Chapter 6 Complete Chemistry PDF

CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos CHAPTER 6 COMPLETE Q1. Define the solution. Write its properties. What are its components? Ans. Solution: A homogeneous mixture of two or more substances is called solution. Solution exists as single...

CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos CHAPTER 6 COMPLETE Q1. Define the solution. Write its properties. What are its components? Ans. Solution: A homogeneous mixture of two or more substances is called solution. Solution exists as single phase. It means boundaries of components cannot be distinguished. For example, sea water, air, soft drinks, brass all are solutions. Properties of Solution i) It shows the properties of its components. It means individual components retain their properties. ii) Its components is variable. It means amounts of solute and solvent can be changed according to the requirement. Components of Solution Solute i) The component of solution which is present in smaller quantity is called solute. For example, in aqueous salt solution, salt is solute and water is solvent. Solvent i) The component of solution which is present In large quantity and dissolves the solute in it is called solvent. For example, in soft drinks, water is solvent and other substances carbon dioxide, salt, sugar etc. are solutes. Q2. What is saturated solution and how is it prepared? Ans. Saturated Solution A solution In which amount of solute cannot be dissolved at particular temperature is called saturated solution. (OR) A solution containing maximum amount of solute at a given temperature is called saturated solution. In saturated solution the dissolved and un-dissolved solute in equilibrium with each other. Solute (crystallized) or (un-dissolved) Solute(dissolved) Preparation of Saturation Solution Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Take definite amount (say 100 ml) of solvent like water in a beaker at some room temperature. Now add small amount of solute time with constant stirring. Solute will continue to dissolve. But a stage will come after some time. When no more amount of solute will dissolve. At this state the excess of solute will settle down at the bottom of container. Solute + Solvent dissolve Solution crystalline This is saturated solution. Q3. Define the terms, unsaturated and super saturated solutions. Ans. Unsaturated Solution The solution in which further amount of solute can be dissolved at a given temperature is called unsaturated solution. A solution which contains lesser amount of solute than that which is required to saturate it at a given temperature is called unsaturated solution. Super Saturated Solution The solution which is more concentrated than saturated solution at a particular temperature is known as super saturated solution. Q4. Differentiate between dilute and concentrated solutions with a common example. Ans. Dilute Solution The solution which contain relatively small amount (smaller number of moles/ grams) of dissolved solute is called dilute solution. For example, a solution containing 5g of sodium chloride (NaCl) in 1 dm3 of water is dilute solution. Concentration Solution The solution which contain relatively large amount (greater number of moles/grams) of dissolved solute is called concentrated solutions. For example, brine is concentrated solution of common salt. Similarly 100 g of common salt (NaCl) in 1 dm3 of water form concentrated solution. Q5. Define aqueous solution. Ans. Aqueous Solution Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos The solution in which water is solvent (or) the solution which is formed by dissolving a solute in water is called aqueous solution. For example, sugar in water, salt in water etc. Q6. How many physical states of solution are? Write one example of each state. Ans. Generally there are three physical states of solution depending upon the physical state of solvent and solute. i) Gaseous solution: For example, air is gaseous solution. ii) Solid solution: For example, alloy is solid solution. Dental amalgam for filling of tooth is solid solution. iii) Liquid Solution: For example, sea water, soft drinks are liquid solutions. Q7. You are provided with two liquids. How will you distinguished between solution and pure liquid? Ans. The simplest way to distinguished between a solution and a pure liquid is evaporation. The liquid which evaporates completely leaving no residue, is a pure liquid. While a liquid leaves behind a residue on evaporation is solution. Q8. Soft drinks are frequently used in our daily life. Write common components of a soft drink. Ans. Usual components of a soft drink are water, sugar, salts, carbon dioxide and flavors. Q9. How many types of solutions are formed? Ans. Nine (9) different types of solutions are formed depending upon physical state of solute and solvent. Q10. Why water is called universal solvent? Ans. Because it dissolves many of the compounds present in earth crust. It is also the reason that liquid solution is more common. Test yourself 6.1 I. Why is a solution considered as mixture? Ans. A solution is considered as mixture because: I. It shows the properties of its components. It means individual components retain their properties. Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos II. Its composition is variable. It means amounts of solute and solvent can be changed according to requirement. II. Distinguish between following pairs as a compound or solution. A. Water and salt solution. B. Vinegar and benzene. C. Carbonated drinks and acetone. A. Ans. Water is a compound and salt solution is a solution B. Vinegar is a solution and benzene is a compound. C. Carbonated drinks are solution and acetone is a compound. III. What is major difference between solution and mixture? Ans. Solution is always homogeneous mixture of two or more substances. Whereas mixture may be homogeneous or heterogeneous. Each solution mixture but each is not a solution. IV. Why are alloys considered as solutions? Ans. Alloys are homogeneous mixture of two or more metals in which components retain their original properties. Therefore, these are called solutions. For example, brass(Zn-Cu alloy) is a solution. V. Dead Sea is so rich with salts that it forms crystals when temperature is lower in water. Can you comment why it is named as “Dead sea”? Ans. Dead Sea is a solution of water and different salts. The concentration of salts is very high. Therefore, sea animals like fishes and plants like algae etc. cannot survive in it. There will die there. It means no life can exit on it. That is why it is named as “ Dead sea”. VI. Dead sea is so rich with salts that it forms crystals when temperature lowers in water. Why? Ans. Dead Sea is saturated solution of water and different salts. Therefore, when temperature lowers in water excessive salts are formed. VII. When you jump in Dead Sea you will not sink. Why? Ans. Density of aqueous solution in Dead Sea is very high due to high concentration of salts. Therefore, nobody sink in it. Q11. Define concentration. What do you mean by percentage composition of solution? How can you express It In different ways? Ans. Concentration The ratio of amount of solute to the amount of solution or solvent is called concentration. Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Percentage Composition The number of parts of solute present in 100 parts of solution is called percentage composition. Ways to express percentage composition It can be expressed in four different ways. i) Percentage mass/ mass (% m/m) it is the number of grams of solute in 100 grams of solution. It can be calculated by the following formula; %age by mass = mass of solute in grams/ volume of solution in cm3 x100 ii) Percentage – mass/volume (% m/v) It is number of grams of solute dissolved in 100 cm3 of solution. %age m/v = mass of solute in grams/volume of solution in cm3 x 100 iii) Percentage volume/ mass (% v/m) It is volume in cm3 of solute dissolved in 100 g of solution. % age v/m = volume of solute in cm3/ Mass of solution in grams x 100 iv) Percentage volume/ volume (%v/v) It is the volume in cm3 of a solute dissolved per 100 cm3 of solution. %age by volume= volume of solute in cm3/ volume of solution in cm3 x 100 Q12. What is the most commonly used concentration unit in chemistry and sallied sciences? Define it and give its formula. Ans. Molarity is the most commonly used unit in chemistry and allied sciences. Molarity It is defined as “ the number of moles of solute dissolved in one dm3 of solution Is called molarity. It is represented by M. Molarity (M) = number of moles of solute/ volume of solution in dm3 (OR) M = mass of solute in grams/ (molar mass of solute) x (volume of solution in dm3) Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Q13. What is the molarity and give its formula to prepare molar solution. Ans. Molarity The number of moles of solute dissolved in one dm3 of solution is called Molarity. It is represented by M. Molarity(M) = number of moles of solute/volume of solution in dm3 (OR) M = Mass of solute in grams/ (molar mass of solute) x (volume of solution in dm3) Molar Solution or one Molar Solution ( 1M) The solution containing one mole of solute in one dm3 of solution is called Molar or one molar solution. Preparation of one Molar solution Take one dm3 measuring flask. Fill half of the flask with water. Add in it one mole of solute like NaOH. ( 1 mole of NaOH = 40 g). Shake it until the solute (NaOH) completely dissolves. Now add more water in the flask up to the mark of 1 dm3. This is one molar solution of NaOH because we have dissolved one mole of solute in one dm3 of solution. Q14. Explain how dilute solutions are prepared from concentrated solutions? Ans. Preparation of dilute solution from concentrated solutions If we want to prepare dilute solution from concentrated solution. Then use the following dilution formula. Concentration solution dilute solution M 1V 1 = M 2V 2 Where M1 = molarity of concentrated solution V1 = volume of concentrated solution taken M2 = required molarity of dilute solution V2 = required molarity of dilute solution Q15. What do you mean by volume / volume %? Ans. Percentage – Volume / volume (% v/v) Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos It is the volume in cm3 of a solute dissolved per 100 cm3 of solution. For example, 30 % v/v alcohol solution in water means 30 cm3 of alcohol are dissolved in 100 cm3 of solution. % age by volume = volume of solute in cm3/ volume of solution in cm3 x 100 Test yourself 6.2 I. Does the percentage calculation require the chemical formula of solute? Ans. No, there is no term in %age calculations which require chemical formula of solute. II. Why is the formula of solute necessary for calculation of molarity of the solution? Ans. The term number of moles of solute is involved in the formula of molarity. M = number of moles of solute/ volume of solution in dm3 In order to calculate number of moles of solute, we need formula mass which is calculated from chemical formula of solute. No. of moles = mass in grams/ formula mass Thus formula of solute is necessary for calculation of molarity of the solution. III. You are asked to prepare 15% (m/m) solution of common salt. How much amount of water will be required to prepared this solution? Ans. 15% w/w solution means that 15g of sodium chloride is dissolved in 85g of water to make 100g solution. Thus, 85g of water is required. IV. How much water should be mixed with 18 cm3 of alcohol so as to obtain 18% v/v alcohol solution? Ans. 18% v/v alcohol solution means; Volume of alcohol = 18cm3 Volume solution (alcohol + water) = 100cm3 Volume of water = 100-18 = 82cm3 Thus 82cm3 of water is required. V. Calculate the concentration % (m/m) of a solution which contains 2.5g of salt dissolved in 50g of water? Ans. Mass of solute = 2.5 g Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Mass of solvent = 50g % age m/m = ? Mass of solution = 2.5 + 50=52.5g %age m/m = mass of solute in grams/ mass of solution in grams x 100 = 2.5/52.5 x 100 = 4.762% VI. Which one of the following is more concentrated one molar or three molar? Ans. One molar solution contains one mole of solute in 1 dm3 of water. Three molar solutions contain 3 moles of solute in 1 dm3 of water. So 3 molar solution is more concentrated because of greater amount of solute. Q16. Define solubility. Name the factors affecting solubility? Ans. Solubility The number of grams of solute dissolved in 100 g of solvent to prepare a saturated solution at a particular temperature is called solubility of solute. For example, solubility of sodium chloride at 0oC is 37.5g per 100g of water. Factors affecting solubility 1. Nature of solute and solvent Polar solutes are soluble in polar solvents and non-polar solutes are soluble in non-polar solvents. 