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Published in: Chemistry
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Chapter Solution detailed notes

Amrita P / Chandigarh

12 years of teaching experience

Qualification: M.Sc (THAPAR UNIVERSITY (TU), Patiala - 1987)

Teaches: Chemistry, IIT JEE Advanced, IIT JEE Mains, NEET

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  1. CONTENTS 1. Types of solution 2. Concentration 3. Solubility of gas Henry Law 4. Raoult s law 5. Ideal and non-ideal solution 6. Azeotropic mixture Colligative properties 7. 8. Van t Hoff factor CHAPTER STRATEGY Solution — 5 Marks Definition or Conceptual Or Numericals (1+1 or 2 Marks) 3 Marks Method of Expre ng Concentration of solution Molarity Molality Types of solution Raoulüs Colligative Law properties RLVP PPM Mole Fraction Solid Gas in in Liquid Liquid Vanmt Hoff factor and Abnormal Molar mass DFP EBP Osmosis & Osmotic Pressure Hennas Law Solubility & Factor affecting
  2. INTRODUCTION Solutions are homogenous mixtures of two or more substances in a single phase. DIFFERENT TYPES OF SOLUTIONS T e of Solution Gaseous Solutions Gas in as Li uid in as Solid in as Li uid Solutions Gas in li uid Li uid in li uid Solid in li uid Solid Solutions Gas in solid Li uid in solid Solid in solid CONCENTRATION UNITS Common Exam le a mixture of ox en and nitro en ases. chloroform va ours mixed with nitro en as. cam hor ox en dissolved in water ethanol dissolved in water sucrose dissolved in water solutions of h dro en in alladium Amal am of mercur with sodium co er dissolved in old The concentration of a solute is the amount of solute dissolved in a given quantity of solvent or solution. The quantity of solvent or solution can be expressed in terms of volume or in terms of mass or molar mass. Thus there are several ways of expressing the concentration of a solution. (a) Molarity (M): Moles of solute present in one litre solution. no. of moles of solute volume of solution in litre (b) Molality (m): Moles of solute present in one kilogram of solvent. no. of moles of solute mass of solvent in kg (c) Normality (N): No. of equivalents present in one litre solution. no. of equvalents of solute volume of solution in litre (d) Mole fraction: The mole fraction of a component substance A(XA) in a solution is defined as the moles of component substance divided by the total moles of solution. moles of substance A total moles of solution (e) Mass percent: The mass percent of a component A in solution is defined as mass of A Mass % of A = x 100 Total mass (f) Part per million (PPM): It is defined as the parts of given component in one million parts of solution. Mathematically 1. 2. mass of solute x106 ppm mass of solution WORKSHEET : 1 5 g of NaCl is dissolved in 1000 g of water. If the density of the resulting solution is 0.997 g per cc, calculate molality, molarity, normality and mole fraction of the solute. If 20 ml of ethanol (density 0.7893 gm / ml) is mixed with 40 ml (density = 0.9971 gm/ml) at 250C, the final solution has density of 0.9571 gm / ml. Calculate
  3. the percentage change in total volume of mixing. Also calculate the molality of alcohol in the final solution. SOLUBILITY OF GASES We are familiar that gases are completely miscible with each other. Gases also dissolve in liquids and solids. For example, soda-water contains carbon dioxide dissolved in water under high pressure. Oxygen is sufficiently soluble in water to allow survival of aquatic life in lakes, rivers and oceans. An example of dissolution of gas in a solid is the solubility of hydrogen gas in palladium. The solubility of a gas in a liquid is determined by several factors. In addition to the nature of the gas and the liquid, solubility of the gas depends on the temperature and pressure of the system. The solubility of a gas in a liquid is governed by Henry' s law which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas. Dalton, a contemporary of Henry, also concluded independently that the solubility of a gas in a liquid solution is a functional of the partial pressure of the gas. If we use the mole fraction of the gas in the solution as a measure of its solubility, then: Mole fraction of the gas in a solution is proportional to the partial pressure of the gas. Or, partial pressure of the gas in solution = KH x mole fraction of the gas in solution Here KH is Henry' s law constant Or p = KHX If we draw a graph between partial pressure of the gas versus mole fraction of the gas in solution, then we should get a plot of the type as shown in figure. 