The point of minimum vapour pressure in the curve means that the bp of this composition will be max.3 351.30 Azeotropes with Min. In many cases.3 332.72 CHCl3 C2 H 5OH 67.0 350. A solution at certain composition which continues to boil at constant temperature without change in the composition of the solution and its vapour is called an AZEOTROPE or constant boiling mixture.LIQUID SOLUTIONS Example acetone + aniline. Azeotropes are of two types: Example of Minimum Boiling Azeotrope Components A 2. H2O + HNO3. boiling point is called the normal boiling point. The difference Tsb – Tb0 is called the boiling point elevation and denoted by Tb. of solute particles in solution. To be more accurate.3 373 188 383 H 2O HNO3 58. In other words.0 373 359 393.5 H 2O HClO 4 71. Greater the no.StudySteps. te (i) Lowering of Vapour Pressure: Vapour pressure lowering of a solution has already been explained under Raoult’ Law. The decreased vapour-pressure means that the solution would have to be heated to a higher temperature so that its vapour pressure becomes equal to the atmospheric pressure. of particles of solute in solution. www.6 373 383 476 6. the colligative property depends upon the fraction of solute and solvent particles in solution. the boiling point of the solution Tb is higher than the boiling point of the pure solvent Tb0. It was derived that the relative lowering of vapour pressure is given by the equation yS PAo – PA nB XB o PA nA nB Page 16 of 33 St ud (ii) Elevation of Boiling Point: The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure.LIQUID SOLUTIONS Examples of Max. 1 atm vapour pressure t en olv s re Pu ion l ut So Tb0 Tb Temperture Plot of variation of vapour pressure elevation of boiling point The vapour pressure of a liquid decreases when a non-volatile solute is dissolved in it.in . greater is the extent to which colligative property is affected. When the atmospheric pressure is 1 atm. Type of colligative properties (i) Lowering of vapour pressure (ii) Elevation of Boiling point (iii) Depression of Freezing point (iv) Osmotic Pressure ps . (1) Nature of the solvent (2) Independent of the nature of the solute (3) Extent of association and dissociation of solute particles in solution.in Factors that Affect the Colligative Property: The no. Colligative Properties The property of a solution which depends upon the fraction of solute particles and solvent particles and not upon the chemical nature of solute is called a colligative property. Boiling Azeotrope Components A Boiling point (K) B Mass% of B A B Azeoterope H 2O HCl 20. it is assumed that crystals of pure solvent always separate out first.in .445°C Tb = 5.2°C = 66..(1) It is found that the elevation of boiling point is directly proportional to the number of moles of the solute in a given amount of the solvent (m). www.445°C + Tb0 = 5.63 × 1.2°C.0g 0..(3) For water Kb 2 (373) 2 0. The boiling point of pure chloroform is 61.445°C + 61... Estimate the boiling point of a solution of 25...63 Km–1. T m Tb = Kb.515 K-kg/mol 1000 540 . K (mol/kg)–1 = K kg mol–1.50 = 5.0 0. When a dilute solution is cooled to freezing point. Determination of Kb of solvent : Kb RTb2 1000L v ..33 = 075 76 g/mol ud Solution: te ps Illustration 9....33mol Total moles of solute = 0..0g of urea NH 2CONH2 plus 25.75mol St Molality.. At freezing point. Tb is the boiling point of the solvent on Kelvin scale and Lv the latent heat of vaporization of solvent in calories per gram. the unit of Kb is Km–1. Molal Elevation Constant or Ebullioscopic Constant: When molality of the solution is 1m..645°C Page 17 of 33 Depression in Freezing Point: The freezing point of a liquid is the temperature at which it begins to freeze and the crystallized solid and liquid are in equilibrium.0g of thiourea NH2CSNH2 in 500g of chloroform.(2) where m is the molality of the solution and Kb is a constant for a given solvent known as boiling-point elevation constant or molal boiling point elevation constant or ebullioscopic constant of solution. Kb of chloroform = 3.StudySteps.50m Tb = Tb – Tb0 = Kbm = 3.m . Moles of thiourea = Mass of urea 25. the vapour pressure of the solid is equal to the vapour pressure of the liquid..44K = 5... CHCl3.in where R is molar gas constant. Solid solvent t ven sol on uti sol vapour pressure (iii) Temperature Tf Tf0 Plot of variation of vapour pressure of a solution with temperature and depression in freezing point.42mol Molecular mass of urea 60g / mol yS Moles of urea = 25.LIQUID SOLUTIONS Tb = Tb – Tb0 . Moles of solute 0. (1 mole of the solute is dissolved in 1 kg of the solvent) the above equation reduces to Tb K b 1m K b This indicates that molal elevation constant of a liquid (Kb) is equal to elevation of boiling point when molality of the solution is 1 m. m = Mass of solvent in kg (500g /1000g)kg 1.42 + 0. .