SOLUTIONS

1. Bass is a mixture of _____________ and _________________. 2. German silver is a mixture of _________, __________ and _________. 3. Bronze is a mixture of ___________ and __________. 4. Blood is a mixture of ______________, __________ , _________ and ___________. 5. Milk is a mixture of _______, __________, ____________, _____________, ___________ and _____. 6. 1 ppm = _____________ mg/L 7. ____________ ppm of fluoride ions prevents tooth decay. 8.  ____________ppm of fluoride ions causes mottled teeth. 9. NaF is used as a _____________ poison. 10. Almost all process in the body occurs in some kind of liquid solution. ___________ (TRUE / FALSE).

11. Solutions are ___________ mixtures of two or more than two components. 12. The component which is present in large quantities is called __________. 13. The component which is present in small quantities is called___________.

14. The solution which contains two components , one as solute & the other as solvent is termed as ______

TYPE OF SOLUTION SOLUTE SOLVENT EXAMPLE GASEOUS SOLUTION SOLID GAS Dust particles in the air.  LIQUID GAS Chloroform mixed with nitrogen gas GAS GAS Air in our atmosphere LIQUID SOLUTION ? LIQUID ? ? LIQUID ? ? LIQUID ? SOLID SOLUTION ? SOLID ? ? SOLID ? ? SOLID ? 15. The type of solution depends on the state of the matter of the solvent like Solid Solution, Liquid Solution or Gaseous Solution. Fill in the table below by various types of common solutions you see in your environment.

16. Quantity of solute dissolved in unit volume or weight of the solvent is called as __________________.

17. Some common methods of calculating concentration of solution are enlisted below, give suitable formula to determine the value of the concentration in the given terms. (a) Mass percentage (w/w) : (b) Volume percentage (v/v) : (c) Mass / volume percentage (w/v) : (d) Parts per million (ppm) : (e) Mole fraction (x) : (f) Morality (M) : (g) Molality (m) :

18. Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2 by mass.

19. Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.

20. Calculate molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of benzene.

21. Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

22. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride

23. Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

24. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.

25. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1

26. Solubility of a __________________ is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature.

27. Factors on which solubility of a solute depends are : (a) ____________________________ (b) ____________________________ (c ) ____________________________

28. Solution in which no more solute can be added is called _______________________.

29. Solution in which some more solute can be added is called _______________________.

30. Solution in which more solute is dissolved by warming the solution above room temperature is called as _______________________.

31.  In a saturated solution the solute and solvent remain in dynamic equilibrium but when temperature is raised, solubility of the solute increases, whereas at room temperature more solute is added into the saturated solution then solute particles settle down at the bottom of the vessel. Explain that the above phenomena obeys Le-Chatelier’s Principle that – ‘if the equilibrium of a system is disturbed by a change in one or more of the determining factors (as temperature, pressure, or concentration) the system tends to adjust itself to a new equilibrium by counteracting as far as possible the effect of the change’

32.  Le- Chatelier’s Principle is also termed as Law of ____________________________________.

33.  State Dalton’s law of partial pressure.

34.  Henry’s law :  The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution. (i) Why are carbonated drinks bottled at high pressure ? (ii) What is bends syndrome? (iii)  Why tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen). (iv) Why mountain climbers suffer from anoxia?

35. How is mole fraction of a gas dissolved in a liquid related to its partial pressure. What is Henry’s law constant?

36.  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 76.48 kbar

37.  H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.

38.  Henry’s law constant for CO2 in water is 1.67x108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K

39. What do you mean by the term Vapour Pressure ?

40. Ethanol is ___________(more/less) volatile than water. At a given temperature the Vapour pressure of Ethanol will be _________________ (more/less) than water.

41. Vaporization or evaporation is a _______________ (surface/bulk) phenomenon. If a solution contains non volatile impurity as solute then at the surface lesser number of solvent will be there as compared to the pure solvent. As a result at normal boiling point lesser amount of vapor will be formed, but to produce as much vapor as much it can resist atmospheric pressure and the solution can be called to be boiling. But due to addition of Non – volatile solute some of the surface is occupied by such substance which can’t produce vapor and hence more temperature is required to bring the solution to boiling. There fore addition of non volatile solute elevates the boiling point of the solution.

42. Raoult’s law states that For a solution of volatile liquids, the partial pressure of each component of the solution is directly proportional to its mole fraction. Express the statement in the form of mathematical expression.