2. Solute – Solvent interaction If the forces of attractions among solute and solvent particles are strongest, then solubility will be highest and vice versa. Temperature Usually solubility increase with the rise of temperature. There are three possibilities, i) If heat is absorbed during solution formation, solubility will increase with the rise of temperature. ii) if heat is evolved during solution formation, solubility will decrease with the rise of temperature. Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos iii) If no heat is absorbed or evolved during solution formation, solubility will not affected with the change of temperature. Note: Solubility of gases decrease with the rise of temperature. Q17. How do solute-solvent interaction concern with the preparation solution? Ans. Solute- solvent interaction and solution formation Solution formation depends upon the relative strength of attractive forces between solute- solute-solvent-solvent and solute-solvent. There are two possibilities i) If attractive forces between solute-solute particles are strong enough than solute-solvent interactions, solutions will not form. For example, benzene and water will not form solution. ii) If attractive forces between solute-solute particles are weak than solute-solvent interactions, solution will form and solubility will be greater. For example, sodium chloride and water will form solution. Q18. How can you explain the solute-solvent interactions to prepare NaCl solution? Ans. Solute-Solvent interaction and formation of NaCl solution When NaCl is added in water, it dissolves readily and form solution. NaCl is an ionic compound. When it is added in water, the attractive forces among the ions(Cl- and Na+ ) and water molecules are stronger enough to overcome attractive forces between Na + and Cl- ions in solid NaCl crystals. In this process positive end of polar water molecule is oriented towards the Cl- ions and negative end is oriented towards the Na+ ions. These interactions between solute (NaCl) and solvent (water) are so strong that these forces pull or detach ions from the crystal and thus NaCl dissolve. Q19. Discuss the effect of temperature on solubility? Ans. Effect of Temperature on solubility In majority of solutes, solubility increase with the increase of temperature. But this is not always true. Sometimes solubility decreases with the rise of temperature. There are three possibilities. i) If Heat absorbed during dissolution If heat is absorbed during formation of a solution (endothermic process), solubility will increase with the rise of temperature. For example, solubility of KNO3, NaNO3, and KCl etc increase with the rise of temperature. Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Solvent + Solute solution ii) If heat is evolved during dissolution If heat is evolved during solution formation, solubility will decrease with the rise of temperature. For example, solubility of Lithium sulphate (Li2SO4) and cesium sulphate, Ce2(SO4)3 decrease with the rise of temperature because solution formation is exothermic process Solvent + Solute solution + heat iii) If no heat evolve or absorb If no heat is evolve or absorbed during solution formation, temperature has minimum effect on solubility. For example, solubility of NaCl is slightly affected with the rise of temperature. Q20. What do you mean by statement, like dissolves like? Explain with examples. Ans. General principle of solubility is “ like Dissolves like” It means that: i) The polar solutes are soluble in polar solvents. Ionic solids and polar covalent compounds are soluble in water. For example, KCl, Na2CO3, CuSO4, sugar, alcohol all are soluble in water. ii) Non polar solutes are soluble in non-polar solvents. For example, Bromine, Grease, Paints, Naphthalene are non-polar solutes. Therefore, these are soluble in non-polar solvents like ether or carbon tetra chloride. iii) Non polar substances are not soluble in polar solvents because these are not like. Non polar compounds, ether, benzene and petrol are not soluble in water (polar solvent). Q21. How does nature of attractive forces of solute-solute and solvent-solvent affect solubility? Ans. Solute- solvent interaction and solution formation Solution formation depends upon the relative strength of attractive forces between solute- solute-solvent-solvent and solute-solvent. There are two possibilities i) If attractive forces between solute-solute particles are strong enough than solute-solvent interactions, solutions will not form. For example, benzene and water will not form solution. ii) If attractive forces between solute-solute particles are weak than solute-solvent interactions, solution will form and solubility will be greater. For example, sodium chloride and water will form solution. Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Q22. Justify with an example that solubility of a salt increases with the increase in temperature? Ans. In majority of salts, solubility increase with the increase in temperature because increase in temperature weak the attractive forces among the salt particles. Similarly increase in temperature also weak the attractive forces among the solvent particles. But forces of attractions increase between solute and solvent particles. Thus solubility increase with the increase in temperature. For example, solubility of KNO3, NaNO3 and KCl etc. increase with the rise of temperature. Solvent + solute + heat Solution Q23. What is general principle of solubility? Ans. General principle of solubility is “ like Dissolves like” It means that: i) The polar solutes are soluble in polar solvents. Ionic solids and polar covalent compounds are soluble in water. For example, KCl, Na2CO3, CuSO4, sugar, alcohol all are soluble in water. ii) Non polar solutes are soluble in non-polar solvents. For example, Bromine, Grease, Paints, Naphthalene are non-polar solutes. Therefore, these are soluble in non-polar solvents like ether or carbon tetra chloride. iii) Non polar substances are not soluble in polar solvents because these are not like. Non polar compounds, ether, benzene and petrol are not soluble in water (polar solvent). Q24. Why are non-polar solutes not soluble in polar solvents? Ans. Because strong solute-solvent interactions are not developed between them to detach the solute particles from its crystal. Hence solution does not form. For example, benzene will not dissolve in water due to weak solute and solvent interactions. Q25. How will you guess from solute-solvent interactions that whether solution will form or not? Ans. If forces of attraction are greater between solute and solvent particles than solute-solute interactions. Solution will form. On the other hand, if solute-solvent interactions are not strong as compared to solute-solute interactions, the solution will not form. Q26. What events must occur during solution formation? Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Ans. To dissolves solute in solvent following three events must occur. i) Solute particles must separate from each other. ii) Solvent particles must separate to provide space for solute particles. iii) Solute and solvent particles must attract and mix up. Test yourself 6.3 I. What will happen if solute- solute attractive forces are stronger than solute- solvent forces? Ans. Solution will not form because solute particles will not separate from its crystal. II. When solute- solute forces are weaker than those of solute- solvent forces, will solution form? Ans. Yes, the solution will form. III. Why is iodine soluble in CCl4 and not in water? Ans. “Like dissolve like”. Iodine is non-polar solute and CCl4 is non-polar solvent. Therefore, Iodine will dissolve in CCl4 due to greater solute-solvent interactions. On the other hand, water is polar solvent. Therefore, Iodine will not dissolve in it due to weaker solute-solvent interactions. IV. Why test tube become cold when KNO3 is dissolved in water? Ans. During dissolution of KNO3, heat is absorbed. That is why test tube becomes cold. Q27. What are true solutions? Ans. true solutions are homogeneous mixture of two or more compounds which are not visible. For example, a drop of ink in water forms true solution. Q28. Define colloids, Suspension, and Tyndall Effect. Ans. Colloid These are solutions in which the solute particles are larger than those present in the true solution but not large enough to be seen by naked eye. The particles in such a system dissolve and do not settle down for long time. Suspension Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Suspension is heterogeneous mixture of un-dissolved particles are big enough to be seen by naked eye. Examples, Chalk in water, paints, milk of magnesia etc. Tyndall Effect The particles of colloids are big enough to scatter the beam of light. The light is emmited in scattered form. This is called tyndall effect. This test distinguishes colloids from true solutions. Examples of colloids Starch, albumin, soap solution, blood, milk, jelly, tooth paste etc. Q29. Give few characteristics of colloids. Ans. i) The particles are large consisting of many atoms, ions or ,molecules. ii) A colloids appears to be a homogeneous for a long time, therefore, colloids are quite stable. iii) Particles do not settle down for a long time, therefore, colloids are quite stable. iv) Particles are large but can’t be seen with naked eye. v) Although particles are big but they can pass through a filter paper. vi) Particles of colloids scatter the path of light rays, thus exhibit the tyndall effect. Q30. Give at least five characteristics of suspension. Ans. i) The particles are of large size. They are large than 10-5 cm in diameter. ii) Particles remain un-dissolved and form a heterogeneous mixture. Particles settle down after sometime. iii) Particles are big enough to be seen with naked eye. iv) Solute particles cannot pass through filter paper. v) Particles are so big that light is blocked and difficult to pass. Q31. What is the difference between colloid and suspension? Ans. Difference between Suspension and colloid Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Colloid Suspension 1. The particles are large consisting of The particles are of largest size. many atoms, ions or molecules. They are larger than 10-5 cm in diameter. 2. A colloid appears to be a Particles remain un-dissolved and form a homogeneous but actually it is a heterogeneous mixture. Particles settle down heterogeneous mixture. after sometime. Hence, they are not true solution. Particles do not settle down for a long time, therefore colloids are quite stable. 3. Particles are big but they can pass Particles are big enough to be seen with through a filter paper. naked eye. 4. Particles are large but can’t be seen Solute particles cannot pass through filter with naked eye. paper. 5. Particles scatter the path of light rays Particles are so big that light is blocked and thus emitting the beam of light, difficult to pass. exhibit the tyndall effect. Q32. Why suspension and solution do not show tyndall effect, which colloids do? Ans. Solution cannot show tyndall effect because particles are so small that they cannot scatter the rays of light. On the other hand suspension particles are so big that light is blocked and difficult to pass. Therefore, cannot show tyndall effect. Particles of colloid scatter the path of light rays thus emitting the beam of light i.e, exhibit the tyndall effect. Q33. What is the reason for difference between solutions, colloids and suspension? Ans. Solutions, colloids and suspension differs from each other on the basis of size of solute particles. In solution particles exist in their simplest form (as molecules or ions. Their diameter is 10-8cm). In colloids, the particles are large consisting of many atoms, ions or molecules, In suspension, the particles are of large size. They are large than 10-5 cm in diameter. Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Q34. Why does not suspension form homogeneous mixture? Ans. In suspension particles remain un-dissolved and settle down after some time. Therefore, it cannot form homogeneous mixture. Instead heterogeneous mixture is formed. Q35. How will you test a given solution is a colloidal solution or not? Ans. Pass the light rays through the solutions. If beam of light scattered, then it will be colloidal. If solution shows tyndall effect, then it will be colloidal solution. Q36. How will you test that a given solution is a colloidal solution. Blood, starch solution, glucose solution, soap solution and milk of magnesia. Ans. Blood, starch solution, tooth paste, soap solution all are colloidal solutions. Glucose solution and copper sulphate solution are true solutions. Milk of magnesia is suspension. Q37. Why are stir paints thoroughly before use? Ans. Paints are suspensions in which particles settle down after some time. Therefore, to mix the