100 o o 500 0.010 0.020 Mole fraction of HCI in its solution in cyclohexane Experimental results for the solubility of HCI gas in cyclohexane at 93 K. The slope of the line is the Henry's law constant KH Different gases have different KH values at the same temperature. This suggests that KH is a function of the nature of the gas. Table gives KH values of some common gases at specified temperature Values of Henry' s law constant (KH) for some selected gases in water Gas N2 02 02 Tem 1K KH/kbar 293 293 293 303 293 393 144.97 69.16 76.48 88.84 34.86 46.82 It is obvious from figure that the higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid. It can be seen from table that KH value for both N2 and 02
  4. increases with increase in temperature indicating that solubility of gases decreases with increase of temperature. It is due to this reason that aquatic species are more comfortable in cold waters rather than warm waters. Henry' s law finds several applications in industry and explains some biological phenomenon. Notable among these are: (i) To increase the solubility of C02 in soft drinks and soda water, the bottle is sealed under high pressure. (ii) To minimize the painful effects accompaynig the decompression of deep sea divers, oxygen diluted with less soluble helium gas is used as breathing gas. (iii) In lungs, where oxygen is present in air with high partial pressure, haemoglobin combines with oxygen to form oxyhaemoglobin. In tissues where partial pressure of oxygen is low, oxyhaemoglobin releases oxygen for utilization in cellular activities. SOLID SOLUTIONS Solid solutions are formed by mixing two solid components in the molten state in appropriate proportion and then cooling the molten mass. Solid solutions are of two types: substitutional solid solutions and interstitial solid solutions. In a substitutional solid solution, atoms, molecules or ions of one substance take the place of similar species of other substance in its crystal lattice as shown in figure (a). Brass, bronze, monel metal and steel are familiar examples of this type of solid solution. Interstitial solid solutions constitute the other type and are formed by placing atoms of one kind into voids or interstices, that exist between atoms in the host lattice. This is illustrated in figure (b). 000000 000000 000000 000000 O Solute (a) Substiutional solid solution 000000 O,OOOOP 000000 oooobo O Solvent Solute (b) Interstitial solid solution Two types of solution (a) substitution solid in which particles of the solute replace interstitial solid, solution in which the solute particles fit in spaces between particles of the host lattice (the solvent) Tungsten carbide. WC. an extremely hard substance, is an example of interstitial solid solution. Here tungsten atoms are arranged in a face-centred cubic pattern with carbon atoms in the octahedral holes where these are surrounded by six tungsten atoms placed at the vertices of the octahedron. Tungsten carbide has many industrial uses in making of cutting and grinding tools.
  5. VAPOUR PRESSURE OF A SOLUTION Relative lowering of vapour pressure of a solvent is a colligative property equal to the vapour pressure of the pure solvent minus the vapour pressure of the solution. For example, water at 200C has a vapour pressure of 17.54 mmHg. Ethylene glycol is a liquid whose vapour pressure at 200C is relatively low, an aqueous solution containing 0.010 mole fraction of ethylene glycol has a vapour pressure of 17.36 mmHg. Thus the vapour pressure lowering, AP = 17.54 mmHg 17.36 mmHg = 0.18 mmHg. RAOULT' S LAW Vapour pressures of a number of binary solution of volatile liquids such as benzene and toluene at constant temperature gave the following generalization which is known as the Raoult' s law. The partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution. Suppose a binary solution contains nA moles of a volatile liquid A and moles of a volatile liquid B, if PA and PB are partial pressure of the two liquid components, the according to Raoult' s law and Where XA is the mole fraction of the component A given by nA/nA+ 11B; XB is the mole fraction of the component B, given by nB/nA+nB and PAO and P? are the vapour pressures of pure components A and B respectively. If the vapour behaves like an ideal gas, then according to Dalton' s law of partial pressures, the total pressure P is given by p _I_PB = x A p: + x B p: n A .PAO + n A + 11B o X p total Mole Fraction The relationship between vapours pressure and mole fraction of an ideal solution at constant temperature. The dashed lines I and Il represent the partial pressure of the components. (It can be seen from the plot that PA and are directly proportional to x A and XB respectively). The total vapour pressure is given by line marked Ill in the figure. WORKSHEET : 2
  6. 3. If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 litre of water. Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry' s law constant for N2 at 293 K is 76.84 kbar. 4. The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40g of methanol. Calculate total vapour pressure of the solution. IDEAL SOLUTION An ideal solution of substance A and B is one in which both substances follow Raoult' s law for all values of mole fractions. Such solutions occurs when the substances are chemically similar so that the intermolecular forces between A and B molecules are similar to those between two A molecules or between two B molecules. The total vapour pressure over an ideal solution equals the sum of the partial vapour pressures, each of which is given by Raoult' s law. For example: Solution of benzene C6H6 and toluene C6H5CH3 are ideal. Note the similarly in their structural formula. Benzene Toulene Suppose a solution is 0.70 mole fraction benzene and 0.30 mole fraction of toluene. The vapour pressure of pure benzene and pure toluene are 75 mmHg and 22 mmHg respectively. Hence the total vapour pressure is P mmHg Ethylene bromide and ethylene chloride, n-Hexane and n-heptane, n-butyl chloride and n — butyl bromide, carbon tetrachloride and silicon tetra chloride are few other examples of ideal solutions. Condition for a solution to be ideal. (a) It should obey Raoult' s law i.e. PA = x APR and PB = XBPBO (b) AH = 0 i.e. no heat should be absorbed or evolved during mixing. mixing (c) AV = 0 i.e. no expansion or contraction on mixing mixing NON - IDEAL SOLUTIONS Those solution which do not obey Raoult' s law over entire range of composition and deviate from ideal behaviour, are real or non — ideal solution. They are divided into two types (a) The solution showing positive deviations from ideal behaviour for those type of solutions, (i) PA and PB (ii) AH mixing (iii) AV > O mixing
  7. Positive deviation Solution of ethyl alcohol in water shows positive deviation. Attractive forces between A-AorB-B>B-B. CD CD o vapour pressure of solution MOLE FRACTION Positive deviations occur when the interaction between unlike molecules is weaker than the interaction between like molecules. The solutions of this type are characterized by positive enthalpy of mixing, AHmix and positive volume of mixing, AVmix. Example are —acetaldehyde —carbon disulphide, water —propyl alcohol, ethyl alcohol —chloroform and cyclohexane —carbon tetrachloride. (b) The solution showing large negative deviations from ideal behaviour and the vapour pressure of each component is considerably less than that predicted by Raoult' s law, for these type of solutions. (i) PA and PB (ii) AH mixing (iii) AV < O mixing Negative deviation The solution of HN03 in water shows negative deviation. Attractive forces between A Vapour pressure of solution O x MOLE FRACTION Negative deviations occurs when the interaction between unlike molecules is stronger than the interaction between like molecules. The solutions of this type are characterized by negative enthalpy of mixing and negative volume of mixing. For example - acetone - chloroform, water - sulphuric acid and water - nitric acid.