(2) where m is the molality of the solution and Kf is a constant for a given solvent known as molal depression constant or cryoscopic constant.. a solid-liquid equilibrium exists only at a temperature lower than the freezing point of the pure solvent. ud Illustration 10.. In other words.in . . the unit of Kf is Km–1.. K (mol/kg)–1 = K kg mol–1.86 K-kg/mole 80 yS Kf te where Tf is the freezing point of solvent in absolute scale and Lf the latent heat of fusion in calories per gram of the solvent.12 × m 2 molal 0. Molal Depression Constant or Cryoscopic Constant: When molality of the solution is 1m.LIQUID SOLUTIONS The vapour pressure of a liquid decreases when a non-volatile solute is dissolved in it.(1) It is found that the depression in freezing point is directly proportional to the number of moles of the solute in a solute given amount of the solvent...5°C – 3... Tf = Kfm .. If Tf0 is the freezing point of the pure solvent and Tf that of the solution. Page 18 of 33 St Solution: Tf = Tf0 – Tf = Kf .5°C = 5.in RTf2 1000L f .. (1 mole of the solute is dissolved in 1 kg of the solvent) the above equation reduces to Tf = Kf × 1m = Kf This indicates that molal depression constant of a liquid (Kf) is equal to depression of freezing point when molality of the solution is 1 m.002 (273) 2 1.5°C and 5.39 mol/kg × 254 g/mol = 99. For water.StudySteps.39m 5. m 5..01% 1099.06 g/kg 1000g + 99.06g solution contains 99.(3) ps Kf 0.. the difference Tf0 – Tf is called the freezing point depression and denoted by Ts Ts = Tf0 – Tf .06g www.06g I2 m 100g solution contains 99. Therefore...5°C? The freezing point and cryoscopic constant of pure benzene are 5... for a solid to havethe samevapour pressure as that of the solution. the freezing point should lower down.06g 100 9.12 Mass of iodine needed for 1000g of benzene = m × molecular mass of iodine I2 = 0. Determination of Kf of solvent Kf is characteristic of a particular solvent and can be calculated from the thermodynamical relationship.12 K/ m respectively.. What is the percent by mass of iodine needed to reduce the freezing point of benzene to 3. 9 K. Is a 28% (by mass) aqueous solution of ethylene glycol suitable for a car radiator ? Kf for water = 1.04º C (B) –3.06ºC (C) 100.6 The amount of ice that will separate out from a solution containing 25 g of ethylene glycol in 100 g of water and is cooled to –10ºC. pt.25 gm of non-electrolyte and 20 gm of water is 271.5 An aqueous solution containing 5% by weight of urea and 10% by weight of glucose.103ºC. What is the relation between molecular weights of the two substance (A) MA = 4MB (B) MA = MB (C) MA = 0. will be [ Given : Kf for H2O = 1.5 mg of octa atomic sulphur is added to 78 g of bromine.544ºC (B) -0.402ºC.78 kg (B) 22. Elevation of b. is [ Kf(water) = 1.2 K mol–1 kg and b.9 The temperature at which ice will begin to separate from a mixture of 20 mass percent of glycol (C2H6O2) in water. so that it does not freeze. the normal temperature in Kullu valley was found to be –11ºC.62 g of an unknown solute dissolved in same amount of CCl4 produced an elevation by 0. The amount of glycerine to be added to 40 dm3 water used in car radiator.StudySteps.53ºC.65 kcal mol–1 Q.62 Q.96 kcal mol–1 (B) 7.86 K kg mol–1.96º C Q.86 K mol–1 kg ] (A) 50.5 K (C) 276.65ºC.2 0.12 The boiling point of an aqueous solution of a non-volatile solute is 100.272ºC (D) -1.86 K molality-1 ] (A) -0.5 K (D) 269. While 4 g of another substance B.65 g naphthalene (C10H8) was dissolved in 100 g methyl acetate.512 and 1.66 kg Q.265ºC (D) 99. Freezing point of solution is [Kf for H2O is 1. pt.LIQUID SOLUTIONS 3.01 kJ mol–1] (A) 39.68 (B) 106. of water by 0.10 The amount of urea to be dissolved in 500 ml of water (Kf = 18.47ºC Q. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water ? [ Kb and Kf for water are 0. by 0. Daily Practice Problem Sheet A solution containing 0.53 (B) 181. pt of methyl acetate solution was 0. the boiling point of bromine is [Given Kb for Br2 is 5.3 (D) 93.3 When 174.96º C (D) 5. Molecular weight of solute is (A) 85.8 In winter.38 (D) 160. If molar depression constant is 1. When 2 mole of glucose are dissolved in 4000 gm of water. pt.in .53ºC (B) 101. Then molar mass of the solute will be (A) 105.1ºC.in Q.86ºC Q. (A) Yes (B) No (C) can’t predict Q.11 The molal boiling point constant of water is 0.52 g of C10H8 in CCl4 produced an elevation in boiling point by 0.[ Hfusion = 6.86 K kg mol–1 ] (A) 280.15ºC.23 K (B) 330 K (C) 220.00 g of substance A. If b. its molar heat of vaporisation will be (A) 8.3 gm (B) 3 gm (C) 6 gm (D) 9 gm Q.5 K St ud yS te ps . On the other hand a solution of 0.512ºC (C) -0.92 K (D) 332. dissolved in 100 g H2O depressed the f.9 Page 19 of 33 www.19 K Q.7 It has been found that minimum temperature recorded in a hill station is –10ºC.946 kcal mol–1 (C) 6.186ºC in freezing point is (A) 0.89 kg (D) 42.04º C (C) –5.45 kg (C) 19.2 ºC.5MB (D) MA = 2MB Q.51 (C) 94.24 kcal mol–1 (D) 15.6 K mol-1 100 g solvent) to produce a depression of 0.0 g (B) 25.5 gm (D) 30. the solution will boil at (A) 100.0 gm Q.13 The freezing point of a solution prepared from 1. of pure methyl acetate is 57ºC. pt.15 K ] (A) 415.86 K mol–1 kg] (A) 3.0 g (C) 12. is .5 K (B) 265.4 4. depressed the f.86 K molality-1.1 Q.7 (C) 115. of Br2 is 332. 