particles it is necessary to stir the paint before use. Q38. Which of the following scatter light and why? Sugar solution, soap solution and milk of magnesium? Ans. Only soap solution will scatter the light because it is colloidal solution. Sugar solution is true solution. Therefore, light will pass through it un-scattered. Milk of magnesia is suspension. Its particles are big and will not allow the light to pass through it. Test yourself 6.4 I. What is the difference between colloid and suspension? Ans. Colloid and suspensions differ from each other due to size of solute particles. In colloids, the particles are large consisting of many atoms, ions, or molecules. In suspension, the particles are of large size. They are larger than 10-5 cm in diameter. II. Can colloids be separated by filtration? If not why? Ans. Colloids cannot be separated by filtration because particles are not so big and pass through filter papar. III. Why are the colloids not true solutions and still quite stable? Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Ans. A colloid appears to be homogeneous but actually it is heterogeneous mixture. Hence they are not true solutions. In colloids particles do not settle down for long time, therefore, colloids are quite stable. IV. Why does the colloids show tyndall effect? Ans. Particles of colloids are big enough to scatter the light, hence can show tyndall effect. V. What is tyndall effect and on what factors it depends? Ans. Tyndall Effect: The scattering of light beam into different colors by the particles of colloids is called tyndall effect. Factors: 1. Size of particles of colloids 2. Wave length of light used. VI. Identify as colloids or suspension from the following: Paint, milk, milk of magnesia, soap solution. Ans. Paints and milk of magnesia are suspension whereas milk and soap solutions are colloids VII. How can you justify the milk is colloids? Ans. When light is passed through the milk, it is scattered. This show milk is colloids. SMART MCQ’S WITH ‘KEY’ FOR ABOVE TOPICS 1) Dental amalgam is used for the filling of tooth in human. It is a __________. a) Solid-solid solution b) liquid-liquid solution c) gas-gas solution d) solid-solid solution 2) Homogeneous mixture of two or more substances is called solution. The most common solutions are: a) solid solution b) liquid solution c) gas solutions d) solid and liquid solutions 3) The simplest way to distinguish between solution and pure liquid is: Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos a) note boiling points b) perform chemical analysis c) evaporation d) note the freezing point 4) Alloy likes Brass or Bronze is: a) Pure compound b) Homogeneous mixture c) Colloid d) suspension 5) The solution made in alcohol or benzene is called ______________. a) aqueous solution b) solid in solid solution c) colloidal d) non aqueous solution 6) On practical level, a saturated solution is one, in which un-dissolved solute is in equilibrium with dissolved solute. This is dynamic equilibrium and is represented as: a) Solute (crystallized) c)solute (dissolved) b) solvent (dissolve) d)solute(crystallize) 7) In order to dilute a solution add: a) more solute b) more solvent c) add both solute and solvent d) catalyst 8) The solution which is more concentrated than saturated solution is called super saturated solution. A super saturated solution is always prepared at: a) lower temperature b) lower pressure c) high pressure d) high temperature 9) Crystallization of solute will start automatically from: a) saturated solution b) unsaturated solution c) super saturated solution d) dilute solution 10) The formula uused to prepare dilute solution from concentrated solution is: a) M1n1 = M2n2 b) M1n1 = M2v1 c) M1/V1 = M2/V2 d) M1V1 = M2V2 Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos 11) There are different concentration units used to express the concentration of solution. You are provided with solid solute and liquid solvent. Which of the following units you will prefer to express the concentration of solution? a) % m/m b) % m/v c) % v/m d) % v/v 12) In order to prepare a solution, you are provided with solute as well as liquid solvent. Which of the following units you will prefer to express the concentration of solution? a) % m/m b) % m/v c) % v/m d) % v/v 13) You are provided with solid solute and solid solvent. Which of the following units you will prefer to express concentration of solution? a) % m/m b) %m/v c) % v/m d) % v/v 14) If number of moles of solute are known to you, which of the following units should be used to express concentration of solution? a) %m/m b) % m/v c) Molarity d) %age composition 15) If we add 5.5 cm3 of acetone in water to prepare 90cm3 of aqueous solution. What is v/v of this solution? a) 5.7% b) 6.5% c) 2.5% d) 90% 16) Water is universal solvent. The solution in which water is used as solvent is called: a) non aqueous solution b) aqueous solution c) dilute solution d) concentrated solution 17) Molarity is represented by ‘M’. 3M solution of NaOH means: a) 3 g of NaOH dissolved in 1 kg of water b) 3 moles of NaOH are dissolved in 1 kg of water Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos c) 3 moles of NaOH are dissolved in 1000cm3 of water. d) 3 moles of NaOH are dissolved in 1000 cm3 of solution 18) You have four different concentrations of same solute like glucose. That is 1M, 2M,3M and 4M, which one of them is dilute? a) 4M b) 2M c) 1M d) 3M 19) You dissolve 8g of NaOH in 500 cm3 of solution.Its molarity is: a) 0.4 M b) 0.1 M c) 0.3 M d) 0.5 M 20) Like dissolved like therefore, polar solute is soluble in polar solvent because: a) solute-solute sources are stronger than solute – solvent forces b) solute-solvent forces are stronger than solute-solute forces c) solute-solute and solute-solvent forces are equal d) solute-solvent forces weaker than solvent-solvent forces 21) Every solute has its own solubility in a specific solvent. The solubility decreases with the rise of temperature if solution formation is_____________. a) endothermic b) temperature remains same c) exothermic d) both a and b 22) Solubility of solute increase with the rise of temperature. If solution formation is________. a) endothermic b) no heat change c) exothermic d) not affected by T 23) Many of the gases dissolve in water. The solubility of gases _______________ with the rise of temperature. a) increase b) remains constant c) both a & b