  8. WORKSHEET : 3 5. Which pair will show positive and negative deviation from Raoult's Law. (i) Methanal and Chloroform (ii) Methanol and Cyclopentane 6. Cyclohexane and ethanol at a particular temperature have vapour pressure of 280 mm and 168 mm of Hg respectively. If these two solutions having mole fraction value of cyclohexane equal to 0.32 are mixed and the mixture has a total vapour pressure of 376 nm, will the mixture be an ideal solution? AZEOTROPIC MIXTURE Some liquids on mixing from azeotropes which are binary mixtures having same composition liquid and vapours phase and boil at a constant temperature. In such cases, it is not possible to separate the components by fractional distillation. There are two types of azeotropes called as minimum boiling azeotrope and maximum boiling azeotrope, respectively. Solutions of ethanol and water show such a large deviation from Raoult' s law that there is a maximum in the vapour pressure curve and hence a minimum in the boiling point. Ethanol-water mixtures (obtained by fermentation of sugars) are rich in water. Fractional distillation is able to concentrate the alcohol to at best, the azeotropic composition of approximately 95% by volume of ethanol. Once this composition has been achieved, the liquid and vapour have the composition, and no additional fractional occurs. Other methods of separation have to be used for preparing 100% C2H50H. There are also solutions that show large negative deviation from ideality and, therefore, have a minimum in their vapour pressure curves. This leads to a maximum on the boiling point diagram. HN03 and 1-120 form examples of this class of the azeotrope. This azeotrope has the approximate composition, 68% nitric acid and 32% water by mass, with a boiling point of 393K. COLLIGATIVE PROPERTIES The properties that depends on the concentration of solute molecule or ions in solution but not on the chemical identity of the solute. For example, addition of ethylene glycol to water lowers the freezing point of water below OOC. The magnitude of freezing point lowering is directly proportional to the number of solute molecules added to a quantity of solvent. If 0.01 mole of ethylene glycol is added to 1 kg of water, the freezing point is lowered to —0.0190C while on adding 0.020 mole of ethylene glycol to 1 kg of water, the freezing point is to —0.0380 C. The various colligative properties are (i) Relative lowering of vapour pressure (ii) osmotic pressure (iii) elevation of boiling point (iv) depression of freezing point
  9. RELATIVE LOWERING OF VAPOUR PRESSURE The addition of a non — volatile solute to a solvent at a given temperature and pressure results in the lowering of the vapour pressure of the solvent. If a solution is formed by dissolving n moles of a non volatile solute in N moles of a volatile solvent. Then Mole fraction of solvent, Xl = n Mole fraction of solute = Since the solute is non — volatile, so it would have negligible vapour pressure and thus the vapour pressure of the solution, is inversely the vapour pressure of the solvent. According to Raoult' s law Since Xl + = 1 SO Xl Now PI = (1 Or in equation (ii) -Plis the vapour pressure lowering and (PIO -PI)/PIO is the relative vapour pressure lowering. From equation (ii) it is clear that the relative vapour pressure lowering is proportional to the mole fraction of the non — volatile solute in solution. It is independent of the nature of the solute. The relative vapour pressure lowering is, therefore a colligative property. n The mole fraction, = For a dilute solution, the number of moles of the solute (n) can be neglected as compared to the number of moles of the solvent (N). Hence n WI W2Ml N Ml M2 WIM2 Where WI is the amount of the solute dissolved in WI amount of solvent, Ml is the molar mass of the solvent and M2 is the molar mass of the solute. WIM2 Vapour pressure in terms of molality of the solution: m Ml where m is the molality 1000gm OSMOSIS AND OSMOTIC PRESSURE There is a natural tendency of solutes in a solution to diffuse from a higher concentration to a lower concentration so as to bring about a uniform distribution throughout. Certain membranes allow solvent molecules to pass through them but not solute molecules, particularly not those of large molecular weight. Such a membrane is called semipermeable and might be an animal bladder, a vegetable tissues or a piece of cellophone. Osmosis is the phenomenon of solvent flow thorough a semi permeable membrane to equalize the solute concentrations on both sides of the membrane.
  10. Osmotic pressure is a colligatvie property of a solution equal to the pressure that, when applied to the solution just stop osmisos. It is denoted by lt. The osmotic pressure Tt of a solution is related to the molar concentration of solute M. Tt = MRT , Where R is the gas constant and T is absolute temperature. Van' t Hoff equation for dilute solution is similar to ideal gas equation, = nRT Where It = osmotic pressure R = gas constant Isotonic solutions A pair of solutions having same osmotic pressure is called isotonic solution. When two Solutions having the same osmotic pressure are put into communication with each other through a semi permeable membrane, there will be net transference of solvent from one solution to the other, but there is a dynamic equilibrium between two solutions. According to Van' t Hoff equation, it is evident that isotonic solutions must have the same molar concentration, if both are at same temperature. For isotonic solution, 2 r12 Or ml VI m2V2 Reverse osmosis and water purification The direction of osmotic can be reversed if a pressure larger than the osmotic pressure is applied to the solutions side. That is, now the pure solvent flows out of the solution through the semipermeable membrane (SPM). This phenomenon is called reverse osmosis and is of great practical utility. Reverse osmosis is used in the desalination of sea water. A schematic set in the desalination of sea water. A schematic set up for the process is shown in figure. When pressure more than osmotic pressure is applied pure water is squeezed out of the sea water through the membrane. A variety of polymer membrane are available for this purpose. Fresh Water Piston Pressurr > 17 Salt Water Water outlet SPM The pressure required for reverse osmosis are quite high and a workable porous membrane is a film of cellulose acetate placed over a suitable support. Cellulose acetate is permeable to water but impermeable to impurities and ions present in sea water. These days many countries use desalination plants to meet their water requirement.
