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Que. from IIT-JEE 8. 10 Yrs. Que. from AIEEE Student’s Name :______________________ Class :______________________ Roll No. :______________________ ADDRESS: R-1, Opp. Raiway Track, New Corner Glass Building, Zone-2, M.P. NAGAR, Bhopal (cid:1) : (0755) 32 00 000, 98930 58881, www.tekoclasses.com KEY CONCEPTS S N O I T Vapor Pressure. The pressure exerted by the vapors of a liquid which are in equilibrium with it at aU L given temperature. O S D I U Note: It depends only on temperature and on nature of the liquid. It does NOT depend on the surfaceQ I area 0 L 2 of 2 Raoult's Law. The equilibrium vapor pressure of a volatile component is linearly proportional to thee g mole fraction of that compoent in liquid phase. Pa For non-volatile solutes : P(solution) = x Po solvent or relative lowering of vapor pressure, (Po - P)/Po = x . L solute A A more useful form is (Po - P)/P = n/N OP H m where n = total number of moles of all the free solute species in the solution finally (i.e. at equilibrium). B es.co Three cases arise. 881 , ss(i) Non-electrolyte is dissolved e.g. glucose or urea. These molecules do not dissociate into ions. If 0.1 8 a 5 cl mol of urea is dissolved in 50 moles of water, then n/N = 0.1/50 simply. 0 o 3 k 9 w.te(ii) Strong electrolyte is dissolved e.g. NaCl, Ca(NO ) etc. These dissociate nearly completely into ions. 0 98 ww If 0.1 mol of NaCl is dissolved in 50 moles of water,3 th2en n/N = 0.2/50 since NaCl dissociates completely 0, in 0.1 mol Na+ ions and 0.1 mol Cl- ions. Similarly, for Ca(NO3 )2, n/N = 0.3/50 if 0.1 mol of it 00 : dissociates completely into ions. 0 e 0 sit 2 b 3 we(iii) Weak electrolyte is dissolved e.g. HCOOH, CH3NH2 etc. In such cases, we should determine the 5)- m total number of moles of all the solute species at equilibrium. e.g. i f no moles of formic acid considered 75 0 o non-volatile here) are dissolved in N moles of solvent then, ( e fr HCOOH ⇔ H+ + COOH- PH: kag no(1-α) noα noα Sir) ac Total number of moles at equilibrium = no(1+ α). Hence, n/N = no(1+ α)/N. K. P R. dy Note : This factor, n (at equilibrium)/n(original) is referred to as van't Hoff factor. S. u ( t A S Y d Ideal Solutions. The solutions which obey Raoult's Law are called ideal solution. For ideality : RI loa (i) ∆Hmix= 0, (ii) ∆Vmix = 0 as well for liquid-liquid solutions. KA wn R. o Non ideal solution (Deviations From Raoult's Law) G D A Positive deviation. When the observed vapor pressure is more than that expected by Raoult's law. H E U E S FR This is observed when ∆Hmix> 0 i.e. energy is absorbed on mixing. Usually obtained by mixing of polar or : liquids with non-polar ones. e.g. cyclohexane and ethanol. ct e r Di S, E S S A L C O K E T Negative deviation. When the observed vapor pressure is less than that expected by Raoult's law.S N This is observed when ∆H < 0 i.e. energy is released on mixing. Attractive forces between unlikeO mix TI molecules are greater than the forces of attraction between like molecules. e.g. chloroform and acetone.U L (Curve 3 in Fig. 1 and 2). O S D I U 2 P20 3 LIQ 0 V .↑ P. 1 ↑ 1 3 of 2 3 B. P. ge P0 2 a 1 P 0 x →→→→ 1 x is mole fraction of more 0 x →→→→ 1 volatile component (2) x is mole fraction of more L Fig. 1 volatile component (2) PA Fig. 2 O m Azeotropic Solutions. During distillation, the mole fraction of more volatile component in vapor state is BH s.co hpiagrhtiecru tlahra cno tmhapto ins iltiiqounisd o sft wathei.c Thh tihse m maoklees f draiscttiilolant ioofn c poomspsiobnlee.n Htso inw leiqvueirs, athnedr ev aepxoisrt s stoatme eis s soalmuteio. nTsh fuosr, 81 , e 8 ss no advantage is derived by distilling such a mixture and it is termed as azeotropic. 