5. Exploring Mixtures and their Separation - Question Answer | Class 9 Science Exploration

[00:00:00]Chapter 5, exploring mixtures and their separation, question answers from grade nine subject science book exploration.

[00:00:08]Think it over. Question one, why do suspended particles settle in muddy water over time but not in milk? Answer, suspended particles in muddy water are larger and heavier. When muddy water is left undisturbed, these particles settle down at the bottom due to gravity. Such a mixture is called a suspension.

[00:00:30]Milk is a colloid. In milk, tiny fat droplets and other particles are very small and remain uniformly dispersed.

[00:00:39]These particles do not settle down easily on standing. Therefore, suspended particles settle in muddy water over time but not in milk.

[00:00:49]Question two, how is evaporation different from boiling? Answer, evaporation and boiling are both processes in which a liquid changes into vapor but they are different.

[00:01:01]Evaporation, it occurs at any temperature. Boiling, it occurs at a fixed temperature called the boiling point. Evaporation, it takes place only at the surface of the liquid. Boiling, it takes place throughout the liquid.

[00:01:16]Evaporation, it is a slow process.

[00:01:19]Boiling, it is a fast process.

[00:01:21]Evaporation, no bubbles are formed.

[00:01:24]Boiling, bubbles are formed inside the liquid. Example, water in a wet cloth evaporates slowly at room temperature but water boils at 100° C. Question three, why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?

[00:01:43]Answer, we see bright rays of sunlight because of the Tyndall effect. When sunlight passes through small gaps between leaves, it gets scattered by tiny dust particles, water droplets, and other particles present in air. Due to this scattering, the path of sunlight becomes visible as bright rays.

[00:02:03]Pause and ponder, page 76, question one.

[00:02:07]A common talcum powder contains 4% m/m zinc oxide which acts as an antiseptic.

[00:02:15]How much zinc oxide is present in 300 g of the talcum powder? Answer, given, talcum powder contains 4% m/m zinc oxide. This means 100 g talcum powder contains 4 g zinc oxide. So, 300 g talcum powder contains 4 / 100 * 300 = 12 g. Therefore, 12 g of zinc oxide is present in 300 g of talcum powder.

[00:02:47]So friends, understand 100 g contains 4 g zinc oxide.

[00:02:55]So, 1 g contains 4 / 100.

[00:03:01]And 300 g, so we multiply 300 on both sides.

[00:03:06]And here also.

[00:03:08]So, that's how we get 12 g. Question two, your mother gives you a bottle of orange juice concentrate to mix with water and serve it to your visiting friends. She asks you to mix two tablespoons of the concentrate with water in a glass tumbler. If each tablespoon measures 15 ml and you make 150 ml of juice per person, what is the percentage v/v of orange juice concentrate in the mixture you prepared?

[00:03:38]Answer, given, one tablespoon 15 ml. Two tablespoons is equals to 2 * 15 = 30 ml.

[00:03:48]Volume of orange juice concentrate 30 ml. Total volume of the juice 150 ml.

[00:03:54]So, formula, volume by volume percentage is equals to volume of solute divided by volume of solution multiplied by 100.

[00:04:03]So, volume of solute is 30 ml, so we write here 30. Volume of solution 150 ml, so we write here 150 multiplied by 100, so we get 20%.

[00:04:15]Therefore, the concentration of orange juice concentrate is 20% v/v, volume by volume.

[00:04:23]Question three, vinegar used as a food preservative and addictive contains 5% volume by volume acetic acid. Glacial acetic acid is a liquid that is 100% acetic acid. If you want to make vinegar from glacial acetic acid, how would you proceed? Answer, vinegar contains 5% volume by volume acetic acid. This means 100 ml of vinegar contains 5 ml of acetic acid. To prepare vinegar from glacial acetic acid, take 5 ml of glacial acetic acid and add enough water to make the final volume 100 ml. So, the mixture will contain 5 ml acetic acid plus 95 ml water is equals to 100 ml vinegar. Therefore, vinegar can be prepared by diluting glacial acetic acid with water in the ratio 5:95.

[00:05:20]Note, glacial acetic acid is corrosive, so it should be handled carefully under adult or teacher supervision.