  11. BOILING POINT ELEVATION BY NON VOLATILE SOLUTE The boiling point Tb of a liquid is the temperature at which its pressure is equal to the atmospheric pressure. When a non — volatile solute is added to a liquid, the vapour pressure of the liquid is decreased. Hence it must be heated to a higher temperature in order that its vapour pressure becomes equal to that of the atmospheric pressure. This means that the addition of a non volatile solute to a liquid raises its boiling point. If Tb0 is the boiling point of the solvent and Tb is the boiling point of the solution, then Tb -T? = A Tb (Elevation of boiling point) A Tb u molality (m) .0 ATM o B D X Y/ Vapour pressure curve 1
  12. 7. 8. 9. WORKSHEET : 4 The pressure of the water vapour of a solution containing a non volatile solute is 2% below that of the vapour of pure water. What is the molality of the solution in mol/g ? Calculate the molecular weight of cellulose acetate if its 0.5% (wt./vol) solution in acetone (sp. gr. = 0.9) shows an osmotic rise of 23 mm against pure acetone at 270C. Calculate the osmotic pressure of sucrose solution containing 1.75 gms in 150 ml of solution at 1 -pc. 10. Osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose that will be isotonic with blood. What is the weight/volume ? 11. Y g of non - volatile organic substance of molecular mass M is dissolved in 250 g benzene. Molal elevation constant of benzene is Kb. What is Elevation in boiling point ? 12. Molal elevation constant (Kb) values of following alcohols are in the order: CH3CH2CH2CH20H > (CH3 ) 2 _ CH20H > (CH3 C _ OH Explain in brief 13. An aqueous solution of glucose boils at 100.01 oc. The molal elevation constant for water is 0.5 K Kg/mol-l Kg. What is the number of glucose molecule in the solution containing 100 gm of water? (Mglucose = 180). 14. An aqueous solution containing 288 gm of a non— volatile compound having the stochiometric composition CxH2xOx in 90 gm water boils at 101.240C at 1.00 atmospheric pressure. What is the molecular formula? Kb(H20) = 0.512 K mol-I kg, Tb (1-120) = loooc. DEPRESSION OF FREEZING POINT BY A NON - VOLATILE SOLUTE The temperature at which solid and liquid states of a substance have the same vapour pressure is called freezing point. If a non volatile solute is mixed with a pure solvent, the freezing point of pure solvent always greater than the impure one. Tf0 -Tf = ATf where TP is the freezing point of the pure solvent and Tf is the freezing point of a non — volatile solute containing solvent and ATf is the depression in freezing point. ATf molality or ATf = K Frozen solvent 1 c Freezing point depression TfO Temperature (K) -----...................................> Kf = molal freezing point depression constant of the solvent (cryoscopic constant) m = molality of the solution Molal freezing point depression constant of the solvent or cryoscopic constant is defined as the depression in freezing point which may be theoretically produced by dissolving 1 mole of any Non volatile non electrolyte solute in 1000 gm of solvent. 1000 X W or ATf ml X W Where ml = molecular weight of solute