86ºC Q.18º (C) 0.6 g (D) 6.0 g (C) 9.2467 atm at 338 K and Kb for water =0. pressure) (A) 0.943 (D) 0. Kb (CCl4) = 5.16 Elevation in boiling point of an aqueous urea solution is 0.19 Relative decrease in V. For dilute solutions = CRT = hg (hydrostatic pressure) Where C is the total molar concentration of all the free species present in the solution. the weight of glucose added is (A) 180 gm (B) 18 gm (C) 1. solution freezes at (A) 1.39 º (C) 3.0567 (C) 0.58ºC (B) -1.372ºC. Page 20 of 33 www.982 (B) 0. St Osmotic pressure Semi permiable membrane Piston Solution Pure liquid Solvent Osmotic Pressure (): The hydrostatic pressure built up on the solution which just stops osmosis. p. Alternatively. P.018 Q.in .34 atm (D) 0.00 kg mol–1 K . (A) 0º (B) 5.15 A solution of a non-volatile solute in water has a boiling point of 375. w is (A) 4. 3 K.86 K kg mol-1. (Kb = 0.18 When a solution containing w g of urea in 1 kg of water is cooled to -.86 K mol-1 kg.14 Elevation in b.52K kg mol-1). p of a solution of non-electrolyte in CCl4 is 0. Hence mole fraction of urea in this solution is (A) 0.18 g/mL.86ºC (C) -3. On the basis of osmotic pressure. it may be defined as the pressure which must be applied to the concentrated solution in order to stop osmosis.59º (D) 2.018. elevation in boiling point is : (it is given 1 molal aq.49º Q. of an aqueous glucose dilute solution is found to be 0. h is the height developed by the column of the concentrated solution and is the density of the solution in the column.52º.0 g .03º yS te ps Q.in Q. 200 g of ice is separated. Hence.54ºC at 1 atm.18 atm (B) 0.02 kg mol–1 K. for the same solution ? Kf (CCl4) = 30.16ºC (D) 1. the solutions can be classified in three classes.52 K kg mol-1] (A) 0.20 Glucose is added to 1 litre water to such an extent that Tf/Kf becomes equal to 1/1000.8 g (B) 12.23 atm (C) 0.StudySteps.018º (B) 0.8 gm (D) 0. urea solution boils at 100. The vapour pressure of water above this solution at 338 K is [Given p0 (water) = 0. If Kf for water is 1.54º (D) 0. If Kf (H2O) is 1.18 gm (iv) Osmosis: ud Spontaneous flow of solvent molecules through a semipermeable membrane from a pure solvent to the solution (or from a dilute solution to a concentrated solution) is termed as osmosis.17 Density of 1 M solution of a non-electrolyte C6H12O6 is 1.60.42 atm Q. What is depression in f.LIQUID SOLUTIONS Q. StudySteps. then the solution with lower osmotic pressure is termed as hypertonic.325 h = 375 cm = 3. Solution: Page 21 of 33 Let the molecular mass of urea be m2 Molarity of sugar = 5 w1 = m1 V1 342 0.877% soltuion of urea.100 mL = 0. What will be the height of the liquid level in the tube at equilibrium? The density of solution may be taken as 1g/cm 2. Note:-Osmotic pressures can be determined quite accurately.325 KPa h 100 101.082 × 288 = 3.8 = 36.7 KPa [Ans. (This implies C1 = C2).082. The bottom of the tube is closed with a semipermeable membrane and 1 g glucose is placed in the tube. 36.92 h 1 9 . mass of sucrose (C12H22O11) = 342 w = 5g V =. Illustration 11. Calculate osmotic pressure of 5% solution of cane sugar (sucrose) at 15ºC.LIQUID SOLUTIONS (a) Hypertonic solution:When two solutions are being compared. Reverse Osmosis: If a pressure greater than the osmotic pressure is applied on the concentrated solution.75 m. This is reverse osmosis. Calculate the molecular mass of urea if molecular mass of cansugar is 342.1 litre S = 0. Consider a vertical tube of corss-sectional area of 1cm 2.75 ×1 × 9. The closed end of the tube is immersed in pure water.1 ps Applying the equation PV = St ud yS te Illustration 12. (c) Isotonic solutions: Two solutions having same osmotic pressures at same temperature. then the solution with higher osmotic pressure is termed as hypertonic. hence it is used in the determination molecular weights of large proteins and similar substances. T = (15 + 273) = 288 K .7 KPa] llustration 13.1 www. A 5% soltuion of cane sugar is isotonic with 0.8 = × 1 atm = 101. What is the osmotic pressure at equilibrium at 25ºC? Assume negligible depth of immersion of tube. (b) Hypotonic solution:When two solutions are being compared.in P= w ST.453] 342 0 . Solution: m = mol. 3.in . the solvent starts to flow from concentrated solution to dilute solution (or pure solvent). m 5 1 × = 0.0821 × 298 1000 180 134.92 atm h = h × d × g = 134.. One of its chief uses is desalination of sea water to get pure drinking water. Solution: Let height in tube = h cm V = (h × 1) cm3 Cross-sectional area = 1cm2 V = nRT h 1 1 = × × 0.75 m = h × d × g = 3. 