8 a 5 cl 0 o 3 k Completely Immiscible Liquids : When they are distilled, they distil in the ratio of their vapor pressure 9 www.te at that temperature. e.g. When A and B are distilled wt ratio wwAB is given as wwAB = PPAB°°··MMAB 000, 0 98 : 0 e Completely Miscibile Liquids. They can be handled by Raoult's Law i.e. 0 bsit yP = xPo 32 we where P = Total pressure of vapors in equiilibriumi wiith the liquid solution, 5)- m Po = vapor pressure of component i in pure state 75 fro yii = mole fraction of ith component in vapor state, xi = mole fraction of ith component in liquid state H: (0 e This most fundamental expression may be arranged in many useful forms. e.g. for binary solutions : P g ) ka P = x1(P10 - P20) + P20 Sir ac or 1/P = 1/P 0 + y (1/P 0 - 1/P 0) K. P Note : Vapor pressure of an ideal solution2 is alw1ays b1etween2 P 0 and P 0 (Curve1 in Fig. 1 and 2) R. dy 1 2 S. u Bubble Point. When the first bubble of vapor appears in liquid solution. ( t A S Y d Dew Point. When the first drop of liquid condenses from a mixture of vapors. OR when the last drop of RI oa liquid remains and rest of the liquid is completely vaporised. KA l wn Colligative Properties. The properties which depend only on the number of moles of solute (and not R. o G D on their molecular weights or sizes) are referred to as colligative properties. A H E e.g. Lowring of vapor pressure, depression of freezing point, elevation of boiling point, osmotic pressure U E S FR etc. or : P0−P n w/m ct 1. Relative lowering of vapour Pressure. = = = x e P0 n+N w/m+W/M solute Dir S, E S 2. Elevation in Boiling Point, ∆∆∆∆T . For dilute solutions, ∆∆∆∆T = K m S b b b A where m is molality of the solution (i.e. total number of moles of all the solute particles per kg of solvent). L C K is ebullioscopic or boiling point elevation constant which is given by O b K E T S R(T0)2M ON K = b solvent I T b 1000∆H U vap L O ∆H is the enthalpy of vaporisation of solvent. S vap D I U Q 3. Depression in freezing Point (∆∆∆∆T). For dilute solutions, ∆∆∆∆T = Km I f f f L 0 where, K = R(Tf0)2Msolvent 4 of 2 f 1000∆H ge fusion a P Osmosis. 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. L A P O Reverse Osmosis. If a pressure greater than the osmotic pressure is applied on the concentrated H m B s.co sreovluetrisoen o, tshmeo ssoilsv.eOnnt es toafr tists t och fileofw u fsreosm is cdoenscaelinntartaitoend osof sluetaio wna ttoe rd itlou tgee sto pluutrieo dnr (ionrk ipnugr ew saotlevre.nt). This is 81 , e 8 ss 8 a 5 cl4. Osmotic Pressure (ππππ). The hydrostatic pressure built up on the solution which just stops osmosis. 0 o 3 k Alternately, it may be defined as the pressure which must be applied to the concentrated solution in order 9 w.te to stop osmosis. 0 98 ww Fwohre rdei lcu tise tshoel utotitoanl smolar concentration of all thππππe == ==fr eceR sTp e=c ihesρρρρ pgresent in the solution, h is the height 00, 0 : developed by the column of the concentrated solution and ρρρρ is the density of the solution in the column. 0 e 0 sit On the basis of osmotic pressure, the solutions can be classified in three classes. 2 b 3 m we Isotonic solutions. Two solutions having same osmotic pressures at same temperature. 755)- (This implies c = c ). 