[00:05:28]Think as a scientist, page 79. If a hot saturated solution of copper sulfate is cooled rapidly in ice cold water, smaller and less well-formed crystals will form than if it is cooled slowly at room temperature. How would you design and perform an experiment to test this hypothesis? Hint, prepare a hot saturated solution of copper sulfate and divide it into two equal parts. Answer, to test this hypothesis, we can perform the following experiment. One, prepare a hot saturated solution of copper sulfate in water. Two, filter the hot solution to remove any insoluble impurities.

[00:06:09]Three, divide the hot saturated solution into two equal parts in two clean beakers. Four, label them as beaker A and beaker B. Five, keep beaker A undisturbed at room temperature and allow it to cool slowly. Six, place beaker B in ice cold water so that it cools rapidly. Seven, after crystals form in both beakers, filter and collect the crystals. Eight, observe the size and shape of crystals from both beakers.

[00:06:40]Observation, in beaker A, slow cooling will form larger, shiny, and well-shaped crystals. In beaker B, rapid cooling will form smaller and less well-formed crystals. Conclusion, slow cooling gives particles more time to arrange themselves in a regular pattern, so larger and better shaped crystals are formed. Rapid cooling does not give enough time for proper arrangement, so smaller and irregular crystals are formed.

[00:07:08]Pause and ponder, page 79, question four. Refer to the solubility curves given in activity 5.2. If equal masses of hot saturated solutions of compounds A and B are cooled from 80° C to 60° C, which solution is likely to deposit more solid? Answer, the saturated solution of compound B is likely to deposit more solid.

[00:07:34]This is because the solubility of compound B changes more with temperature than that of compound A. When the saturated solution of compound B is cooled from 80° C to 60° C, a larger amount of solute becomes insoluble and separates out as solid. Compound A shows very little change in solubility between 80° C and 60° C, so it will deposit little or no solid.

[00:08:02]Question five, will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.

[00:08:12]Answer, yes, the size of common salt crystals may change with the rate of evaporation. If evaporation is slow, the salt particles get more time to arrange themselves properly. Therefore, larger and well-formed crystals may be formed.

[00:08:27]If evaporation is fast, the salt particles do not get enough time to arrange themselves in a regular pattern.

[00:08:34]Therefore, smaller and less well-formed crystals may be formed. So, slow evaporation generally produces larger crystals while fast evaporation generally produces smaller crystals.

[00:08:46]Pause and ponder, page 82, question six.

[00:08:50]State whether the following statements are true or false. Also, correct the false statements. Answer, first, salt can be separated from a salt solution by evaporation or distillation. Answer, true. Salt can be separated from salt solution by evaporation because water evaporates and salt remains behind. Salt can also be separated by distillation if we want to recover water along with salt. In distillation, water vaporizes, condenses, and is collected separately while salt remains in the flask.

[00:09:25]Second, distillation can be used for separation of two liquids even when these have the same boiling point.

[00:09:32]Answer, false. Correct statement, distillation can be used to separate two liquids only when they have different boiling points. Simple distillation is suitable when the boiling point difference is about 25° C or more.

[00:09:47]Third, in paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment. Answer, false. Correct statement, in paper chromatography the solvent level should be below the sample spot at the beginning of the experiment.

[00:10:05]If the solvent level is above the spot, the sample may dissolve directly into the solvent instead of moving up the paper and separating properly.

[00:10:14]Fourth, evaporation and crystallization are the same processes. Answer, false.

[00:10:20]Correct statement, evaporation and crystallization are different processes.

[00:10:25]In evaporation, the solvent changes into vapor and leaves the solute behind. In crystallization, pure solid crystals are formed from a saturated solution, usually by cooling or slow evaporation.

[00:10:39]What if, page 83, question two immiscible liquids of the same density are mixed in a separating funnel? How will the layers form? Answer, if two immiscible liquids have the same density, they will not form clear layers based on density in the usual way.

[00:10:57]Normally, immiscible liquids form layers because the denser liquid settles at the bottom and the less dense liquid stays on top. But, if both liquids have the same density, there will be no clear reason for one liquid to settle below the other. So, in a separating funnel, they may remain as droplets or an unstable mixture rather than forming two clearly separated layers. Therefore, separation using a separating funnel would be difficult.