  13. 15. 16. 17. 18. 19. 20. WORKSHEET : 5 The amount of ice that will separate on cooling a solution containing 50g of ethylene glycol in 200g water to — 9.30C is : [Kf= 1.86 K molality- 1] The freezing point of aqueous solution contains 5% by mass urea, 1.0% by mass KCI and 10% by mass of glucose is: ( KfH o = 1.86 K molality 1) 1. If 34.2 g of cane sugar The molal freezing point constant for water is 1.86 K.molarlty (C12H22011) are dissolved in 1000g of water, the solution will freeze at An aqueous solution containing 0.25 g of a solute dissolved in 20g of water freezes at 0.40C. Calculate the molar mass of the solute. 14 = 1.86 K kg mol-I. If for a particular solution, ATf = 0.730, 14 = 5.028 K Kg/mole, what is the molality of the solution. 1 kg of an aqueous solution of Sucrose is cooled and maintained at — 40C. How much ice will be separated out if the molality of the solution is 0.75? Kf (H20) = 1.86 Kg mol- 1K. ABNORMAL MOLECULAR WEIGHT AND VAN'T HOFF FACTOR Since colligative properties depend upon the number of particles of the solute, in some cases where the solute associates or dissociates in solution, abnormal results for molecular masses are obtained. Van't Hoff Factor: Van't Hoff, in order to account for all abnormal cases introduced a factor i known as the Van't Hoff factor, such that Observed colligative property (actual) Theortical molecular mass Theoretical colligative property (expected) observed molecular mass No. of molecules actually present No. of molecules expected to be present i > 1 for dissociation of solute in solution i < 1 for association of solute. i = 1 when solute neither associate non dissociate. WORKSHEET : 6 21. The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is 22. 20 g of a binary electrolyte (mol.wt. = 100) are dissolved in 500 g of water. The freezing of the solution is — 0.740C, Kf= 1.86 K. molality 1. The degree of ionisation of the electrolyte is 23. Acetic acid (CH3COOH) associates in benzene to form double molecules. 1.65 g of acetic acid when dissolved in 100g of benzene raised the boiling point by 0.360C. Calculate the Van't Hoff Factor and the degree of association of acetic acid in benzene (Molal elevation constant of benzene is 2.57). 24. A milimolar solution of potassium ferricyanide is 70% dissociated at 270C. Find out the osmotic pressure of the solution. 25. A storage battery contains a solution of H2S04 38% by weight. At this concentration, Van' t Hoff factor is 2.50. At what temperature will the battery contents freeze? (14 = 1.860 mol-I kg)
  14. 26. The molar heat of vaporization of water at 1000C is 40.585 kJ/mol. At what temperature will a solution containing 5.60 g of per 1000 g of water boil? [Assuming the degree of ionization of salt to be 1] 27. Acetic acid (CH3COOH) associates in benzene to form double molecules. 1.65 g of acetic acid when dissolved in 100g of benzene raised the boiling point by 0.360C. Calculate the Van't Hoff Factor and the degree of association of acetic acid in benzene (Molal elevation constant of benzene is 2.57). 28. A 0.5 percent aqueous solution of Potassium Chloride was found to freeze at 0.240C. Calculate the Van't Hoff factor and degree of dissociation of the solute at this concentration (14 for water is 1.86). 29. 2.0 gm of benzoic acid dissolved in 25 gm of benzene shows a depression of freezing point equal to 1.62 K. Molal depression constant (KD of benzene is 4.9 kg mol-I K. What is the percentage association of the acid? 30. A weak electrolyte, AB is 5% dissociated in aqueous solution of AB having molality 0.1 m. Freezing point of the solution will be (14 for water 1.86 oc /molal) 31. The freezing point of a solution of acetic acid (mole fraction is 0.02) in benzene 277.4 K. Acetic acid exists partly as a dimer 2A A 2. calculate equilibrium constant for dimerisation. Freezing point of benzene is 278.4 k and (Kf for benzene is 5).