0 o 1 2 ( fr H: e P g Hypertonic solution. When two solutions are being compared, then the solution with higher osmotic ) ka pressure is termed as hypertonic. The solution with lower osmotic pressure is termed as hypotonic. Sir ac K. dy P Important. Osmotic pressures can be determined quite accurately, hence it is used in the S. R. u determination of molecular weights of large proteins and similar substances. ( t A S Y d RI a Van't Hoff Factor (i) A o K wnl Since colligative properties depends upon the number of particles of the solute, in some cases where the R. o solute associates or dissociates in solution, abnormal results for molecules masses are obtained. G D A Observed colligative property(actual) H E i = U E Theoretical colligative property S FR or : t c e r Di S, E S S A L C O K E T THE ATLAS S N O I T U L O S D I U Q I L 0 2 of 5 e g a P L A P O H m B s.co 81 , e 8 ss 8 a 5 cl 0 o 3 k 9 w.te 0 98 ww 0, 0 0 : 0 e 0 sit 2 b 3 we 5)- m 75 0 o ( fr H: e P g ) ka Sir ac K. P R. dy S. u ( t A S Y d RI a A o K l wn R. o G D A H E U E S FR or : t c e r Di S, E S S A L C O K E T EXERCISE I S N O I T U Raoult’s law L O S D Q.1 At 25°C, the vapour pressure of methyl alcohol is 96.0 torr. What is the mole fraction of CH OH in aUI 3 Q solution in which the (partial) vapor pressure of CH OH is 23.0 torr at 25°C? I 3 L 0 2 Q.2 The vapour pressure of pure liquid solvent A is 0.80 atm. When a nonvolatile substance B is added to theof 6 solvent its vapour pressure drops to 0.60 atm. What is the mole fraction of component B in the solution?ge a P Q.3 The vapour pressure of pure water at 26°C is 25.21 torr. What is the vapour pressure of a solution which contains 20.0 glucose, C H O , in 70 g water? 6 12 6 L A P Q.4 The vapour pressure of pure water at 25°C is 23.76 torr. The vapour pressure of a solution containing O H m 5.40 g of a nonvolatile substance in 90.0 g water is 23.32 torr. Compute the molecular weight of the B s.co solute. 81 , e 8 ss 8 aRaoult’s law in combinaton with Dalton’s law of P.P. and V.P. lowering 5 cl 0 o 3 k 9 w.teQ.5 Tsohlue tvioanp oisu prr pepreasrseudr aet othf ee sthamaneo tle amnpde mraetuthrea nboy lm airxei n4g4 .650 mg mof aenthda 8no8l. 7w mithm 4 H0 gg oref smpeetchtaivneolly. .C Aanlc iudlaetael 0 98 ww total vapour pressure of the solution. 0, 0 0 : 0 eQ.6 Calculate the mole fraction of toluene in the vapour phase which is in equilibrium with a solution of 0 bsit benzene and toluene having a mole fraction of toluene 0.50. The vapour pressure of pure benzene is 32 we 119 torr; that of toluene is 37 torr at the same temperature. 5)- m 75 0 o ( frQ.7 What is the composition of the vapour which is in equilibrium at 30°C with a benzene-toluene solution H: e with a mole fraction of benzene of 0.40? With a mole fraction of benzene of 0.60? P g ) ka P°=119 torr and P°= 37 torr Sir ac b t K. P R. dyQ.8 At 90°C, the vapour pressure of toluene is 400 torr and that of σ-xylene is 150 torr. What is the S. u ( t composition of the liquid mixture that boils at 90°C, when the pressure is 0.50 atm? What is the composition A S Y d of vapour produced? RI a A o K l wnQ.9 Two liquids A and B form an ideal solution at temperature T. When the total vapour pressure above the R. o solution is 400 torr, the mole fraction of A in the vapour phase is 0.40 and in the liquid phase 0.75. What G D A are the vapour pressure of pure A and pure B at temperature T? H E U E S FRQ.10 Calculate the relative lowering in vapour pressure if 100 g of a nonvolatile solute (mol.wt.100) are or : t dissolved in 432 g water. c e r Di Q.11 What weight of the non−volatile solute, urea needs to be dissolved in 100 g of water, in order to decrease S, E the vapour pressure of water by 25%? What will be the molality of