[00:11:26]Pause and ponder, page 84, question seven. Why do immiscible liquids form two separate layers in a separating funnel? Answer, immiscible liquids form two separate layers because they do not dissolve in each other.

[00:11:41]They also usually have different densities. The liquid with higher density forms the lower layer, while the liquid with lower density forms the upper layer. For example, in a mixture of mustard oil and water, water is denser, so it forms the lower layer.

[00:11:58]Mustard oil is less dense, so it forms the upper layer.

[00:12:02]Question eight, is sublimation different from evaporation? Justify. Answer, yes, sublimation is different from evaporation.

[00:12:11]Sublimation, it is the change of a solid directly into vapor. Evaporation, it is the change of a liquid into vapor.

[00:12:19]Sublimation, the liquid state is not formed. Evaporation, the substance is already in liquid state. Sublimation, it occurs in sublimable solids.

[00:12:29]Evaporation, it occurs in liquids.

[00:12:32]Sublimation examples, camphor, naphthalene, dry ice. Evaporation examples, water, alcohol, acetone.

[00:12:41]Therefore, sublimation and evaporation are different processes because sublimation involves solid to vapor, while evaporation involves liquid to vapor.

[00:12:52]Pause and ponder, page 88, question nine. Clouds are made up of tiny water droplets or ice crystals floating in the air. Based on what you know about solutions, suspensions, and colloids, what type of mixture do you think clouds are and why? Answer, clouds are colloids. Clouds contain tiny water droplets or ice crystals dispersed in air. These particles are small enough to remain suspended in air and do not settle quickly. They can also scatter light. Therefore, clouds are colloidal mixtures in which water droplets or ice crystals form the dispersed phase and air acts as the dispersion medium.

[00:13:37]Question 10, why do cities with a lot of smoke and dust in the air often look hazy? Answer, cities with a lot of smoke and dust look hazy because smoke and dust particles scatter light. This scattering of light by suspended particles is called the Tyndall effect.

[00:13:55]Due to this effect, light spreads in different directions, reducing visibility and making the air appear hazy.

[00:14:03]Page 88, activity 5.9. Complete table 5.1 and review what you have learned about solutions, suspensions, and colloids. Answer, properties of different types of mixtures.

[00:14:17]First, nature, solution homogeneous, suspension heterogeneous, colloid appears homogeneous but actually heterogeneous.

[00:14:25]Particle size, solution very small, less than 1 nanometer. Suspension, large, more than 1,000 nanometer. Colloid, intermediate, 1 to 1,000 nanometer.

[00:14:38]Visibility, solution, particles are not visible. Suspension, particles are visible to naked eye. Colloid, particles are not visible to naked eye.

[00:14:49]Separation by filtration, solution, cannot be separated by ordinary filtration. Suspension, can be separated by filtration. Colloid, cannot be separated by ordinary filtration.

[00:15:02]Settling, solution, particles do not settle down. Suspension, particles settle down on standing. Colloid, particles do not settle down. Tyndall effect, solution, does not show Tyndall effect. Suspension, shows Tyndall effect. Colloid, shows Tyndall effect.

[00:15:21]Next, revise, reflect, refine. Question one, which of the following mixtures are correctly classified as homogeneous HM and heterogeneous HT? Choose the correct option. Answer, the correct option is fourth, muddy water heterogeneous, milk heterogeneous, blood heterogeneous, brass homogeneous.

[00:15:43]Reason, muddy water is a heterogeneous mixture because mud particles are suspended in water and can settle down.

[00:15:51]Milk is a colloid, it appears uniform, but it is actually heterogeneous because tiny fat droplets are dispersed in water.

[00:16:00]Blood is also a colloid and hence a heterogeneous mixture. Brass is an alloy of copper and zinc. It is a homogeneous mixture because its components are uniformly mixed. Therefore, option four is correct.

[00:16:15]Question two, choose the correct options and explain the reason for the correct and incorrect options.

[00:16:22]Which among the following mixtures show the Tyndall effect? A mixture of A, air and dust particles. B, copper sulfate and water. C, starch and water. D, acetone and water. Answer, the correct option is third, A and C.

