Class 12 Solutions ppt for the best preparation

Solutions ppt

1. CLASS 12, CHEMISTRY CHAPTER -2, SOLUTIONS

2. Bronze = Cu + Sn If the concentration of Fluoride ions 1 ppm in water, prevents tooth decay, While if concentration of Fluoride ions 1.5 ppm cause the tooth to become mottled. High concentration of fluoride ions can be poisonous. Ex: Sodium fluoride is used as rat poison. Glucose in water Brass = Cu + Zn German Silver = Cu + Zn + Ni SOLUTIONS IN OUR DAY TO DAY LIFE

3. Solutions are homogeneous mixtures of two or more than two pure substance whose composition may be altered within certain limits. In homogeneous mixture its composition and properties are uniform throughout the mixture. Definition of Solution The solution may be homogenous in nature, but it retained properties of its constituents. The solution of two components is referred to as binary solution. (Water + Glucose)

4. Components of binary solution: Solution = Solvent + Solute Solvent : It is the component of the solution which is present in a relatively large proportion in the solution whose physical state is same as that of the resulting solution. Solute : It is the second component of a binary solution which is in present relatively small proportion.

5. Types of Solutions In a binary solution components can be solid, liquid or gas. There are three types of the Solutions: Gaseous Solution Liquid Solutions Solid Solutions

6. Expressing concentration of Solutions: The Concentration of the solution refers to the amount of solute present in the given quantity of solution or solvent. 1. Mass percentage (w/w) : It is defined as the number of parts by mass of solute per hundred parts by mass of solution. Mass % of a component =   If W B be the mass of solute (B) and W A be the mass of solvent (A), then Mass % of B =  

7. Example : Calculate the mass percentage of glucose in the solution if 10 g of glucose is dissolved in 90 g of water. Solution : Mass of glucose/solute (W B ) = 10 g Mass of water/solvent (W A ) = 90 g mass % of glucose =   =   =   = 10 %

8. 2. Volume percentage (V/V): It is defined as the number of part by volume of solute per hundred parts by volume of solution. Volume % of a components =   If V B be the volume of solute (B) and V A be the volume of solvent (A), Then Volume % of B =   Note : 35 % (v/v) solution of ethylene glycol, an antifreeze, is used in cars for cooling the engine. At his concentration the antifreeze lowers the freezing point of water to 255.4 K (-17.6 ).  

9. 3. Mass by volume percentage (w/v) : It is the mass of solute dissolved in 100 mL of the solution. 4. Parts per million (ppm) : It is defined as the number of parts by mass of solute per million parts by mass of solution. ppm =  

10. 5. Mole Fraction (x) : Mole fraction is defined as the ratio of number of moles of one component to the total number of moles of all the components present in the solution. Mole fraction of a component =   Let us consider a solution that contains the components A and B. If the number of moles of A and B are and respectively, then   Mole fraction of A ( ) =   We know Number of moles =   if the mole fraction of is and mole fraction of B is , then     The sum of mole fractions of all the components is in a solution is always unity.

11. Example : Calculate the mole fraction of ethylene glycol (C 2 H 6 O 2 ) in a solution containing 20% of C 2 H 6 O 2 by mass. Solution : Assume that total mass of solution = 100 g Mass of = 80 g   Mass of = 20 g   Molar mass of =   Molar mass of =   Moles of =   =   = 4.444 mol Moles of =   =   = 0.322 mol =   =   = 0.068 =   =   = 0.932 As we know that the sum of mole fractions of all the components is in a solution is always unity, so mole fraction of water we can also be calculate as : 1 - 0.068 = 0.932

12. 6. Molarity (M) : It is defined as the number of gram mole of the solute present in one litter (or one cubic decimetre) of the solution. Molarity =   As we know Mole =   Molarity =   =   Here –      

13. Example : Calculate the molarity of a solution containing 20 g of NaOH in 500 ml solution. Solution : Mass of solute ( NaOH ) = 20 g Molar mass of NaOH = 23 + 16 + 1 = 40 g   Mole of NaOH =   = 0.5 mol Molarity of solution =   =   = 1 mol   Note : Molarity is a function of temperature, Because Volume depends on Temperature. If one mole of solute present in one litter of solution then solution known as molar solution.