the solution? S S A L Q.12 The vapour pressure of an aqueous solution of glucose is 750 mm Hg at 373 K. Calculate molality and C O mole fraction of solute. K E T Q.13 The vapour pressure of pure benzene at 25° C is 639.7 mm of Hg and the vapour pressure of a solutionS N of a solute in C H at the same temperature is 631.7 mm of Hg. Calculate molality of solution. O 6 6 TI U L Q.14 The vapour pressure of pure benzene at a certain temperature is 640 mm of Hg. A nonvolatile nonelectrolyteO S solid weighing 2.175 g is added to 39.0 of benzene. The vapour pressure of the solution is 600 mm ofD I U Hg. What is molecular weight of solid substance? Q I L 0 Q.15 The vapour pressure of water is 17.54 mm Hg at 293 K. Calculate vapour pressure of 0.5 molalof 2 solution of a solute in it. e 7 g a P Q.16 Benzene and toluene form two ideal solution A and B at 313 K. Solution A (total pressure P ) contains A equal mole of toluene and benzene. Solution B contains equal masses of both (total pressure P ). The B vapour pressure of benzene and toluene are 160 and 60 mm Hg respectively at 313 K. Calculate the L A value of P /P . P A B O H m B s.coBoiling point elevation and freezing point depression 81 , e 8 ssQ.17 When 10.6 g of a nonvolatile substance is dissolved in 740 g of ether, its boiling point is raised 0.284°C. 8 a 5 cl What is the molecular weight of the substance? Molal boiling point constant for ether is 2.11°C·kg/mol. 0 o 3 k 9 w.teQ.18 A solution containing 3.24 of a nonvolatile nonelectrolyte and 200 g of water boils at 100.130°C at 0 98 ww 1atm. What is the molecular weight of the solute? (Kb for water 0.513°C/m) 0, 0 0 : Q.19 The molecular weight of an organic compound is 58.0 g/mol. Compute the boiling point of a solution 0 e 0 sit containing 24.0 g of the solute and 600 g of water, when the barometric pressure is such that pure water 2 b 3 we boils at 99.725°C. 5)- m 75 oQ.20 An aqueous solution of a nonvolatile solute boils at 100.17°C. At what temperature will this solution (0 fr freeze? [K for water 1.86°C/m] H: e f P g ) kaQ.21 Pure benzene freeze at 5.45°C. A solution containing 7.24 g of C H Cl in 115.3 g of benzene was Sir ac 2 2 4 K. P observed to freeze at 3.55°C. What is the molal freezing point constant of benzene? R. dy S. uQ.22 A solution containing 6.35 g of a nonelectrolyte dissolved in 500 g of water freezes at – 0.465°C. ( t A d S Determine the molecular weight of the solute. RIY a A o K wnlQ.23 The freezing point of a solution containing 2.40 g of a compound in 60.0 g of benzene is 0.10°C lower R. o than that of pure benzene. What is the molecular weight of the compound? (K is 5.12°C/m for benzene) G D f A H E U EQ.24 The elements X and Y form compounds having molecular formula XY and XY . When dissolved in 20 gm S FR of benzene, 1 gm XY2 lowers the freezing point by 2.3°, whereas 1 gm2 of XY4 lo4wers the freezing point by or : 1.3°C. The molal depression constant for benzene is 5.1. Calculate the atomic masses of X and Y. t c e r Di Q.25 Calculate the molal elevation constant, Kb for water and the boiling of 0.1 molal urea solution. Latent S, heat of vaporisation of water is 9.72 kcal mol–1 at 373.15 K. E S S A L Q.26 Calculate the amount of ice that will separate out of cooling a solution containing 50g of ethylene glycol C in 200 g water to –9.3°C. (K for water = 1.86 K mol−1 kg) O f K E T Q.27 A solution of 0.643 g of an organic compound in 50ml of benzene (density ; 0.879 g/ml) lowers itsS N freezing point from 5.51°C to 5.03°C. If K for benzene is 5.12 K, calculate the molecular weight of theO