[00:16:39]Reason, the Tyndall effect is shown by colloids and suspensions because their particles are large enough to scatter light. A, air and dust particles shows Tyndall effect because dust particles suspended in air scatter light. B, copper sulfate and water does not show Tyndall effect because it forms a true solution. C, starch and water shows Tyndall effect because starch in water forms a colloidal mixture.

[00:17:08]D, acetone and water does not show Tyndall effect because acetone and water form a homogeneous solution. Therefore, only A and C show the Tyndall effect.

[00:17:20]Question three, a mixture can be categorized as a solution, a suspension, or a colloid, each possessing distinct properties. Utilize the words or phrases provided in the box to fill the table 5.2. Words and phrases may be used more than once.

[00:17:37]So, friends, these are the words and phrases and we have to complete the table 5.2. Answer, solution properties, small-sized particles less than 1 nanometer diameter, particles remain evenly distributed, transparent, does not settle down, cannot be separated by filtration. Examples, salt solution, brass.

[00:18:02]Next, suspension properties, large-sized particles more than 1,000 nanometer in diameter, settles down when left undisturbed, separates by filtration, scatters light, heterogeneous mixture. Examples, sand in water, mud. Colloid properties, moderate-sized particles 1 to 1,000 nanometer, particles remain evenly distributed, does not settle down, scatters light, heterogeneous mixture.

[00:18:32]Examples, milk, smoke, butter.

[00:18:36]So, friends, you can write any two to three properties from the given five.

[00:18:41]Question four, solve the following problems. First, a cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of all-purpose flour and 5 g of sodium hydrogen carbonate. Express the concentration of each component in the mixture using an appropriate method.

[00:19:01]Answer, since all the components are solids, the appropriate method is mass by mass percentage. Percent m/m.

[00:19:10]Total mass of mixture, 75 g plus 420 g plus 5 g is equals to 500 g. So, 75 g is sugar, 420 g all-purpose flour, and 5 g sodium hydrogen carbonate. So, when we add them, it becomes 500 g. Sugar, 75 divided by 500 multiplied by 100. So, 75 is the weight of sugar. So, 75 is for sugar divided by 500 total solution multiplied by 100 is equals to 15%.

[00:19:43]All-purpose flour, 420 divided by 500 multiplied by 100. 420 g for all-purpose flour. So, we get by solving this 84% sodium hydrogen carbonate five 5 g of sodium hydrogen carbonate divided by total solution 500 x 100 is equals to 1%. Therefore, the composition of the mixture is sugar 15% m/m, all purpose flour 84% m/m, sodium hydrogen carbonate 1% m/m.

[00:20:19]So friends, I hope you have understood this.

[00:20:22]Second, a brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.

[00:20:32]Answer given mass of brass 120 g, copper 70% by mass, mass of copper 70 / 100 x 120 is equals to 84 g.

[00:20:47]So friends, brass is made up of copper and zinc. Copper by mass is 70%.

[00:20:54]So the remaining zinc is 30%.

[00:20:57]Because total is 100%.

[00:21:00]So total mass of brass is given 120 g.

[00:21:03]So 100% is equals to 120 g. So we can write 1% is equals to 120 / 100.

[00:21:17]Now we have to find out copper 70% and zinc 30%. So 70% we multiply 70 on both sides. 70 x 1% and 70 here.

[00:21:30]So we get 70% is equals to 84 g. That is the same thing written here.

[00:21:39]Now for mass of zinc 120 - 84 g is equals to 36 g. So friends, we know the total weight of brass is 120 g and we have already found mass of copper 84 g. So if we subtract mass of brass by mass of copper, we will get mass of zinc. That's what we have done here.

[00:22:04]Therefore copper is equals to 84 g, zinc is equals to 36 g.

[00:22:10]I hope you have understood it. Question five, the label on a cooking oil pack says 1 L 910 g. If this oil is mixed with water, will it form a separate layer? If so, which substance will be on top? How will you separate the two layers? Also, draw the diagram of the apparatus used. Answer, yes. Cooking oil will form a separate layer when mixed with water because oil and water are immiscible liquids.

[00:22:40]Mass of 1 L oil is equals to 910 g. So density of oil is equals to 910 g per liter is equals to 0.91 g per milliliter.