14. 7. Molality (m): It is defined as the number of moles of the solute present in one kilogram (1000 g) of solvent. Molality =   As we know Mole =   Molality =   =   Here - = mass solute = molar mass of solute    

15. Example : Calculate the molality of a solution if 100 g of NaOH present in 1 kg of solution. Solution : Mole of solute =   Mass of Solute = 100 g = 2.5 mol Mass of solvent = Mass of solution mass of solute   = 1000 g 100 g   = 900 g Molality =   =   = 2.77 mol   Note : Molality does not depend on temperature. If one mole of solute present in one kilogram of solvent then solution known as molal solution .

16. Assessment : 1. Define the following terms a. Solution. b. Molal Solution. c. Mole fraction d. Molarity 2. When temperature increase, Molarity of a solution: (a)Increases (b) Decreases (c) remains same (d) first increases then decreases 3. Calculate the mass percentage of benzene (C 6 H 6 ) and carbon tetrachloride (CCl 4 ) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. 4. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. 5. Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO 3 ) 2 . 6H 2 O in 4.3 L of solution (M of Co= 59 u) (b) 30 mL of 0.5 M H 2 SO 4 diluted to 500 ml. NCERT Exercise : 2.2, 2.4, 2.5, 2.6, 2.7, 2.8 & 2.9 1.(a) A homogenous mixture of 2 or more components. 1.(b) When 1 mole solute is present in 1000 kg of solvent. 1.(c) Ratio of moles of any one component of a solution to the total moles of all the components. 1.(d) Total number of moles of solute present in 1000 ml of solution is called the molarity of the solution. 2. Molarity is inversely proportional to volume hence it decrease on increasing temperature. Answers : 4. 3.

17. Solubility of a Solid in a Liquid

18. How to Make a Supersaturated Solution The way to make a supersaturated solution is to add heat, but just a little heat won't do the job. You have to heat the water close to the boiling point. When the water gets this hot, the water molecules have more freedom to move around, and there is more space for solute molecules between them. You can keep stirring in salt, sugar or any other solute, and it will continue to dissolve, even though the saturation point has been reached. Take away the heat and let the solution gradually cool, and the solute will remain dissolved, at least for a time. This is, in essence, the supersaturated definition. A supersaturated solution is highly unstable, and strange things can happen. Strange, sudden crystallization in super saturated solutions.

19. QUESTION

20. ANSWER

21. Solubility of a Solid in a Liquid 1. Nature of solute and solvent : polar solutes dissolve in polar solvents and non polar solutes in nonpolar solvents. In general, a solute dissolves in a solvent if the intermolecular interactions are similar in the two or we may say like dissolves like. 27 FACTORS AFFECTING SOLUBILITY OF SOLIDS IN LIQUIDS

22. The above is in accordance with Le Chatelier’s Principle

23. Solubilities of Solids vs Temperature

24. 30 Solid solubility and Pressure Pressure does not have any significant effect on solubility of solids in liquids . It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

25. FACTORS AFFECTING THE SOLUBILITY OF GASES IN LIQUIDS 1. Effect of Pressure:

26. conclusion The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution ( Henry’s law ). low P low c high P high c 32

27. From the above table it is obvious that He gas is least soluble in water at 293 K as it has highest K H value.

28. APPLICATIONS OF HENRY’S LAW

29. 1. On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid? ( i ) Sugar crystals in cold water. (ii) Sugar crystals in hot water. (iii) Powdered sugar in cold water. (iv) Powdered sugar in hot water. 2. At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________. ( i ) less than the rate of crystallisation (ii) greater than the rate of crystallisation (iii) equal to the rate of crystallisation (iv) zero 3. A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes place when small amount of ‘A’ is added to the solution. The solution is _________. saturated supersaturated unsaturated concentrated 4. Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon ____________. Temperature Nature of solute Pressure Nature of solvent 5. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ____________. low temperature low atmospheric pressure high atmospheric pressure both low temperature and high atmospheric pressure TEST YOURSELF Answers: (iv) powdered sugar in hot water 2. (iii) equal to the rate of crystallisation 3. ( i ) saturated 4. (iii) Pressure 5. (ii) low atmospheric pressure