1 0.25 (D) 6.72 atm (D) 24.7 100 mL aqueous solution of glucose with osmotic pressure 1.1 A solution containing 4 gm of a non-volatile organic solute per 100 ml was found to have an osmotic pressure equal to 500 cm of mercury at 27ºC. Daily Practice Problem Sheet Q. Osmotic pressure of the mixture is (A) 1.33 (D) 0.4 atm at 25ºC. If the density of this solution is 0.in Q.94 mm Q.5 Two solutions each in 100 mL having 4 g glucose and 10 g sucrose respectively.3 Q.2 (D) 86.67 g of A and 30 g of B present in same volume of solution at same temperature.6% solution (w/v) of cane-sugar (m.96 kg dm–3 at 300 K.9 (B) 34.88 atm (B) 2.1 2 2 2 For isotonic solutions.06 atm (D) 9. when the osmotic pressure is found to be 105.97 (B) 29.22 atm (C) 21. when 10 gm glucose (P1).12 Q.4% solution (w/v) of urea (m.25 (B) 0. w1 w2 = m1 V1 m 2 V2 m2 = 5 0. the height of solution that will represent this pressure is (A) 28. The solution is diluted and the temperature is raised to 25ºC.3 mm.9 At 10ºC.34 g/mL.2 atm at 25ºC is mixed with 300 mL aqueous solution of urea at 2.in .10 The relation ship between osmotic pressure at 273 K.72 g urea in sucrose solution (D) 0.877 342 = 59.6 5 g of a polymer of molecular weight 50kg mol–1 is dissolved in 1 dm3 solution.LIQUID SOLUTIONS w2 0.877 Molarity of urea = m V = m 0. The molecular weight of solute is (A) 348. The molecular weight of solute is (A) 14.52 mm (D) 24. The ratio of molar masses of solute A and B will be (A) 0.44 atm (D) 124.4 The osmotic pressure of a solution containing 100 ml of 3.75 Q.5 (C) 1.13 atm Q. wt 342) at 27ºC is (A) 10. The solution has the same osmotic pressure as 6.987 5 4.48 atm (C) 12.55 atm www. 10 gm urea (P2) and 10 gm 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 Q.2 The osmotic pressure of decimolar solution of glucose at 30ºC is (A) 24.85 mm (C) 26.4218 g urea in glucose solution (B) 0.59 atm (C) 1.66 (C) 0.25 molal sucrose solution exhibit at 25ºC ? The density of this solution is 1.70 atm (B) 30.1 m 2 0. wt.6 gm urea in one litre was found to be isotonic with a 5% (wt/vol) solution of an organic volatile solute.877 = 342 0.2 Q.70 atm Q.98 atm (C) 17.89 (C) 861.7 (D) 137.8 10 g of solute A and 20 g of solute B are present in 500 mL of solution. (A) 28. Find out the extent of dilution (A) 5 (B) 2.37 atm (B) 2.77 g urea in glucose solution (C) 0.11 Page 22 of 33 What osmotic pressure would the 1.StudySteps.13 mm (B) 20.56 atm (B) 8. 60) and 50 ml of 1. How much urea should be added to one of them in order to make them isotonic ? (A) 0.44 (C) 149. the osmotic pressure of urea solution is 500 mm.421 g urea in sucrose solution Q.4 atm A solution containing 8.5 St ud yS te ps .85 atm (D) 2. a colligative property shows an increased effect. nAB (AB)n 2C 6 H5 COOH (C6 H 5COOH) 2 St ud The molecular mass of a solute is inversely proportional to its molality. How much glucose is used per lit for an intravenous injection that is to isotonic with blood ? (A) 180 gm (B) 342 gm (C) 58. r F m o a r l e m x o a l m e p c u l l a e . lowering of vapour pressure shows an increased effect.86 K kg mol-1 & molality is equal to molarity] (A) 12.3 atm (B) 1.12 The osmotic pressure of blood is 8. the calculated molecular mass is one third of the actual molecular mass of the solute.14 At certain temperature. freezing point depression and osmotic pressure are colligative properties which depend upon the fraction of solute and solvent particles in solution and not upon the chemical nature of the solute.15 atm (D) 2. If colligative molality is 3 m.46 atm 7.0072 atm. the molecular mass obtained by a colligative property is half of its normal molecular mass.23 atm (C) 6. Hence. The extent of dissociation and colligative property. height of water column due to this pressure is [given d (Hg) = 13.06 gm (D) 55.21 atm at 37ºC.93ºC then the osmotic pressure of aqueous solution of the given nonelectrolyte at 27ºC is [Given : Kf for water = 1. molecular masses obtained of strong acids. yS te – NaCl(s) Na +(aq) + Cl(aq) If the solute molecules associates in solution. www.in Vapour pressure lowering. If solute molecules dissociates in solution. one particle of potassium chloride on dissociation in water gives two particles. We can summarize the results as: 1. Page 23 of 33 1mol 1 mol The extent of association and colligative property: A solute that associates in solution provides less particles that would otherwise be present in solution and therefore. As an example. If colligative molality is 2 m. boiling point elevation. Benzoic acid associates to form a dimer and therefore its colligative molality is one-half of the molality of benzoic acid. the colligative p m r o o p l e e r c u t y l s a r h m o w a s s s t w h h e i c d h e i s c j r u e s a t t s w e d i c e e f i t f s e c n t o . r b m e a n s z s . K Cl(s) nH 2 O K (aq) Cl(aq) 1 mol 2. The molecular mass of benzoic acid is 122 g/mol.28 mm (C) 74 mm (D) 760 mm Q.StudySteps.13 Osmotic pressure of insulin solution at 298 K is found to be 0. As molecular mass of a solute is inversely proportional to molality. Abnormal Molecular Masses ps . there are more particles in solution and therefore. the calculated molecular mass is one-half of the actual molecular mass of the solute. But the molecular mass of benzoic acid dissolved in benzene is found to be 244 g/mol by using a colligative property. How many times the solution should be diluted in order to exhibit the osmotic pressure of 81 mm at the same temperature ? (A) 2 times (B) 4 times (C) 8 times (D) 5 times Q.15 If depression in freezing point is 0.LIQUID SOLUTIONS Q. K+ and Cl– and therefore.6 g/mL] – (A) 0.760 mm (B) 70. and therefore lowering of vapour pressure shows a decreased effect. the molecular mass of benzoic acid determined using a colligative property is double the actual molecular mass of benzoic acid. For example. the osmotic pressure of an aqueous solution of urea was found to be 405 mm.55 gm Q.in o i c a c i d i n b e n z e n e i s f o u n d t o h a v e . there are less particles in solution. A solute dissociates completely or partially in solution makes available more particles than would otherwise be present in solution and therefore. bases and salts are much less than their normal values. 