[00:22:52]Density of water is about 1 g per milliliter. Since oil is less dense than water, oil will float on top and water will form the lower layer.

[00:23:03]The two layers can be separated using a separating funnel.

[00:23:07]Method So friends, this is the diagram.

[00:23:11]Pour the mixture of oil and water into a separating funnel.

[00:23:15]So friends, this is a separating funnel.

[00:23:17]Allow it to stand undisturbed. Two layers will form, oil on the top, this one, cooking oil, and water at the bottom. Open the stopcock and collect the lower water layer. This is the stopcock.

[00:23:32]Close the stopcock before the oil layer comes out.

[00:23:36]Collect the oil separately. So friends, I have created a labeled diagram for you for better understanding. This is a stopper, this is separating funnel, this is cooking oil layer less than 0.91 g per ml. This is water layer more dense 1 g per ml. This is stopcock or you can say tap. Oil and water are immiscible liquids. So water more dense flows out first and is collected in the receiving flask.

[00:24:05]Key information, 1 L of cooking oil is equals to 910 g. So density of oil is equals to 0.91 g per ml. Since oil is less dense than water, oil floats on top. Collecting the oil, after collecting the water, close the stopcock. Place a clean beaker or flask below the funnel. Reopen the stopcock to let the oil top layer flow out and collect it separately. Question six, assertion A, solutions do not exhibit the Tyndall effect. Reason R, the particles in solutions are larger than 100 nm, so they cannot scatter light.

[00:24:41]Answer, the correct option is three, A is true but R is false. Explanation, assertion is true because solutions do not show the Tyndall effect. The particles in a true solution are very small, usually less than 1 nm, so they cannot scatter light. Reason is false because it says that particles in solutions are larger than 100 nm. This is incorrect. Particles in solutions are very small, not larger than 100 nm.

[00:25:11]Therefore, A is true but R is false.

[00:25:14]Question seven, how would you separate the mixture given in table 5.3? Mention the reason for choosing your method. If a mixture cannot be separated, explain why.

[00:25:24]Answer mixture mud from muddy water, method of separation sedimentation, decantation, filtration or coagulation. Reason for selection, mud particles are insoluble and can settle or be filtered. Alum can be used to coagulate fine particles.

[00:25:43]Next, plasma from other components in the blood sample, method of separation centrifugation.

[00:25:50]Reason for selection, blood components have different densities. On spinning, heavier cells settle and plasma remains above. Next, naphthalene and sand, method of separation sublimation. Reason for selection, naphthalene sublimes on heating but sand does not.

[00:26:08]Next, chalk powder and common salt, method of separation dissolution in water, filtration, then evaporation crystallization.

[00:26:17]Reason for selection, common salt dissolves in water, chalk powder does not. Chalk is removed by filtration and salt is obtained from filtrate.

[00:26:28]Next, common salt and water, method of separation evaporation or distillation. Reason for selection, evaporation gives a salt, distillation gives both salt and water separately.

[00:26:41]Next, oil from water, method of separation separating funnel. Reason for selection, oil and water are immiscible liquids and have different densities.

[00:26:51]Next, pigments of the flower, method of separation paper chromatography. Reason for selection, different pigments move at different rates on paper with the solvent.

[00:27:03]So friends, you can note it down.

[00:27:05]Question eight, two immiscible liquids A and B are present in a mixture. The boiling point of A is 60°C and the boiling point of B is 90°C.

[00:27:16]Suggest a method to separate them. Also, draw a labeled diagram of the method suggested. Answer, the mixture can be separated by simple distillation. The boiling point of liquid A is 60°C and that of liquid B is 90°C.

[00:27:33]The difference in boiling points is 90°C - 60°C is equals to 30°C.

[00:27:41]Since the difference is more than 25°C simple distillation can be used.

[00:27:47]Liquid A has a lower boiling point, so it will vaporize first. Its vapor will pass through the condenser, cool down and collect as liquid A in the receiving flask. Liquid B will remain in the distillation flask.

[00:28:01]So friends, this is the simple diagram.

[00:28:03]Simple distillation of liquids A and B.

[00:28:06]This is the thermometer, this is delivering tube, distillation head. Here is mixture of liquid A and liquid B.