30. Q. Calculate the solubility of gaseous oxygen in water at a temperature of 293 K when the partial pressure exerted by O 2  is 1 bar. (Given: k H  for O 2  34840 bar.L.mol -1 ) Ans: As per Henry’s law, P = k H C Substituting, k H  = 34840 bar.L.mol -1  and P = 1 bar, the equation becomes C = 1/34840 mol.L -1  = 2.87 x 10 -5   mol /L Therefore, the solubility of oxygen in water under the given conditions is 2.87 x 10 -5  M.

31. Q. The value of k H  for carbon dioxide at a temperature of 293 K is 1.6 x 10 3  atm.L.mol -1 . At what partial pressure would the gas have a solubility (in water) of 2 x 10 -5  M? ANS: Substituting the given values k H  = 1.6 X 10 3  atm.L.mol -1  and C = 2 x 10 -5  M into the Henry’s law formula: P = k H x C = (1.6 X 10 3  atm.L.mol -1 ) x (2 X 10 -5  mol.L -1 ) = 0.032 atm. Q. If N 2 gas is bubbled through water at 293 K, how many millimoles of N 2 gas would dissolve in 1 litre of water. Assume that N 2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N 2 at 293 K is 76.48 kbar . ANS:

32. HOME WORK Examples 2.1, 2.2, 2.3, 2.4 In text 2.1 to 2.7

33. VAPOUR PRESSURE OF LIQUID - LIQUID SOLUTIONS Consider a binary solution of two volatile liquids A and B . When taken in a closed vessel, both the components would evaporate and eventually an equilibrium would be established between vapour phase and the liquid phase. Let the total vapour pressure at this stage be p total and p A and p B be the partial vapour pressures of the two components A and B respectively. These partial pressures are related to the mole fractions x A and x B of the two components A and B respectively.

34. Graph for a binary solution containing non volatile solutes

35. Video explaining Raoult’s law : https://www.youtube.com/watch?v=QRYT-QfPBK4

36. Draw the following graph to verify Raoult’s law:

37. Q

38. Henry’s law Raoult’s law P = P x Applicable to solute(gas) Applicable to solvent(liquid) The proportionality constant in Henry’s law is called Henry’s law constant. The Raoult’s law does not use a proportionality constant. In Henry’s law solute is a gas. In Raoult’s law solute is a volatile liquid or volatile or non volatile solid. Solvent is liquid Solvent is liquid Henry's law  takes care of what happens in the solution when you have gas over it. Means when you make changes in the vapour phase, what happens in solution. In other words, we study the change in solubility of a gas(solute) in the solution by changing partial pressure of the solute(gas) above the solution. Raoult’s law looks at what is happening over the solution when you mix a solute to a solvent that has a known vapor pressure when it's pure. In other words, we study the change in P of pure solvent by changing the concentration of solvent in the solution. Comparison between Henry's law and Raoult’s law: If we compare the equations for Raoult’s law and Henry’s law, it can be seen that the partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant K H differs from p 1 . Thus, Raoult’s law becomes a special case of Henry’s law in which K H becomes equal to p 1

39. Ideal solutions Non ideal solutions Obey Raoult’s Law over the entire range of concentration. Do not obey Raoult’s law over entire range of concentration but only for very dilute solutions. Intermolecular interaction forces between A-A and B-B are nearly equal to those between A-B Intermolecular interaction forces between A-A and B-B are not equal to those between A-B Examples: n-hexane and n- heptane Bromoethane and Chloroethane Benzene and Toluene. Chlorobenzene and bromobenzene Examples: 1. Ethanol + Cyclohexane. 2. Acetone + Carbon disulphide . 3. Acetone + Benzene. 4. Carbon tetrachloride + Chloroform.