  15. ASSIGNMENT : 1 1. The density of 2.0 M solution of acetic acid in water is 1.02 g ml-I. Calculate the mole fraction of acetic acid. 2. Liquid A and B form an ideal solution obeying Raoult' s law. At 500C, the total vapour pressure of a solution containing 1 mole of A and 2 mole of B is 300 torr. When 1 more mole of A is added to the solution, the vapour pressure increases to 400 torr. Calculate the vapour pressure of pure components. 3. An aqueous solution of non — volatile solute boils at 100.170C. At what temperature would it 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. freeze? For water, Kb = 0.52 K Kg mol-I and 14 = 1.86 K Kg mol-I The complex compound is 45% dissociated in 0.1M aqueous solution of the complex at 270C. What would be the osmotic pressure of the solution? The formula for low molecular weight starch is (C6H1005)n where n averages 2.00 x 102. When 0.798 gm of starch is dissolved in 100 ml of water solution. What is the osmotic pressure at 250C? At IOOC, the osmotic pressure of urea solution is 500 mm. The solution is diluted and the temperature is raised to 250C, when the osmotic pressure is found to be 105.3 mm. Determine the extent of dilution. The vapour pressure of an aqueous solution of glucose is 750 mm of Hg at 373 K. Calculate the molality and mole fraction of solute. How much ethyl alcohol must be added to 1.00 litre of water so that the solution will freeze at 140F? (14 for water = 1.86 oc kg/mol) 1000 gm of 1 m sucrose solution in water is cooled to -3.5340C. What weight of ice would be separated out at this temperature? K'fH20 = 1.86 K mol-Ikg A 1% (wt/vol) KCI solution is ionized to the extent of 82%. What would be its osmotic pressure at 180C? Calculate the osmotic pressure of 20% (wt/vol) anhydrous solution at OOC assuming 100% ionization. 17.4% (wt/vol) K2S04 solution at 27 oc is isotonic to 5.85% (wt/vol) NaCl solution at 270C. If NaCl is 100% ionized, what is % ionization of K2S04 in aqueous solution. An aqueous solution of liquid ' X' (mol. Wt. 56) 28% by weight has a vapour pressure 150 mm. Find the vapour pressure of ' X' if vapour pressure of water is 155 mm of Hg. Calculate the osmotic pressure at 1 7 oc of an aqueous solution containing 1.75 gm of sucrose per 150 ml solution. A mixture of two immiscible liquids nitrobenzene and water boiling at 990C has a partial vapour pressure of water 733 mm and that of nitrobenzene 27 mm. Calculate the ratio of weights of nitrobenzene to the water in distillate.
  16. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. ASSIGNMENT : 2 One litre aqueous solution of sucrose (molar mass = 342) weighting 1015 gm is found to record an osmotic pressure of 4.82 atm at 293 K. What is the molality of the sucrose solution? (R = 0.0821 atm mol-I k-l) An aqueous solution freezes at 272.4 K, while pure water freezes at 273.0 K. Determine (i) molality of the solution (ii) boiling point of the solution Given 1.86 k Kg mol-I, Kb = 0.512 K kg mol-I 2 gm of C6H5COOH dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molar depression constant for benzene is 4.9 K Kg mol-I. What is the percentage association of acid if it exists as dimer in solution? Find the molality of a solution containing a non — volatile solute if the vapour pressure is 2% below the vapour pressure of pure water. An aqueous solution of 2% (wt/wt) non — volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute? An aqueous solution of glucose containing 12 gm in 100 gm of water was found to boil at 100.340C. Calculate Kb for water in K mol-ikg . A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The freezing point of pure water is 273.15 K. What is the depression in freezing point of a solution of non — electrolyte if elevation in boiling point is 0.13 K, Kb = 0.52K mol-ikg ; Kf = 1.86K mol-ikg . Calculate O.P of a solution obtained by mixing 100 ml of 3.4% solution (wt/vol) of urea and 100 ml of 1.6% solution (wt/vol) of cane sugar at 200C. A beaker containing 20 gm sugar in 100 gm water and another containing 10 gm sugar in 10 gm water are placed under a bell — jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to other? What will be osmotic pressure of 0.1 M monobasic acid? Its pH is 2 at 250C. Vapour pressure of a saturated solution of a sparingly soluble salt A2B2 is 31.8 mm of Hg at 400C. If vapour pressure of pure water is 31.9 mm of Hg