[00:28:14]These are the vapors of liquid A in orange, liquid A and in blue, flow of cooling water. This is cooling water.

[00:28:24]Cold water in, water out. And this is condenser.

[00:28:29]This is the receiving flask and this is distillate liquid A. Liquid A is collected here. Key idea, simple distillation is used to separate two miscible liquids when the difference in boiling points is 25°C or more. Liquid A boiling point 60°C, liquid B boiling point 90°C.

[00:28:49]On heating, liquid A vaporizes first because it has a lower boiling point 60°C.

[00:28:55]Its vapor passes into the condenser, cools down and is collected as liquid A here. Process, heat the mixture. Liquid A boils first at 60°C.

[00:29:06]Its vapor enters the condenser. Vapor cools and condenses. Liquid A is collected. Liquid B remains in the flask. Question nine, compare evaporation, crystallization and distillation. In which situation would you prefer each of these over the others? Answer Point of comparison meaning, evaporation process of converting liquid into vapor from the surface. Crystallization process of obtaining pure solid crystals from a saturated solution. Distillation process of vaporizing a liquid and condensing it back to liquid.

[00:29:41]Purpose, evaporation to obtain dissolved solid. Crystallization to obtain pure solid crystals, distillation to obtain liquid or separate miscible liquids.

[00:29:53]Solvent recovery, evaporation, solvent is not recovered. Crystallization, solvent is usually not the main product.

[00:30:01]Distillation, solvent or liquid is recovered.

[00:30:04]Purity of solid, evaporation, solid may contain impurities. Crystallization gives purer solid. Distillation used mainly for liquids. Example, evaporation getting salt from salt solution.

[00:30:18]Crystallization getting copper sulfate crystals. Distillation separating acetone and water.

[00:30:26]When to prefer evaporation? Evaporation is preferred when we only want the dissolved solid and do not need to recover the solvent. Example, getting salt from salt water.

[00:30:38]When to prefer crystallization?

[00:30:40]Crystallization is preferred when we want a pure solid in crystal form.

[00:30:44]Example, purifying copper sulfate or obtaining sugar crystals.

[00:30:49]When to prefer distillation?

[00:30:51]Distillation is preferred when we want to recover the liquid or separate two miscible liquids with different boiling points. Example, separating acetone and water or obtaining pure water from salt water.

[00:31:04]Question 10, blood is an example of a colloidal mixture.

[00:31:08]First, what would happen if blood behaved like a true suspension inside the body? Answer, if blood behaved like a true suspension inside the body, its particles would settle down when left undisturbed. This would be harmful because blood cells would not remain evenly distributed in plasma. As a result, blood flow would become uneven.

[00:31:31]Cells could settle in blood vessels.

[00:31:34]Transport of oxygen, nutrients, hormones, and waste materials would be disturbed. It could block small blood vessels and affect body functions.

[00:31:43]Therefore, blood must remain as a colloidal mixture so that its components remain properly dispersed.

[00:31:51]Second, in a blood sample, identify the dispersed phase and the dispersion medium.

[00:31:57]Answer, in blood, dispersed phase, blood cells such as red blood cells, white blood cells, and platelets. Dispersion medium, plasma.

[00:32:07]Question 11, you are given a mixture of sand, common salt, and naphthalene.

[00:32:12]Figure 5.25a, this one.

[00:32:15]The figure 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques. Answer, the correct sequence of separation techniques is sublimation, dissolution, filtration, evaporation, crystallization.

[00:32:35]Explanation, sublimation, heat the mixture gently, naphthalene sublimes and separates from sand and common salt.

[00:32:44]Next, dissolution, add water to the remaining mixture of sand and common salt. Common salt dissolves in water, but sand does not. Filtration, filter the mixture, sand remains as residue on the filter paper while salt solution passes through as filtrate. Evaporation, crystallization, evaporate the water from the salt solution to obtain common salt.

[00:33:09]Therefore, the three components are separated as naphthalene by sublimation, sand by filtration, common salt by evaporation or crystallization.

[00:33:20]So, friends, first one is this, sublimation.

[00:33:25]Second one is third, filtration.

[00:33:29]And the last one is second, evaporation or crystallization.