43. Do you think Raoult’s law is a special case of Henry’s law? If yes, why?

44. Interpret the information depicted in the graph below and jot down your observations.

45.  Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at 298 K and, (ii) mole fractions of each component in vapour phase.

46.  Distinguish between Ideal Solution and Non Ideal Solution. IDEAL SOLUTION NON IDEAL SOLUTION The solutions which obey Raoult’s Law at every range of concentration and at all temperatures are Ideal Solutions. We can obtain ideal solutions by mixing two ideal components that are, solute and a solvent having similar molecular size and structure. Characteristics of Ideal Solutions Ideal Solutions generally have characteristics as follows: They follow Raoult’s Law. This implies that the partial pressure of components A and B in a solution will be PA = PA0 xa and PB = PB0 Xb .  PA0 and PB0 are respective vapour pressure in pure form. On the other hand, XA and XB are respective mole fractions of components A and B The enthalpy of mixing of two components should be zero, that is, Δmix H = 0. This signifies that no heat is released or absorbed during mixing of two pure components to form an ideal solution The volume of the mixing is equal to zero that is, Δmix V = 0. This means that total volume of solution is exactly the same as the sum of the volume of solute and solution. Adding further, it also signifies that there will be contraction or expansion of the volume while the mixing of two components is taking place. The solute-solute interaction and solvent-solvent interaction is almost similar to the solute-solvent interaction. For Example, consider two liquids A and B, and mix them. The formed solution will experience several intermolecular forces of attractions inside it, which will be: A – A intermolecular forces of attraction B – B intermolecular forces of attraction A – B intermolecular forces of attraction The solution is said to be an ideal solution, only when the intermolecular forces of attraction between A – A, B – B and A – B are nearly equal. Examples of Ideal Solutions n-hexane and n-heptane Bromoethane and Chloroethane Benzene and Toluene CCl4 and SiCl4 Chlorobenzene and Bromobenzene Ethyl Bromide and Ethyl Iodide n-Butyl Chloride and n-Butyl Bromide The solutions which don’t obey Raoult’s law at every range of concentration and at all temperatures are Non-Ideal Solutions. Non-ideal solutions deviate from ideal solutions and are also known as Non-Ideal Solutions. Characteristics of Non-ideal Solutions Non-ideal solutions depict characteristics as follows: The solute-solute and solvent-solvent interaction is different from that of solute-solvent interaction The enthalpy of mixing that is, Δmix H ≠ 0, which means that heat might have released if enthalpy of mixing is negative  (Δmix H < 0) or the heat might have observed if enthalpy of mixing is positive (Δmix H > 0) The volume of mixing that is,  Δmix V ≠ 0, which depicts that there will be some expansion or contraction in the dissolution of liquids Non ideal solutions are further of two types : (i) Positively deviated non-ideal solution (ii) Negatively deviated non-ideal solution.  i) Positive Deviation from Raoult’s Law Positive Deviation from Raoult’s Law occurs when the vapour pressure of the component is greater than what is expected in Raoult’s Law. For Example, consider two components A and B to form non-ideal solutions. Let the vapour pressure, pure vapour pressure and mole fraction of component A be PA , PA0 and xA respectively and that of component B be PB , PB0 and xB respectively. These liquids will show positive deviation when Raoult’s Law when: PA > PA0 xA and PB > P0B xB, as the total vapour pressure (PA0 xA + P0B xB) is greater than what it should be according to Raoult’s Law. The solute-solvent forces of attraction is weaker than solute-solute and solvent-solvent interaction that is, A – B < A – A or B – B The enthalpy of mixing is positive that is, Δmix H > 0 because the heat absorbed to form new molecular interaction is less than the heat released on breaking of original molecular interaction The volume of mixing is positive that is, Δmix V > 0 as the volume expands on the dissolution of components A and B Examples of Positive Deviation Following are examples of solutions showing positive deviation from Raoult’s Law: Acetone and Carbon disulphide Acetone and Benzene Carbon Tetrachloride and Toluene or Chloroform Methyl Alcohol and Water Acetone and Ethanol Ethanol and Water Negative Deviation from Raoult’s Law Negative Deviation occurs when the total vapour pressure is less than what it should be according to Raoult’s Law. Considering the same A and B components to form a non-ideal solution, it will show negative deviation from Raoult’s Law only when: PA < PA0 xA and PB < P0B xB as the total vapour pressure (PA0 xA + P0B xB) is less than what it should be with respect to Raoult’s Law The solute-solvent interaction is stronger than solute-solute and solvent-solvent interaction that is, A – B > A – A or B – B The enthalpy of mixing is negative that is, Δmix H < 0 because more heat is released when new molecular interactions are formed The volume of mixing is negative that is,  Δmix V < 0 as the volume decreases on the dissolution of components A and B A Solved Question for You Q: Give some examples of solutions showing negative deviation from Raoult’s Law. Solution: Following are examples of solutions showing negative deviation from Raoult’s Law Chloroform and Benzene Chloroform and Diether Acetone and Aniline Nitric Acid ( HNO3) and water Acetic Acid and pyridine Hydrochloric Acid ( HCl) and water