d) decrease Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos 24) Temperature has very small effect on solubility of solutes. If solution formation involves: a) heat absorb b) heat evolve c) volume increase d) no change in heat 25) The scattering of beam of light on passing through a solution is called tyndall effect. Which one of the following will show tyndall effect. a) colloid b) true solution c) suspension d) paints 26) Suspension do not show tyndall’s effect because of: a) small size solute particles b) big size solute particles c) atomic size of solute particles d) medium size particles 27) Particles of following can be seen with naked eye: a) true solution b) colloid c) suspension d) none of above 28) Particles of which of the following cannot pass through filter paper. a) colloid b) soft drinks c) true solution d) suspension Answer Key A B C B D A B D C D 1 2 3 4 5 6 7 8 9 10 Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos B D A C A B D C A B 11 12 13 14 15 16 17 18 19 20 C A D D A B C D 21 22 23 24 25 26 27 28 Exercise MCQ’S 1) Mist is an example of solution: a) liquid in gas b) gas in liquid c) solid in gas d) gas in solid 2) Which one of the following is a liquid in solid solution? a) sugar in water b) butter c) opal d) fog 3) Concentration is ration of: a) solvent to solute b) solute to solution c) solvent to solution d) both a and b 4) Which one of the following solutions contains more water: a) 2 M b) 1 M c) 0.5 M d) 0.25 M 5) A 5 percent(w/w) sugar solution means that: a) 5 g of sugar is dissolved in 90g of water b) 5 g of sugar is dissolved in 100g of water Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos c) 5 g of sugar is dissolved in 105g of water d) 5 g of sugar is dissolved in 95 g of water 6) If the solute-solute forces are strong enough than those of solutes-solvent forces. The solute: a) dissolves readily b) does not dissolve c) dissolves slowly d) dissolves and precipitates 7) Which one of the following will show negligible effect of temperature on its solubility? a) KCl b) KNO3 c) NaNO3 d) NaCl 8) Which one of the following is heterogeneous mixture? a) milk b) ink c) milk of magnesia d) sugar solution 9) Tyndall effect is shown by: a) sugar solution b) paints c) jelly d) Chalk solution 10) Tyndall effect is due to: a) blocking of beam of light b) non-scattering of beam of light c) scattering of beam of light d) passing through beam of light 11) If 10cm3 of alcohol is dissolved in 100 g of water, it is called: a) % w/w b) % w/v c) % v/w d) % v/v 12) When a saturated solution is diluted it turns into: a) super saturated solution b) saturated solution c) a concentrated solution d) unsaturated solution 13) Molarity is the number of moles of solute dissolved in: Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos a) 1 kg of solution b) 100g of solvent c) 1 dm3 of solvent d) 1 dm3 of solution Answer Key A B B D D B D C C C 1 2 3 4 5 6 7 8 9 10 C D D 11 12 13 Chapter 3 Short Questions Q1. Why suspension and solution do not show tyndall effect, which colloids do? Ans. Solution cannot show tyndall effect because particles are so small that they cannot scatter the rays of light. On the other hand suspension particles are so big that light is blocked and difficult to pass. Therefore, cannot show tyndall effect. Particles of colloid scatter the path of light rays thus emitting the beam of light i.e, exhibit the tyndall effect. Q2. What is the reason for difference between solutions, colloids and suspension? Ans. Solutions, colloids and suspension differs from each other on the basis of size of solute particles. In solution particles exist in their simplest form (as molecules or ions. Their diameter is 10-8cm). In colloids, the particles are large consisting of many atoms, ions or molecules, In suspension, the particles are of large size. They are large than 10-5 cm in diameter. Q3. Why does not suspension form homogeneous mixture? Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Ans. In suspension particles remain un-dissolved and settle down after some time. Therefore, it cannot form homogeneous mixture. Instead heterogeneous mixture is formed. Q4. How will you test a given solution is a colloidal solution or not? Ans. Pass the light rays through the solutions. If beam of light scattered, then it will be colloidal. If solution shows tyndall effect, then it will be colloidal solution. Q5. How will you test that a given solution is a colloidal solution. Blood, starch solution, glucose solution, soap solution and milk of magnesia. Ans. Blood, starch solution, tooth paste, soap solution all are colloidal solutions. Glucose solution and copper sulphate solution are true solutions. Milk of magnesia is suspension. Q6. Why are stir paints thoroughly before use? Ans. Paints are suspensions in which particles settle down after some time. Therefore, to mix the particles it is necessary to stir the paint before use. Q7. Which of the following scatter light and why? Sugar solution, soap solution and milk of magnesium? Ans. Only soap solution will scatter the light because it is colloidal solution. Sugar solution is true solution. Therefore, light will pass through it un-scattered. Milk of magnesia is suspension. Its particles are big and will not allow the light to pass through it. Q8. What do you mean by statement, like dissolves like? Explain with examples. Ans. General principle of solubility is “ like Dissolves like” It means that: i) The polar solutes are soluble in polar solvents. Ionic solids and polar covalent compounds are soluble in water. For example, KCl, Na2CO3, CuSO4, sugar, alcohol all are soluble in water. ii) Non polar solutes are soluble in non-polar solvents. For example, Bromine, Grease, Paints, Naphthalene are non-polar solutes. Therefore, these are soluble in non-polar solvents like ether or carbon tetra chloride. iii) Non polar substances are not soluble in polar solvents because these are not like. Non polar compounds, ether, benzene and petrol are not soluble in water (polar solvent). Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Q9. How does nature of attractive forces of solute-solute and solvent-solvent affect solubility? Ans. Solute- solvent interaction and solution formation Solution formation depends upon the relative strength of attractive forces between solute- solute, solvent-solvent and solute-solvent. There are two possibilities i) If attractive forces between solute-solute particles are strong enough than solute-solvent interactions, solutions will not form. For example, benzene and water will not form solution. ii) If attractive forces between solute-solute particles are weak than solute-solvent interactions, solution will form and solubility will be greater. For example, sodium chloride and water will form solution. Q10. How can you explain the solute-solvent interactions to prepare NaCl solution? Ans. Solute-Solvent interaction and formation of NaCl solution When NaCl is added in water, it dissolves readily and form solution. NaCl is an ionic compound. When it is added in water, the attractive forces among the ions(Cl- and Na+ ) and water molecules are stronger enough to overcome attractive forces between Na + and Cl- ions in solid NaCl crystals. In this process positive end of polar water molecule is oriented towards the Cl - ions and negative end is oriented towards the Na+ ions. These interactions between solute (NaCl) and solvent(water) are so strong that these forces pull or detach ions from the crystal and thus NaCl dissolve. Q11. Justify with an example that solubility of a salt increases with the increase in temperature? Ans. In majority of salts, solubility increase with the increase in temperature because increase in temperature weak the attractive forces among the salt particles. Similarly increase in temperature also weak the attractive forces among the solvent particles. But forces of attractions increase between solute and solvent particles. Thus solubility increase with the increase in temperature. For example, solubility of KNO3, NaNO3 and KCl etc. increase with the rise of temperature. Solvent + solute + heat Solution Q12. What do you mean by volume / volume %? Ans. Percentage – Volume / volume (% v/v) Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos It is the volume in cm3 of a solute dissolved per 100 cm3 of solution. For example, 30 % v/v alcohol solution in water means 30 cm3 of alcohol are dissolved in 100 cm3 of solution. % age by volume = volume of solute in cm3/ volume of solution in cm3 x 100 Chapter 3 Long Questions Q1. What is saturated solution and how is it prepared? Ans. Saturated Solution A solution in which amount of solute cannot be dissolved at particular temperature is called saturated solution. (OR) A solution containing maximum amount of solute at a given temperature is called saturated solution. In saturated solution the dissolved and un-dissolved solute in equilibrium with each other. Solute (crystallized) or (un-dissolved) Solute(dissolved) Preparation of Saturation Solution Take definite amount (say 100 ml) of solvent like water in a beaker at some room temperature. Now add small amount of solute time with constant stirring. Solute will continue to dissolve. But a stage will come after some time. When no more amount of solute will dissolve. At this state the excess of solute will settle down at the bottom of container. Solute + Solvent dissolve Solution crystalline This is saturated solution. Q2. Differentiate between dilute and concentrated solutions with a common example. Ans. Dilute Solution Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos The solution which contain relatively small amount (smaller number of moles/ grams) of dissolved solute is called dilute solution. For example, a solution containing 5g of sodium chloride (NaCl) in 1 dm3 of water is dilute solution. Concentration Solution The solution which contain relatively large amount (greater number of moles/grams) of dissolved solute is called concentrated solutions. For example, brine is concentrated solution of common salt. Similarly 100 g of common salt (NaCl) in 1 dm3 of water form concentrated solution. Q3. Explain how dilute solutions are prepared from concentrated solutions? Ans. Preparation of dilute solution from concentrated solutions If we want to prepare dilute solution from concentrated solution. Then use the following dilution formula. Concentration solution dilute solution M 1V 1 = M 2V 2 Where M1 = molarity of concentrated solution V1 = volume of concentrated solution taken M2 = required molarity of dilute solution V2 = required molarity of dilute solution Q4. What is the molarity and give its formula to prepare molar solution. Ans. Molarity The number of moles of solute dissolved in one dm3 of solution is called Molarity. It is represented by M. Molarity(M) = number of moles of solute/volume of solution in dm3 (OR) M = Mass of solute in grams/(molar mass of solute) x (volume of solution in dm3) Molar Solution or one Molar Solution ( 1M) Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos The solution containing one mole of solute in one dm3 of solution is called Molar or one molar solution. Preparation of one Molar solution Take one dm3 measuring flask. Fill half of the flask with water. Add in it one mole of solute like NaOH. ( 1 mole of NaOH = 40 g). Shake it until the solute (NaOH) completely dissolves. Now add more water in the flask up to the mark of 1 dm3. This is one molar solution of NaOH because we have dissolved one mole of solute in one dm3 of solution. Q5. How do solute-solvent interaction concern with the preparation solution? Ans. Solute- solvent interaction and solution formation Solution formation depends upon the relative strength of attractive forces between solute- solute-solvent-solvent and solute-solvent. There are two possibilities i) If attractive forces between solute-solute particles are strong enough than solute-solvent interactions, solutions will not form. For example, benzene and water will not form solution. ii) If attractive forces between solute-solute particles are weak than solute-solvent interactions, solution will form and solubility will be greater. For example, sodium chloride and water will form solution. Q6. What is general principle of solubility? Ans. General principle of solubility is “ like Dissolves like” It means that: i) The polar solutes are soluble in polar solvents. Ionic solids and polar covalent compounds are soluble in water. For example, KCl, Na2CO3, CuSO4, sugar, alcohol all are soluble in water. ii) Non polar solutes are soluble in non-polar solvents. For example, Bromine, Grease, Paints, Naphthalene are non-polar solutes. Therefore, these are soluble in non-polar solvents like ether or carbon tetra chloride. iii) Non polar substances are not soluble in polar solvents because these are not like. Non polar