[00:33:35]Question 12, why is distillation an effective method for separating a mixture of water and acetone? Answer, distillation is effective for separating water and acetone because they are miscible liquids with different boiling points. Acetone boils at about 56°C, water boils at 100°C.

[00:33:57]The difference in their boiling points is 100°C - 56°C = 44°C.

[00:34:05]Since this difference is more than 25°C, simple distillation can be used. On heating, acetone vaporizes first because it has a lower boiling point. Its vapors are cooled in the condenser and collected as liquid acetone. Water remains in the distillation flask.

[00:34:24]Question 13, answer the following questions with the help of the data given in table 5.4.

[00:34:30]So, friends, this is the table 5.4, solubility of various salts in gram per 100 g of water at different temperatures.

[00:34:39]First, what mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40°C?

[00:34:48]Answer, at 40°C, solubility of potassium nitrate, 62 g per 100 g of water.

[00:34:57]So, friends, here you can see, at 40°C, potassium nitrate, 62 g.

[00:35:04]So, for 50 g of water, 62 / 100 * 50 = 31 g.

[00:35:13]Therefore, 31 g of potassium nitrate is needed to prepare its saturated solution in 50 g of water at 40°C.

[00:35:23]So, friends, don't worry if you don't understand this question. I will prepare a separate video for numericals of this chapter in Hindi explained video so that you can understand the numericals better.

[00:35:36]Second, a student makes a saturated solution of potassium chloride in water at 80°C and leaves the solution to cool at room temperature, 25°C.

[00:35:48]What would she observe as the solution cools? Explain. Answer, at 80°C, solubility of potassium chloride is 54 g per 100 g water. So, this is potassium chloride at 80°C, 54. At around 25°C, its solubility will be between its solubility at 25°C and 30°C.

[00:36:14]From the table, at 20°C, 35 g per 100 g of water. At 30°C, 37.4 g per 100 g water.

[00:36:25]Here it is given, at 30 g, 37.4, at 20 g, 35.

[00:36:30]So, at 25°C, only about 36 g of potassium chloride can remain dissolved in 100 g water.

[00:36:39]As the solution cools, its solubility decreases. Therefore, the extra potassium chloride will separate out as crystals.

[00:36:47]The student would observe crystals of potassium chloride forming in the solution.

[00:36:52]Third, what is the effect of a change in temperature on the solubility of salts?

[00:36:57]Also, compare the changes in the solubility of the four given salts with increasing temperature from 10°C to 80°C.

[00:37:06]Answer, generally, the solubility of salts in water increases with an increase in temperature. However, the increase is not the same for all salts.

[00:37:16]Comparison from 10°C to 80°C, increase, 167 - 21 = 146 g.

[00:37:25]So, here you can see, potassium nitrate, 80°C, 167.

[00:37:30]And at 10°C, 21. So, we differentiate these two, we get 146 g. It shows the maximum increase.

[00:37:40]Sodium chloride, solubility increases from 36 g to 37 g. Sodium chloride, 10°C, 36, 80°C, 37. Increase, 37 - 36 = 1 g. It shows very little increase.

[00:37:57]Potassium chloride, 10°C, 35, 80°C, 54.

[00:38:01]So, solubility increases from 35 g to 54 g. Increase, 54 - 35 = 19 g.

[00:38:09]Ammonium chloride, here, 10°C, 24, 80°C, 66. So, solubility increases from 24 g to 66 g. Increase = 66 - 24 = 42 g.

[00:38:24]Order of increase in solubility, potassium nitrate, ammonium chloride, potassium chloride, sodium chloride.

[00:38:31]Therefore, potassium nitrate is most affected by temperature, while sodium chloride is least effective.

[00:38:39]Question 14, three students, A, B, and C are preparing sugar solutions for an experiment.

[00:38:45]Student A dissolves 20 g of sugar in 80 g of water. Student B dissolves 20 g of sugar in 100 g of water.

[00:38:54]Student C dissolves 30 g of sugar in 80 g of water.

[00:38:58]First, calculate the mass percentage, percent m/m concentration of sugar in each student's solution. Answer, formula, mass percentage = mass of solute divided by mass of solution multiplied by 100.