1859. i = ii) Page 24 of 33 Moles of solute actually present in solution Moles of solute dissolved (1 ) n i 1 1 (n 1) or 1 n 1 Calculation of Degree of Association of solute particles: Let n moles of the solute. If is the degree of association.in The Van’t Hoff factor for a solute can be calculated by the following modified equations: where C is molarity of the solution. carboxylic aids. Van’t Hoff. phenols. Van’t Hoff Factor In 1886.. A. associate to form (A)n. of moles dissolved 1 mol 0 No.StudySteps. of moles after dissociation 1– /n Total number of moles present in solution 1 i 1 1 n Hence = (1 – ) + /n i 1 n (i 1) 1 1 n 1 n www. of moles after dissociation 1– n Total number of moles present in solution = (1 – ) + n St i) ud Application of Van’t Hoff Factor: Van’t Hoff factor. e. O H O H5C 6 C 6H5 O H O A dimer of benzoic acid 8.-1911) introduced a factor ‘i’ known as Van’t Hoff factor to express the extent to association or dissociation of a solute in solution.LIQUID SOLUTIONS It is found that compounds which are capable of forming hydrogen bonds. It can be calculated as: i= = number of solute particles actually present in solution number of solute particles dissolved Observed colligative property observed molality normal molecular weight of solute = = normal colligative property normal molality observed molecular weight of solute PA0 PA PA0 = i XB (ii) Tf = iKfm (iii) Tb = iKbm (iv) = iCRT ps (i) . i = 1 For electrolytes. alcohols: because of association show decreased effect of colligative property.in . of moles dissolved 1 mol 0 No.g. Jacobus Henricus (Dutch chemist. nA An No. yS te Note: For non-electrolytes. i > 1 ( If solute particles undergo Dissociation in the solution) i < 1 ( If solute particles undergo Association in the solution) Calculation of Degree of Dissociation of solute particles: An nA No. Calculate the freezing point of 0. (A) 1.5°C – 0.2 M MgCl2 (C) 0.solute.12 Km –1 respectively. then 0.3g of CdSO 4 in 1000g water. the degree of dissociation of KCl is (A) 75% (B) 83% (C) 65% (D) 92% Q.05 M solution of A will produce an osmotic pressure.6 Tf = Tf0 – 0.5 P (D) 0. associate in benzene to form species A 3. The depression in freezing point of solution was found to be 0.284K Tf = iKfm or i K m 1. Three particles of a solute.1 M urea and 0. Daily Practice Problem Sheet Q.177m 3 3 3 3 0 –1 or Tf – Tf = 5. of moles after dissociation m(1–) Total moles present after dissociation ps Solution: St 5.80. 1 M glucose and 1 M NaCl are in the ratio : (A) 1 : 2 : 3 (B) 3 : 2 : 2 (C) 1 : 1 : 2 (D) none of these Q.6°C = 5.117 m = 0.4 Which is the correct relation between osmotic pressure of 0.1 M Na2SO4 Page 25 of 33 www.1 A pentimolal solution of potassium chloride freezes at -0. A solution is prepared by dissolving 26.75 P Q. assuming that the electrolyte is completely ionised.2 The depressions in freezing point of 1 M urea.1 M NaCl solution and 0. of moles dissolved 3 No.86K kg solvent mol–1.21 f Illustration 15. Solution: Molecular mass CdSO4 = 112.LIQUID SOLUTIONS Illustration 14. the freezing point of benzene and its cryoscopic constant are 5.9°C ud yS te = m(1 ) m Tf = Kfm A3 0 m/3 . A.5 Which one of the following pairs of solutions will be expected to be isotonic under the same temperature? (A) 0.1 M Ca(NO3)2 and 0. If 0.68ºC.25m = 0.1 M Na2SO4 (D) 0.1 M NaCl (B) 0. 2 3 2 0.StudySteps.1 M solution of B produces an osmotic pressure P.25 molal solution.86K / m 0.4 +32 + 4 × 16 = 208. The degree of association of solute A is found to be 0.4 g/mol Mass CdSO 4 Molality CdSO4 = Molecular mass CdSO Mass solvent in kg 4 = 26.5 P (B) P (C) 0.12 Km × 0. The cryoscopic constant of water is 1. If Kf for H2O is 1.in .1 M urea and 0.216m 1000 (208.3 An electrolyte A gives 3 ions and B is a non-electrolyte.in 3A No.3g = 0.8 m 1 m 1 = 0.6°C = 4.1 M Na2SO4 solution ? (A) the osmotic pressure of Na2SO4 is less than NaCl solution (B) the osmotic pressure of Na2SO4 is more than NaCl solution (C) both have same osmotic pressures (D) none of the above Q.126m = 1.1 M NaCl and 0.284K.4g / mol) kg 1000 Tf 0.5°C and 5.86. Calculate the Van’t Hoff factor. is . The dissociation constant for the reaction.5 gm (B) 103 gm (C) 75 gm (D) 37. dissociation and no change respectively.in (A) P2 > P1 ud yS Q.95.0 (C) 0.372ºC (D) +372ºC Q.08 molal solution of NaHSO4 is –0. HSO4– H+ + SO42– .8 The molal freezing point constant for water is 1.1 M NaCl solution in water is expected to be : (A) –1.11 (B) P1 > P2 . The value of Kb for water is – (A) Tb 2 (B) 10 (C) 10TB (D) Tb 10 Q. NaCl solution will exert an osmotic pressure closest to which one of the following (A) 5. The correct relationship between the osmotic pressure is (C) P1 = P2 (D) P1 P2 + P1 P2 P1 P2 ps A 5.