f TI compound. U L O S Q.28 The cryoscopic constant for acetic acid is 3.6 K kg/mol. A solution of 1 g of a hydrocarbon in 100 g ofD I U acetic acid freezes at 16.14°C instead of the usual 16.60°C. The hydrocarbon contains 92.3% carbon.Q I What is the molecular formula? L 0 2 of Osmotic pressure e 8 g a P Q.29 Find the freezing point of a glucose solution whose osmotic pressure at 25oC is found to be 30 atm. K(water) = 1.86kg.mol−1.K. f L A Q.30 At 300 K, two solutions of glucose in water of concentration 0.01 M and 0.001 M are separated by P O m semipermeable membrane. Pressure needs to be applied on which solution, to prevent osmosis? Calculate BH s.co the magnitude of this applied pressure. 81 , e 8 ssQ.31 At 10oC, the osmotic pressure of urea solution is 500 mm. The solution is diluted and the temperature 8 a 5 cl is raised to 25°C, when the osmotic pressure is found to be105.3 mm. Determine extent of dilution. 0 o 3 k 9 w.teQ.32 The osmotic pressure of blood is 7.65 atm at 37°C. How much glucose should be used per L for an 0 98 ww intravenous injection that is to have the same osmotic pressure as blood? 0, 0 0 : Q.33 What would be the osmotic pressure at 17°C of an aqueous solution containing 1.75 g of sucrose 0 e 0 bsit (C12H22O11) per 150 cm3 of solution? 32 we 5)- m Q.34 A 250 mL water solution containing 48.0 g of sucrose, C12H22O11, at 300 K is separated from pure 75 o water by means of a semipermeable membrane. What pressure must be applied above the solution in (0 fr order to just prevent osmosis? H: e P g ) kaQ.35 A solution of crab hemocyanin, a pigmented protein extracted from crabs, was prepared by dissolving Sir dy Pac 0w.a7s5 o0b gs einrv 1e2d5. Tcmhe3 s oofl uatnio anq huaedo uas d menesditiyu mof. 1A.0t 04 °gC/c amn3 o. Dsmeotetrimc pinree stshuer me orilseec uolfa 2r .w6 emigmht o off t thhee s porloutteioinn. S. R. K. u ( t A d SQ.36 The osmotic pressure of a solution of a synthetic polyisobutylene in benzene was determined at 25°C. A RIY a sample containing 0.20 g of solute/100 cm3 of solution developed a rise of 2.4 mm at osmotic equilibrium. A o K wnl The density of the solution was 0.88 g/cm3. What is the molecular weight of the polyisobutylene? R. o G D A Q.37 A 5% solution (w/v) of cane-sugar (Mol. weight = 342) is isotonic with 0.877%(w/v) of urea solution. H E U E Find molecular weight of urea. S FR or : Q.38 10 gm of solute A and 20 gm of solute B are both dissolved in 500 ml water. The solution has the same t c e osmotic pressure as 6.67 gm of A and 30 gm of B dissolved in the same amount of water at the same r Di temperature. What is the ratio of molar masses of A and B? S, E S Van’t Hoff factor & colligative properties S A L C Q.39 A storage battery contains a solution of H SO 38% by weight. What will be the Van't Hoff factor if the O 2 4 K ∆T in 29.08. [Given K = 1.86 mol–1 Kg] E f(experiment) f T Q.40 A certain mass of a substance, when dissolved in 100 g C H , lowers the freezing point by 1.28°C. TheS 6 6 N same mass of solute dissolved in 100g water lowers the freezing point by 1.40°C. If the substance hasO I T normal molecular weight in benzene and is completely ionized in water, into how many ions does itU L dissociate in water? K for H O and C H are 1.86 and 5.12K kg mol−1. O f 2 6 6 S D I U Q.41 2.0 g of benzoic acid dissolved in 25.0g of benzene shows a depression in freezing point equal to 1.62K.Q I Molal depression constant (K) of benzene is 4.9 K.kg.mol−1. What is the percentage association of the acid?L f 0 2 of Q.42 A decimolar solution of potassium ferrocyanide is 50% dissociated at 300K. Calculate the osmotice 9 g pressure of the solution.