47. What are Azeotropes? Azeotropes are defined as a mixture of two liquids which has a constant composition in liquid and vapour phase at all temperatures. Azeotropes can’t be separated by fractional distillation, as the composition of vapour phase remains same after boiling. Because of uniform composition azeotropes are also known as Constant Boiling Mixtures. There are two types of Azeotropes: Maximum Boiling Azeotrope Minimum Boiling Azeotrope  Maximum Boiling Azeotrope Maximum Boiling Azeotrope is formed when we mix two non-ideal solutions at some specific composition, showing large negative deviation from Raoult’s law. Examples: Nitric Acid (HNO3) (68%) and water (32%) form maximum boiling azeotrope at boiling temperature of 393.5 K Hydrochloric Acid (HCl) (20.24%) and water form maximum boiling azeotrope at boiling temperature of 373 K Minimum Boiling Azeotrope:  Minimum Boiling Azeotrope is formed when we mix two non-ideal solutions at some specific composition, which shows large positive deviation from Raoult’s Law. Example: Ethanol ( 95.5 %) and water ( 4.5 %) form minimum boiling azeotrope at boiling temperature of 351.5 K

48. The vapor pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapor pressure is 600 mm Hg. Also find the composition of the vapor phase.

49. What do you understand by the term Colligative Properties?

50. Establish relation between vapor pressure of the solution, mole fraction and vapour pressure of the solvent. Show that relative lowering of vapor pressure is equal to the mole fraction of the solute.

51. The vapor pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol-1). Vapor pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?

52. Why on dissolving non volatile solute in volatile solvent, the boiling point of the solution elevates?

53. Interpret the graph to explain how elevation in boiling point takes place on dissolving non volatile solute in the solution.

54. 18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? Kb for water is 0.52 K kg mol-1

55. The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is 2.53 K kg mol–1

56. Why do you think dissolution of a non volatile solute in a volatile solvent decreases the freezing point of the solution ?

57. Interpret the graph to explain the phenomena of depression in freezing point on dissolution of solute.

58.  Define the following terms and give their units? (i) Ebullioscopic Constant  / molal boiling point elevation constant: (ii) Cryoscopic Constant / molal freezing point depression constant:

59.  45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution.

60. 1.00 g of a nonelectrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol–1. Find the molar mass of the solute.

70.  Define the following terms : (i) Osmosis : (ii) Reverse osmosis : (iii) Osmotic pressure : (iv) Semi permeable membrane : (v) Hypertonic solution : (vi) Hypotonic solution : (vi) Isotonic solution :

71. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

72. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

73. Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol-1.

74. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

75. Why the conditions of abnormal molecular mass arise while calculating the molecular masses by using colligative properties?

76. Give an example to show that association of molecules in the solutions results to abnormal molecular mass obtained by calculations using colligative properties.

77.  Give an example to show that dissociation of molecules in the solutions results to abnormal molecular mass obtained by calculations using colligative properties.

78. How the relations of colligative properties are modified using van’t Hoff Factor to rectify the differences between observed molecular mass and calculated molecular mass?

79. What do you understand by the term ‘van’t Hoff factor’. How ‘van’t Hoff factor helps to find calculated molecular mass correctly ?

80. 2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol–1. What is the percentage association of acid if it forms a dimer in solution?

81.  0.6 mL of acetic acid (CH3COOH), having density 1.06 g mL–1, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van’t Hoff factor and the dissociation constant of acid.