[Kf for water = 1.53ºC (C) 0ºC (D) -0. ? [Kf for water = 1. is a colligative property (D) all of the above Q. p.86ºC (B) –0. elevation of 1 molal glucose solution is half of the 1 molal KCl solution (C) b.00 atm.3 (D) 3.57 (D) 0.94 (D) 0. of 1 molal NaCl solution is twice that of 1 molal sucrose solution (B) b.005 M aqueous solution of KCl is 1.17 The elevation in boiling point of a solution of 10 g of a binary electrolyte (molecular mass 100) in 100 of water is Tb.18 Van’t Hoff factors are x.15 If the observed and normal osmotic pressures of a 1% NaCl solution are 5.16 0.35ºC Q. the degree of dissociation of NaCl is – (A) 0.5 gm St Q.StudySteps. The freezing point of KCl is found to be – 2ºC.7 What is the freezing point of a solution containing 8.1 g HBr in 100 g water.85ºC (B) -3.12 How many grams of NaBr must be added to270 gm of water to lower the vapour pressure by 3.97 (C) 0.3 (B) 5.01 M solution of KCl and BaCl2 are prepared in water.in .p.13 Sea water is found to contain NaCl & MgCl2. If NaCl is 80% ionised and MgCl2 is 50% ionised then van’t Hoff factor is (A) 1.14 The f.96 Q. Increasing order is (A) x < y < z (B) x = y = z (C) y < x < z (D) x < z < y Page 26 of 33 www. z in the case of association.9 Select correct statement : (A) b. p. What freezing point would you expect for BaCl2 solution – (A) –5ºC (B) –4ºC (C) –3ºC (D) –2ºC Q.0 (C) 3.6 The van’t Hoff factor of a 0.8% wt/vol. The degree of ionisation of KCl is(A) 0.372ºC.1 M and osmotic pressure are P1 & P2.86ºC] (A) 2 × 10–4 (B) 4 × 10–4 (C) 2 × 10–2 (D) 4 × 10–2 Q.3 Q.125 mm Hg at temperature at which vapour pressure of water is 50 mm Hg.9 (B) 1.86 K kg mol–1. assuming the acid to be 90% ionised.p of a 0. Assume 100% ionisation of NaBr (A) 51.8% (wt/vol) sucrose solution (B) 5.75 and 3. Therefore.LIQUID SOLUTIONS Q.95 (B) 0.186ºC (C) –0. the freezing point of 0. y.86 K kg mol-1] (A) 0.10 Two solution of KNO3 and CH3COOH are prepared separately molarity of both in 0.8% (wt/vol) glucose solution (C) 2 molal sucrose solution (D) 1 molal glucose solution te Q.8 Q. i = 1.4 × 10-4 bar Q. (A) 1.16ºC (D) 6.1% Q.18ºC (iii) 488.72ºC (C) 11.20 A 0.2% solution (wt.90 (C)= 85%. What would be the depression of freezing point of water when : (i) 120 g of urea is dissolved in 1000 gm of water. The apparent degree of dissociation of KCl at this dilution is (A) 92.2ºC. P. relative decrease in V./ volume) of NaCl is isotonic with 7. i = 1.44ºC (B) 11. i = 1.19 1 mole each of following solutes are taken in 5 moles water.95 (B)= 90%.5 g NH4Br in 100 g of water is –6.86. will be in order (A) i < ii < iii < iv (B) iv < iii < ii < i (C) iv < i < ii < iii (D) equal Q.1 mL of N/10 alkali for complete neutralization.in . the vapour pressure of 0.24 1 g of monobasic acid in 100 g of water lowers the freezing point by 0. If 0.87ºC www.4% (C) 19.85 (D)= 75%.8% (B) 22. /volume) of glucose.75 te ps Q.4 bar (C) 0.74 g of BaCl2.1% (B) 84.168º.72ºC (ii) 117 g of sodium chloride is dissolved in 1000 gm of water (Assuming sodium chloride is fully ionised) (A) 7. The freezing point of a solution of 3.2% Page 27 of 33 St ud yS Q. degree of dissociation of acid will be [Kf for H2O is 1.94 bar (B) 9. Degree of ionisation and van’t Hoff factor of NaCl is (A)= 95%.in Q.87ºC (D) 3.6% (D) 26.1 M solution of urea is 0.LIQUID SOLUTIONS Q.21 At 20ºC.185ºC.23 A 1.86 K mol–1 kg] (A) 16.1 M solution of KCl is 0.42 g cane-sugar in 100 g of water is –0. (Assuming barium chloride is fully ionised) (A) 7.2H2O have been dissolved in 1000 g of water.22 The freezing point of a solution of 20. (i) NaCl (ii) K2SO4 (iii) Na3PO4 (iv) glucose Assuming 100% ionisation of the electrolyte.26ºC (C) 7.0311 mm less than that of water and the vapour pressure of 0.002 molar solution of NaCl having degree of dissociation of 90% at 27ºC has osmotic pressure equal to (A) 0.16ºC (C) 3.2% solution (wt.25 The molal depression of the freezing point in 1000 g of water is 1.094 bar (D) 9.86ºC (B) 8.2 g of same acid requires 15.6% (C) 68.4% (D) 54. The degree of ionisation of salt is (A) 80% (B) 30% (C) 60% (D) 40% .StudySteps.44ºC (B) 3.0574 mm less than that of water.72ºC (D) 5. i = 1. (C) 4. (D) 17.in . (C) 12. (A) 13. (C) 5. (C) 28. (A) 13. (B) Daily Practice Problem Sheet 1.LIQUID SOLUTIONS ANSWERS 1. (D) 19. (A) 25. Daily Practice Problem Sheet 1. (C) 9. (C) 6. (A) 22. (D) 5. (C) 3. (D) 4. St 15. (B) 5. (C) 13.in 3. (C) 23. (A) 27. (D) 26. (B) 10. (i) (D) (ii) (A) (iii) (B) Page 28 of 33 www. (D) 4. (B) 7. (A) 14. (C) 13. (A) 2. (B) 5. (B) 15. (A) 10. (A) 23. (B) 2. (A) 19. (A) 6. (B) 10. (C) 15. (i) (A) (ii) (D) 14. ps 1. (A) 29. (B) 16. (B) 2. (C) te Daily Practice Problem Sheet yS 4. (C) 8. (D) 19. (A) 4. (B) 10. (C) 20. (C) 3. (C) 16. (A) 11. (A) 4. (B) 8. (A) 16. (B) 11. (A) 12. (C) . (D) 5. (A) 12. (D) 15. (B) 17. (C) 14. (C) 13. (A) 7. (B) 8.StudySteps. (B) 12. (C) 30. (C) 2. (C) 7. (C) 14. (C) 19. (A) 7. (C) 9. (C) 21. (D) 24. (D) 11. 