[00:39:13]Student A, mass of sugar, 20 g, mass of water, 80 g. Mass of solution, 20 + 80 = 100 g. So, 20 / 100, mass of solute, sugar is 20, and solution, 100 * 100 = 20%.

[00:39:31]Student A's solution, 20% m/m.

[00:39:35]Student B, mass of sugar, 20 g, mass of water, 100 g. Total mass of solution, 20 + 100 = 120 g.

[00:39:45]So, by applying formula, solute is 20 g, solution 120, multiplied by 100 = 16.67%.

[00:39:53]Student B solution = 16.67% m/m.

[00:39:59]Student C, mass of sugar 30 g, mass of water 80 g, total mass of solution 30 + 80 = 110 g.

[00:40:09]So, mass of solute, sugar, 30 divided by mass of solution 110 multiplied by 100 = 27.27%.

[00:40:19]Student C solution = 27.27% m/m.

[00:40:25]Second, whose solution is the most concentrated? Explain why. Answer, student C solution is the most concentrated because it has the highest mass percentage of sugar, that is 27.27% mass by mass.

[00:40:43]Question 15, examine figure 5.26, this one.

[00:40:47]First, identify the separation technique marked as S.

[00:40:51]Answer, the separation technique marked as S is distillation.

[00:40:56]So, friends, this is distillation.

[00:40:59]Second, label the apparatus A, B, and C.

[00:41:03]A, B, and C. Answer, So, friends, this is distillation. A is distillation flask, B is condenser, C is receiving flask or conical flask.

[00:41:16]Third, which of the following mixtures can be separated by the technique identified above? Use the data given in table 5.5. So, friends, this is the table 5.5, boiling points of some compounds.

[00:41:29]Answer, distillation is used to separate two miscible liquids whose boiling points differ by about 25° C or more, or a liquid from a solution containing dissolved solid. Now, let us check each mixture. A, water acetone. Water 100° C, acetone 56° C, difference 44° C, can be separated by distillation.

[00:41:56]B, water salt. Salt is a dissolved solid and water is a liquid, can be separated by distillation if we want to recover water.

[00:42:06]C, acetone alcohol. Acetone 56° C, alcohol 78° C, difference 22° C, simple distillation is not suitable because the difference is less than 25° C.

[00:42:21]D, sand salt. This is a solid solid mixture, it cannot be separated by distillation.

[00:42:28]E, alcohol chloroform. Alcohol 78° C, chloroform 61° C, difference 17° C, simple distillation is not suitable because the difference is less than 25° C.

[00:42:43]F, alcohol benzene. Alcohol 78° C, benzene 80° C, difference 2° C, simple distillation is not suitable because the boiling points are very close.

[00:42:56]The mixtures that can be separated by the technique shown are A, water acetone, B, water salt.

[00:43:04]Next, the quest continues. Question, can we create artificial blood that works just as real blood for all patients?

[00:43:12]Answer, scientists are trying to develop artificial blood, but creating artificial blood that works exactly like real blood for all patients is very difficult.

[00:43:23]The real blood performs many important functions. It carries oxygen, carbon dioxide, nutrients, hormones, and waste materials. It also helps in fighting infections, clotting during injuries, and maintaining body temperature.

[00:43:39]Artificial blood can be designed mainly to carry oxygen, but it cannot yet perform all the functions of real blood.

[00:43:46]For example, it cannot fully replace white blood cells, platelets, plasma proteins, and immune functions. If scientists create safe and effective artificial blood in the future, it may help in emergencies, accidents, surgeries, and places where blood banks are not easily available. It may also reduce problems related to blood group matching and shortage of donated blood.

[00:44:12]However, artificial blood must be carefully tested to make sure it is safe, non-toxic, long-lasting, and suitable for the human body. Therefore, we may be able to create artificial blood for some uses, especially oxygen transport, but making artificial blood that works exactly like real blood for all patients is still a major scientific challenge.

[00:44:37]So, friends, we have completed all the question answers from chapter five, exploring mixtures and their separation.

[00:44:44]We have already studied notes and summary of this chapter. You can find the link in the description and on the screen. If you like the video, kindly like hype the video and share it with your friends. Also, subscribe to my channel and press the bell icon to receive new video notifications. We will meet in the next video. Thank you for watching.

[00:45:05]We will complete chapter three in the following videos.