(R=8.314 JK−1 mol−1) a P Q.43 The freezing point of a solution containing 0.2 g of acetic acid in 20.0g of benzene is lowered by 0.45°C. Calculate the degree of association of acetic acid in benzene. (K for benzene = 5.12 K mol−1 kg) L f A P O mQ.44 0.85 % aqueous solution of NaNO3 is apparently 90% dissociated at 27°C. Calculate its osmotic BH s.co pressure. (R= 0.082 l atm K−1 mol−1 ) 81 , e 8 ssQ.45 A 1.2% solution (w/v) of NaCl is isotonic with 7.2% solution (w/v) of glucose. Calculate degree of 8 a 5 cl ionization and Van’t Hoff factor of NaCl. 0 o 3 k 9 w.te 0 98 ww 0, 0 0 : 0 e 0 sit 2 b 3 we 5)- m 75 0 o ( fr H: e P g ) ka Sir ac K. P R. dy S. u ( t A S Y d RI a A o K l wn R. o G D A H E U E S FR or : t c e r Di S, E S S A L C O K E T PROFICIENCY TEST S N O Q.1 Fill in the blanks with appropriate items : TI U L 1. Lowering of vapour pressure is ______ to the mole fraction of the solute. O S D 2. The ratio of the value of any colligative property for NaCl solution to that of equimolal solution of sugarI U is nearly______. Q I L 3. Semipermeable membrane allows the passage of ________through it. 0 2 of 4. A binary solution which has same composition in liquid as well as vapour phase is called______. 0 1 e g 5. The molal elevation constant of solvent is also called _____. a P 6. The 0.1 M aqueous solution of acetic acid has boiling point _______ than that of 0.1 M aqueous solution of KCl. L A 7. For ideal solutions, the plot of total vapour pressure v/s composition is ______. P O H m8. A solution of CHCl and acetone shows________deviation. B s.co9. Gases which react w3ith water are generally _________ soluble in it. 81 , e 8 ss10. Assuming complete dissociation, Van’t Hoff’s factor for Na SO is equal to ________. 8 a 2 4 5 ocl11. The osmotic pressure of a solution _______ with increase in temperature. 30 k 9 w.te12. Water will boil at 101.5°C at pressure of _______76 cm of Hg. 0 98 ww13. Vant’s Hoff’s factor ‘i’ for dimerisation of CH3COOH in benzene is_________. 0, 0 n 0 e: 14. π = B RT is known as_________. 00 sit V 2 b 3 we15. The molal elevation constant is the ratio of the elevation in boiling point to _________. 5)- m Q.2 True or False Statements : 75 0 o ( fr16. Relative lowering of vapour pressure is a colligative property. H: e P kag17. Lowering of vapour pressure of a solution is equal to the mole fraction of the non-volatile solute present in it. Sir) ac18. The components of an azeotropic solution can be separated by simple distillation. K. dy P19. Vapour pressure of a liquid depends on the size of the vessel. S. R. tu20. Addition of non-volatile solute to water always lowers it vapour pressure. A ( S Y d 21. Reverse osmosis is generally used to make saline water fit for domestic use. RI a A o K l22. A 6% solution of NaCl should be isotonic with 6% solution of sucrose. wn R. o23. A real solution obeys Raoult’s law. G D A E 24. Boiling point is a characteristic temperature at which vapour pressure of the liquid becomes higher than UH E S FR the atmospheric pressure. or : 25. Molal depression constant is independent of the nature of solute as well as that of solvent. t c e 26. The real solutions can exhibit ideal behaviour at high concentrations. r Di 27. The osmotic pressure decreases on addition of solvent to the solution. S, E S 28. For urea the value of Vant’s Hoff’s factor ‘i’ is equal to 1. S A L 29. The unit of k is kg K–1 mol–1. C b O 30. 0.1 M solution of urea would be hypotonic with 0.1 M solution of NaCl. K E T