compounds, ether, benzene and petrol are not soluble in water (polar solvent). Q7. Discuss the effect of temperature on solubility? Ans. Effect of Temperature on solubility Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos In majority of solutes, solubility increase with the increase of temperature. But this is not always true. Sometimes solubility decreases with the rise of temperature. There are three possibilities. i) If Heat absorbed during dissolution If heat is absorbed during formation of a solution (endothermic process), solubility will increase with the rise of temperature. For example, solubility of KNO3, NaNO3, and KCl etc increase with the rise of temperature. Solvent + Solute solution ii) If heat is evolved during dissolution If heat is evolved during solution formation, solubility will decrease with the rise of temperature. For example, solubility of Lithium sulphate (Li2SO4) and cesium sulphate, Ce2(SO4)3 decrease with the rise of temperature because solution formation is exothermic process Solvent + Solute solution + heat iii) If no heat evolve or absorb If no heat is evolve or absorbed during solution formation, temperature has minimum effect on solubility. For example, solubility of NaCl is slightly affected with the rise of temperature. Q8. Give few characteristics of colloids. Ans. i) The particles are large consisting of many atoms, ions or ,molecules. ii) A colloids appears to be a homogeneous for a long time, therefore, colloids are quite stable. iii) Particles do not settle down for a long time, therefore, colloids are quite stable. iv) Particles are large but can’t be seen with naked eye. v) Although particles are big but they can pass through a filter paper. vi) Particles of colloids scatter the path of light rays, thus exhibit the tyndall effect. Q9. Give at least five characteristics of suspension. Ans. i) The particles are of large size. They are large than 10-5 cm in diameter. ii) Particles remain un-dissolved and form a heterogeneous mixture. Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Particles settle down after sometime. iii) Particles are big enough to be seen with naked eye. iv) Solute particles cannot pass through filter paper. v) Particles are so big that light is blocked and difficult to pass. Numericals: Q1. A solution contains 50g of sugar dissolved in 450g of water. What is concentration of this solution? Ans. Mass of solute(sugar) = 5.0 g Mass of solvent(water) = 400g Concentration of solution = ? Mass of solution = 50 + 50 = 500g %age of mass = mass of solute in grams/mass of solution in grams = 50.500 x 100 = 10% The concentration of solution is 10% m/m. 2. If 60cm3 of alcohol is dissolved in 940 cm3 of water, what is concentration of this solution? Ans. Volume of solute(alcohol) = 60 cm3 Volume of solvent(water) = 940 cm3 Volume of solution = 60 + 940 = 1000cm3 Concentration of solution = ? %age v/v = volume of solute/volume of solution x 100 = 60/1000 x 100 = 6% Concentration of solution is 6% v/v. 3. how much salt will be required to prepare following solutions (atomic mass : k = 39; Na= 23; S=32; O=16; and H=1) Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos (A) 250 cm3 of KOH solution of 0.25 M (B) 600 cm3 of NaNo3 solution of 0.25 M (C) 800 cm3 of Na2SO4 solution of 1.0 M Ans. (A) volume of KOH solution = 250cm3 = 0.25 dm3 Molarity of KOH solution = 0.5 Mass of KOH =? Molar mass of KOH = 39 + 16 + 1 = 56 gmol-1 Molarity = mass of KOH/ (Molar mass of KOH) (volume of solution in dm3) Mass of KOH = Molarity x Molar mass of KOH x volume of solution in dm3 = 0.5 x 56 x 0.25 = 7g (B) Molarity of NaNo3 solution = 0.25 M Volume of solution = 600 cm3 = 600/1000 = 0.6 dm3 Molar mass of NaNO3 = 23 + 14 + 16 + 3 = 85 gmol-1 Mass of solute = ? Molarity = mass of solute/ (Molar mass of solute) ( volume of solution in dm3) Mass of solute = Molarity x Molar mass of solute x volume of solution = 0.25 x 85 x 0.6 = 12.75 g Thus mass of NaNO3 required = 12.75 g (C) Molarity of Na2SO4 solution = M Volume of solution = 800 cm3 = 800/1000 dm3 Molar mass of Na2SO4 = 23 + 2 + 32 + 16 + 4 = 142 gmol-1 M = mass of solute/ (Molar mass of solute) x ( volume of solution in dm3) Mass of solute = M x Molar mass of solute x vol. of solution = 1 x 142 x 0.8 = 113.6g Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Thus mass of Na2SO4 required = 113.6 g 4. When we dissolve 20g of NaCl in 400 cm3 of solution, what will be its molarity? Ans. mass of solute (NaCl) = 20 g Vol. of solution = 400 cm3 = 400/ 1000 dm3 Molarity of solution = ? Molar mass of NaCl = 23 + 35.5 = 58.5 gmol-1 Number of moles of NaCl = 20/ 58.5 = 0.342 M = no. of moles of NaCl/ Vol. of solution in dm3 = 0.32/0.4 = 0.85 M So molarity of NaCl solution = 0.85 M 5. We desire to prepare 100 cm3 0.4 M solution of MgCl2, how much MgCl2 is needed? Ans. Volume solution = 100 cm3 = 100/ 1000 = 0.1 dm3 Molarity of solution = 0.4 Molar mass of MgCl2 = 24 + 35.5 x 2 = 95 gmol-1 Mass of MgCl2 = ? M = mass of solute in grams/ molar mass of solute x volume of solution in dm3 Mass of solute = M x Molar mass of solute x vol. of solution in dm3 = M x Molar mass of solute x vol. of solution in dm3 = 0.4 x 95 x 0.1 = 3.8 g Mass of MgCl2 required = 3.8 g 6. 12 M H2SO4 solution is available in the laboratory. We need only 500 cm3 of 0.1 M solution, how it will be prepared? Ans. Given molarity of H2SO4 = M1 = 12M Volume of H2SO4 = V1 = ? Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos Required molarity of H2SO4 = M1 = 0.1 M Required volume of H2SO4 = V2 = 500 cm3 Now apply dilution formula Given Required Concentration solution dilute solution M 1V 1 = M 2V 2 V1 = M2V2/M1 V1 = 0.1 x 500/ 12 = 4.17 cm3 V1 = 4.17 cm3 Take 4.17 cm3 of 12 M H2SO4 with the help of stoppered pipette and transfer to 500cm3 flask containing some quantity of water, shake well then more quantity of water to make up volume 500cm3. This is 0.1 M H2SO4. Email us if you find any mistake [email protected] Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com CH 6 Solutions Visit GS Academy You Tube Channel 4 More Videos WE PROVIDE NOTES, LECTURES, TEST YOUR SELF, SELF ASSESMENTS, EXERCISE MCQS, EXERCISE SHORT QUESTIONS, SOLUTION TO NUMERICAL PROBLEMS, EXERCISE LONG QUESTIONS, PAST PAPERS CHAPTER WISE, CHAPTER WISE IMPORTANT LONG QUESTIONS CONTACT CELL: +92344 4789022 WATS APP: 034447899022 Prepared by SIR MOHAMMAD KHURRAM RAJPUT https://youtube.com/c/GSACADEMYKHURRAM www.gsacademies.com