20. (C) 12. (B) 18. (B) 3. (C) 9. (C) 20. (C) 15. (A) 16. (A) 18. (A) ud 1. (A) 24. (C) 3. Daily Practice Problem Sheet 1. (A) 18. (D) 5. (B) 18. (A) 8. (D) 17. (i) (A) (ii) (D) (iii) (B) Daily Practice Problem Sheet 2. (B) 22. (A) 9. (B) 6. (A) 11. (C) 17. (B) 6. (D) 6. (C) 14. (B) 7. (D) 11. (B) 10. (C) 25. (B) 2. (C) 21. (B) 3. (A) 9. (A) 8. water) msolution = mwater + msucrose = 1000 + . How much ps ice will be separated out if the molality of the solution is 0. Solution: 256.5 gm 204.5 + 8. hence .5% MgCl 2. 70% ionization of MgCl2 and 50% ionization of NaHCO3 (Kb for water = 0.4 = 0. n NaHCO3 = i MgCl2 = 1 + 2 = 1+ 0.5 ud msucrose (in kg) = yS te Since molality of solution is .5 gm of cellulose acetate is dissolved in 100 ml solution.2. gr.in = Problem 2: 1kg of an aqueous solution of Sucrose is cooled and maintained at –4°C.75 × 342 = 1256.94°C 70.1 84 (i NaCl n NaCl i MgCl2 n MgCl2 i NaHCO3 n NaHCO3 ) K b 1000 Weight of solvent (1.95°C www.4%.5 2 = 2.1 M = 61575 .86 – 277.3 Cm of pure acetone = = 2.1) 0. and 8.7 2 = 2. 58.14 / 342 W / 1000 amount of ice = 795.75 .082 lit atm mol–1 K–1. i NaHCO3 = 1+ 2 = 1+ 0.4 g Tb = = 8.9 cm of Hg = 0.9.5 × 1000 = 204.14 gm .4 Boiling point of solution = 100 + 5.7% NaCl.7 = 0. 9. = 0.9) shows an osmotic rise of 23 mm against pure acetone at 27°C.86 gm 1256.15 St Tf = Kf × m Problem 3: River water is found to contain 11.1.4) = 70. NaHCO3 by weight of solution.7 + 9.StudySteps.9 = 1.5 = 0.0821 300 v M 0. Calculate its normal boiling point assuming 90% ionization of NaCl.5% (wt.52) Solution: Page 29 of 33 nNaCl = 11.3 0.5 n MgCl2 = 9.in . / vol) solution means 0.94 = 105.9 0.86 × w = 277.75 moles of sucrose are present in 1000 g of solvent (i..5 M Volume = v = 100 ml = 0.e. T = ( 27 + 273) = 300 K n 0.31 gm 4 = 1. Osmotic pressure = 23 mm of pure acetone = 2.LIQUID SOLUTIONS SOLVED PROBLEMS SUBJECTIVE Problem 1: Calculate the molecular weight of cellulose acetate if its 0.4 .4 0.14 = 795.2 2.0 Weight of solvent = 100 – (11.1 2 0.1522 cm of Hg 13. 95 iNaCl = 1+ = 1+ 0.002 = 0.002 atm 76 Let the molecular weight of the cellulose acetate be M ncellulose acetate = 0.15 = 518.1522 atm = 0. mwatrer = 1000 – 204. Solution: 0.1 lit R = 0.75? K f (H2O) = 1.6 0.52 1000 = 5.86 Kg mol–1K.5 1 RT 0.5% (wt./vol) solution in acetone (sp. 91°C = –19. What is the complex? Solution: Tf = i Kf m.9984 gm cm–3) are mixed at 25°C to form a solution of density 0. What is the molecular formula? Kb(H2O) = 0.9 mol of benzene is cooled until Page 30 of 33 some benzene freezes out.558 =3 0.1 mol of naphthalene and 0.9575 gm cm–3.94 + 69. Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. CoCl3.5NH3 [Co(NH3)5Cl]++ + 2Cl– 1 Cation 2 anions So.558° Kf(H2O) = 1. i 3 indicates that complex ionize to form three ions since coordination number is 6 hence x = 5 i.7980 gm Cm –3 ) and 70 ml of H 2 O (d = 0. = 101.5 K and 353 K.94 1000 = 10.512 K mol–1 kg Tb (H2O) = 100°C Solution: Elevation in B. where its vapour pressure is found to be 670 torr.9984 gm cm3 = 69. 0.94 1000 mol lit–1 = 7.24°C at 1.86 mol–1 K and assume 100% ionization and co–ordination number of Co(III) is six .24°C Problem 4: An Tb = Kb i molality 1.StudySteps.99 ml 0. Its 0.512 288 1000 m 90 ( i = 1) m = 1321.828 te volume of the solution = density of the solution = = 97.99 yS M (molarity) = ud Problem 6: A complex is represented as CoCl3.7046°C = 19. Calculate the freezing point of the solution.xNH3.LIQUID SOLUTIONS aqueous solution containing 288 gm of a non–volalite compound having the stochiometric composition CxH2xOx in 90 gm water boils at 101.63 M 32 97. respectively.888 = 93.91°C Weight of solution = weight of solute + weight of solvent = 23.e.00 atmospheric pressure .7046 molal 32 69. www.888 gm m= 23.2 gm mol–1 molar mass of CxH2xOx = 12x + 1 2x + 16x = 30x 30x = 1321. the complex is [Co(NH3)5Cl]Cl2 Problem 7: A solution comprising 0.in Tf = Kf i molality.91°C Freezing point of the solution = 0– 19. of ps .95751 23.1 i or. Kf (H2O) is 1.86 0.P.88 wt.24 = 0.24 – 100 = 1.86 10.. Also calculate its molarity Solution: Weight of CH3OH (w1) = 30cm3 0.186 St Tf = 0.94 gm Weight of solvent (H2O) (w2) = 70 cm3 0.1 m solution in aqueous solution shows i= 0.828 gm the solution 93.7980 gm/cm3 = 23.558 = 1. Kf for benzene = 5K kg mole–1. The solution is then decanted off from the solid and warmed to 353 K. Assume conditions of ideal solution. The freezing and normal boiling point of benzene are 278. (for CH3OH.in .2 x = 44 Hence the molecular formula is = C44H88O44 Problem 5: 30 ml of CH 3OH (d = 0.86 Kg mol–1 K. i = 1) Tf = 1. 1 2 0. 