40. There are two types of non ideal solutions: Which show positive deviation Which show negative deviation The total vapour pressure of such solutions is higher than that predicted by Raoult’s law. Means they show positive deviation from Raoult’s law. The total vapour pressure of such solutions is lower than that predicted by Raoult’s law. Means they show negative deviation from Raoult’s law. In these solutions, the molecular interaction between A-B is weaker than those between A-A and between B-B In these solutions, the molecular interaction between A-B is stronger than those between A-A and between B-B The molecules of A or B can escape easily in the vapour phase to increase the vapour pressure more than expected Examples: Ethanol + Acetone, CS 2 + Acetone The molecules of A or B can not escape easily in the vapour phase as compared to their pure states. Thus they decrease the vapour pressure than expected VP. Example: Phenol + Aniline, Chloroform + Acetone

41. A-A intermolecular forces are very strong in ethanol due to intermolecular H- bonding. Hexamer and pentamer are formed in pure ethanol. When acetone is added into ethanol, the acetone molecules hinders the formation of H- bonds in ethanol hence ethanol can not form hexamers or pentamers. So its molecules become free to escape out easily to increase the vapour pressure morethan expected.

42. BP water = 100 C BP Ethanol = 78.4 C BP Nitric acid = 83 C BP water + ethanol azeotrope = 78.2 C BP water + Nitric acid azeotrope = 120.4 C

43. Deviation from ‘ideality’ means deviation from the values corresponding to Raoult’s law. If the deviation is such that minima or maxima in the vapour pressure of the mixture are formed, then the mixture is an azeotrope at those concentrations.

44. Q. On mixing liquid X and liquid Y, the volume of the resulting solution increases. What type of deviation from Raoult’s law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y? Answer: Positive deviation. The temperature will decrease.

45. They depend on the number of particles dissolved.

46. RELATIVE LOWERING OF VAPOUR PRESSURE

47. The  freezing point  is defined as "the  temperature  at which  the vapor pressure  of the substance in its  liquid phase  is  equal  to its  vapor pressure  in the  solid phase " which means that at that  temperature  the  same  amount of atoms/molecules are bonding/breaking between  solid and liquid  at the  same  rate. When we add non volatile solute to a solvent, its vapour pressure is lowered. The lowering of the vapor pressure leads to the lowering of the temperature at which the vapor pressures of the liquid and frozen forms of the solution will be equal. Freezing point depression is the phenomena that describes why adding a solute to a solvent results in the lowering of the freezing point of the solvent. When a substance starts to freeze, the molecules slow down due to the decreases in temperature, and the intermolecular forces start to take over. The molecules will then arrange themselves in a pattern, and thus turn into a solid. For example, as water is cooled to the freezing point, its molecules become slower and hydrogen bonds begin to “stick” more, eventually creating a solid. If salt is added to the water, the Na +  and Cl –  ions attract to the water molecules and interfere with the formation of the large network solid known as ice. In order to achieve a solid, the solution must be cooled to an even lower temperature. Depression in Freezing Point

48. through a Semipermeable membrane

49. DIFFERENCE BETWEEN OSMOSIS AND DIFUSION

50. Pig bladder  (also  pig's bladder ) is the  urinary bladder  of a domestic  pig , similar to the human  urinary bladder . Today, this hollow organ has various applications in medicine, and in traditional cuisines and customs. Parchment: a stiff, flat, thin material made from the prepared skin of an animal, usually a sheep or goat, and used as a durable writing surface in ancient and medieval times. Cellophane  is a thin, transparent sheet made of regenerated cellulose. Its low permeability to air, oils, greases, bacteria, and water makes it useful for food packaging Pig’s bladder Parchment membrane Cellophane A  semipermeable membrane  is a layer that only certain molecules can pass through.  Semipermeable membranes  can be both biological and artificial.  Examples:

51. Osmotic Pressure

52. TYPES OF SOLUTION ON THE BASIS OF OSMOSIS

53. EXAMPLE OF HYPER, HYPO AND ISOTONIC SOLUTIONS

54. ABNORMAL MOLAR MASSES