at 400C. Calculate Ksp of A2B2 at 400C. A complex is represented as CoC13.XNH3. It is 0.1 molal solution in aqueous solution shows ATf = 0.5580C Kf for 1420 is 1.86 K molality 1. Assuming 100% ionization of complex and co-ordination number of Co as six, calculate formula of complex. A storage battery contains a solution of H2S04 38% by weight. At this concentration, Van' t Hoff factor is 2.50. At what temperature will the battery contents freeze? Kf = 1.860Ckg/mol 1 gm of mono basic acid in 100 gm of water lowers the freezing point by 0.168 oc. If 0.2 gm of same acid require 15.1 ml of N/ 10 alkali for complete neutralization, calculate degree of dissociation of acid Kf for 1-120 is 1.86 K mol-ikg . 0.025 M solution of monobasic acid had a freezing point of -0.060C. Calculate Ka for the acid. Kf for 1-120=1.86 oc molality l. Assume molality equal to molarity? Phenol associate in C6H6 to form dimer. A solution of 2 gm of phenol in 100 gm C6H6 has its f.pt lowered by 0.72 K. Kf for C6H6 is 5.12 K mol-I kg. Calculate degree of association of phenol. What approximate proportion by volume of water (d - 1 gm/mL) and ethylene glycol (d = 1.2 gm/mL) must be mixed to ensure protection of an automobile radiator to cooling ICC. H20 = 1.86 K mol-Ikg
  17. 2. 3. 4. 5. 6. 7. 8. 9. PREVIOUS YEARS QUESTIONS Write the colligative property which is used to find the molecular mass of macromoleucles. In non-ideal solution, what type of deviation shows the formation of minimum boiling azerotropes ? Calculate the boiling point of solution when 2 g of Na2S04 (M = 142 g mol-I) was dissolved in 50 g of water , assuming Na2S04 undergoes complete ionization. (Kb for water = 0.52 K kg mol-I) Vapour pressure of water at 200C is 17.5 mm Hg. Calculate the vapour pressure of water at 200C when 15 g of glucose (Molar mass = 180 g mol-I) dissolved in 150 g of water. (a) Define the following terms : (i) Molairty (ii) Molal elevation constant (Kb) (b) A solution containing 15 g urea (molar mass = 60 g mol-I) per litre of solution in water has the same osmotic pressure (isotonic) as solution of glucose (molar mass = 180 g mol-I) in water. Calculate the mass of glucose present in one litre of its solution. (i) State Raoult' s law for a solution containing components. How does Raoult' s law become a special case of Henry' s law ? (ii) 1.00 g of non-electrolyte solute when dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (Kf for benzene = 5.12 kg mol-I) Explain the Henry' s law about dissolution of a gas in a liquid. Define the following terms : (i) Mole fraction (ii) Isotonic solution (iii) Van' t Hoff factor (iv) Ideal solution. 15.0 g of unknown molecular material is dissolved in 450 g of water. The resulting solution freezes -0.340C. What is the molar mass of the material ? (Kf for water = 1.86k mol-I) A solution glycerol (C3H803) in water was prepared by dissolving some glycerol in 500 g of water. The solution has a boiling point of 100.420C, what mass fo glycerol was dissolved to make this solution ? (Kb for water = 0.512 kg mol-I) 10. Define Raoult' s law in its general form in reference to solution. 11. A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 250C. Assumign the gene fragment is non-electrolyte, determine its molar mass. 12. A solution is prepared by dissolving 43 gm of naphthalene in 117 gm of benzene. Calculate the mole fractions of the two components of the solution. 13. 1.2 gm of a non-volatile substance was dissolved in 100 gm of acetone at 200C. The vapour pressure of the solution was found to be 182.5 torr. Calculate the molar mass of the substance (vapour pressure of acetone at 200C is 185.0 torr) 14. A solution containing 8.6 gm per dm2 of urea (molar mass - 60 gm/mole) was found to be isotonic with a 5 percent solution of an organic non volatile solute. Calculate the molar mass of the latter. 15. Acetic acid associate in benzene to form dimer 1.65 gm of acetic acid when dissolved in 100 gm of benzene raised the boiling point by 0.360C. Calculate the Van' t Hoff factor and 2.57 K Kg mol-I) 16. At 250C, the osmotic solutes in the blood is the freezing point of blood. the degree of association of acetic acid in benzene. (Kb pressure of human blood due to the presence of various 7.65 atm. Assuming that molarity equals molality, calculate 1.86 K Kg mol-I.

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