1 C Tf = Kf i Cm 0.1M.21 = 1.1 g 1.88 78 X naph 1000 1000 0.72 = 8.StudySteps.964 2 www. (760 670) 760 = 1.205 = 1. after the benzene freezes out) can be determined on the basis Raoult’s law.86 kg mol–1 K.036) 2 Ka = = = = 1.109 = 0.in HCOO– (aq) 0 C .1 1 0.11 10–3 ud Problem 9: The freezing point depression of a 0.21 = 1.100 molal aqueous solution of tartaric acid freezes at –0. It ionizes as AH A + H+ Initially conc. Solution: ps C(1 ) C C C(1 ) . where = degree of dissociation = = 1+ C C Molal concentration = 0. i = . m X benzene 78 0.9 K Problem 8: Find Ka.109 ( for dilute solution molality molarity) 1+ = 0.0358 = 0. Page 31 of 33 St Calculate the equilibrium constant for the reaction.1 = 1.109 M aq. Assume that only the first ionization is of importance and that 0.41 × 10–5 [HCOOH] 1 0.5 – 8.205°C.1 Tf = Kf Cm i 0. C C C [A ][H ] = C(1 ) 1 [AH] Ka = yS 2 Ka = 1.036 C [H ][HCOO ] 0.72 mol kg–1. solution of formic acid is –0. Kf = 1.21°C.86 (1+) 0.11 10–3 Ka = 0.LIQUID SOLUTIONS Solution: Molality of the resulting solution (i.60 K Hence the temp to which the solution was cooled = 278. after dissociation C(1–) C C.60 = 269.9 78 – 58.86 kg mol–1 K Solution: HCOOH (aq) H+(aq) + Initially moles C 0 Moles after dissociation (1–) C C i (1 ) C C C = 1 + .1 (1+) te Here.72 Weight of benzene in the resulting solution = Amount of benzene frozen = 0.1 = 58. HCOOH (aq) H+ (aq) + HCOO– (aq) Kf for water = 1.1 m = 0. C 0 0 Conc.86 0.0358 1.1 = 12. the ionization constant of tartaric acid if a 0.in Assuming that the tartaric acid be a monobasic as AH.1 = 0.86 0.e.109 (0.1 g Tf = Kfm = 5 1. 02.4 K.68°C (C) 101.48) initial molality= 0.05 × 1. (A) –3.763. n = 2 i = 1 + (2–1) = 1 + = 1.47 Hence the molality of A after dimer is formed = (1–) initial molality = (1 – 0. 20 = Po XB Hence.1. 1 = 5 i 0. The mole fraction of solute in solution is 0. mole fraction of solvent = 1 – 0. AB.036 = 1.262 = 0.1 = .LIQUID SOLUTIONS Problem 10: The freezing point of a solution of acetic acid (mole fraction is 0.52 0. What is the freezing point of a Problem 2: Calculate St ud 0.48 0.02 1000 Molality of A in B = m X = = 0. Freezing point of benzene is 278. Atomic masses of K = 39. Tf = Kf i molality 278.4–277.86 × .6 Page 32 of 33 www. Acetic acid exists partly as a dimer 2A A2.4 (D) 0.763 = 0. Kb for water = 0.04 g mL ) of potassium chloride. 1 = 1 – /2 XB = 0.7°C Solution: Ans (b) ForAB. Cl = 35. (c) –1 m= 1000 1 =. = 1 (if not given) Tb = 101.86 deg/molal.4 = 0.1953°C (C) –1.: (b) Po – Ps = Po XB 10 = Po 0.5 Tb = 1 × Xb × m Tb = 2 × 0. is 5% dissociated in aqueous solution.4. what would be mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg.8°C (B) –0.4 k and (Kf for benzene is 5) Solution: 2A A2 1 0 initially moles 1– /2 moles after dimer is formed i= 1 2 XA = 0.StudySteps.262 1– /2 = 0.in .2 Solution: Ans.2 Again.13624) 2 = 3.05 Tf = i × Xf × m = 1. Calculate equilibrium constant for dimerisation.26 = 0.1953ºC the boiling point of a one molar aqueous solution (density = 1. i = 1 = 0.78°C Tf = – .24 0. so.262 .28°C (B) 103.in 0.52 kg mol–1.04 1 74.5 (A) 107.4 = 5 i 0.6 (C) 0.100 molal aqueous solution of AB? Kf for water is 1.262 Molality of A2 after dimer is formed A 2 0. (A) 0.39 kg mol–1 te The equilibrium constant .078°C (D) None of these Solution: Ans.8 (B) 0.2. 5 0.262 or.06288 molality = 2 2 ps = yS OBJECTIVE Problem 1: A weak electrolyte.262 mol kg–1 of Benzene 78 0.036 1000 1.078ºC Problem 3: The vapour pressure of a solvent decreased by 10 mm Hg when a non volatile solute was added to the solvent.06288 Keq = A 2 = (0.078 (KCl is an electrolyte having n = 2. n = 0.98 B B Since.1953 (D) – 0.52 × 1.02) in benzene 277.98 xA 1000 0. 86 K molality–1] (A) 38. Elevation in its boiling point is given by Solution: M KbY (B) Ans.wt 170 =1+= –1 = 0. = 100) are dissolved in 500 g of water. The freezing point of the solution is –0.71 g (B) 38.: (a) P = 119XA + 135.:(b) i for AgNO3 = (D) 60.64 yS Solution: te respectively.86 K.3°C is : [Kf = 1. P = 119XA + 135 where XA is mole fraction of methyl alcohol.23% Normal mol.wt. Kf = 1.71 g www.in (A) ps Problem 7: The values of observed and calculated molecular weights of silver nitrate are 92.29 62 W Ice separated = 200 – 161.in .: (a) Page 33 of 33 T = 1000 K f w WM 9.86 = –1. Molal elevation constant of benzene is Kb.3 = (D) 42 mg 1000 1.86 34. (C) 119 torr Lim X A 1 (D) 140 torr PA X A = 119 + 135 = 254 Problem 9: The amount of ice that will separate on cooling a solution containing 50g of ethylene glycol in 200g water to –9.86°C (B) 1.LIQUID SOLUTIONS Problem 4: The molal freezing point constant for water is 1.2 g of cane sugar (C12H22O11) are dissolved in 1000g of water.5% (C) 46.: (a) Tf = 1000 K f w 1000 1.29 = 38.74 = (1 + ) × 20 / 100 500 / 1000 1+=1 =0 Problem 6: Y g of non .wt.86° C Problem 5: 20 g of a binary electrolyte (mol. of methyl alcohol .: (d) Tf = i ×Xf × m . The degree of dissociation of silver nitrate is (A) 60% (B) 83.86 K.835 = 83. molarlity–1.42°C Solution: Ans.92°C (D) 2.ethy alcohol solution is represented X A 1 PA X A is (A) 254 torr Solution: St Lim ud by the equation.molality–1.volatile organic substance of molecular mass M is dissolved in 250 g benzene.64 and 170 Ans.7% Problem 8: At 40°C.: (b) T = 4K b Y M (C) KbY 4M 1000 K b Y 4K b Y = 250 M M (D) KbY M .StudySteps. If 34.86 50 W = 161.2 = 1. the solution will freeze at (A) –1.86° MW 100 342 Tf = 0 – 1.86°C (C) –3.5 % Observed mol.71 mg (C) 42 g Solution: Ans. 92. the vapour pressures in torr. The degree of ionisation of the electrolyte is (A) 50% (B) 75% (C) 100% (D) 0 Solution: Ans. then the value of (B) 135 torr Ans.74°C.