Exploring Mixtures and their Separation Class 9

20 Questions | 46 Marks

Define homogeneous and heterogeneous mixtures with two examples for each. 2marks

Explain the concept of concentration of a solution. Give any two methods to express the concentration of a solution. 3marks

Calculate the mass by mass percentage of a solution prepared by dissolving 50 g of common salt in 200 g of water. 2marks

What is a saturated solution? Explain the effect of temperature on the solubility of a solid in a liquid. 2marks

Briefly explain the process of crystallization with an example. 3marks

What is distillation? Under what conditions is it used to separate miscible liquids? 2marks

(Competency-based/Situational) A student observed that a black sketch pen ink, when placed on chromatographic paper and exposed to water vapor rising from the bottom, separated into different colors. Explain the principle behind this observation and state what type of mixture this ink is likely to be. 3marks

Describe the process of separating two immiscible liquids using a separating funnel. Draw a labelled diagram of the apparatus. 3marks

What is sublimation? Give two examples of substances that sublime. 2marks

(Competency-based/Application) A milkman adds a little water to milk to increase his profit. Which property of milk, that makes it a colloid, would he be exploiting, and how does this property help in his deception? 3marks

Explain the Tyndall effect with a suitable example. Why do solutions not exhibit this effect? 3marks

Differentiate between a solution, a suspension, and a colloid based on particle size and settling of particles. 3marks

(Competency-based/Data Interpretation) Table 5.4 shows the solubility of four salts at different temperatures. If you cool a saturated solution of potassium nitrate from 60°C to 40°C, how much solid will be deposited? 2marks

(Competency-based/Scenario) A village has access to a well that provides muddy water. The villagers often add 'fitkari' (alum) to the water before using it. Explain the scientific principle behind this practice and what happens to the mud particles. 3marks

Why is it necessary to mention the temperature when stating the solubility of a substance? 1mark

What is an alloy? Give two examples of alloys and mention their components. 2marks

(Competency-based/Application) You want to prepare pure copper sulfate crystals from an impure sample. Which method would you use and why? Briefly outline the steps involved. 3marks

(Competency-based/Situational) In a laboratory, a student tries to separate a mixture of sand and camphor by heating. What will be observed, and what is the scientific term for this process? 2marks

Name the process used to separate components of blood like plasma from red blood cells. Explain the principle behind this process. 2marks

(Competency-based/Data Interpretation) Refer to Table 5.4. Which of the four salts shows the least change in solubility with a change in temperature? Justify your answer. 2marks

Answers

Homogeneous Mixture: A mixture that has a uniform composition throughout. The components are uniformly distributed and indistinguishable.

Example 1: Salt solution
Example 2: Air (a mixture of gases like nitrogen, oxygen, etc.)
Heterogeneous Mixture: A mixture that does not have a uniform composition throughout. The components are not uniformly distributed and are often visible.
Example 1: Sand and water
Example 2: Oil and water

Concentration of a Solution: The concentration of a solution refers to the amount of solute dissolved in a given amount of solvent or solution. It expresses how much of a substance is present in a mixture.Methods to express concentration:

Mass by mass percentage (% m/m): It tells us how many grams of solute are present in 100 grams of the total solution.

Mass by mass percentage = \frac{Mass of solute}{Mass of solution} \times 100

Mass by volume percentage (% m/v): It tells us how many grams of solute are present in 100 millilitres of the solution.

Mass by volume percentage = \frac{Mass of solute}{Volume of solution} \times 100

Volume by volume percentage (% v/v): It tells us how many millilitres of solute are present in 100 millilitres of the solution.

Volume by volume percentage = \frac{Volume of solute}{Volume of solution} \times 100

Mass of salt (solute) = 50 g Mass of water (solvent) = 200 g Total mass of solution = Mass of solute + Mass of solvent = 50 g + 200 g = 250 g

Mass by mass percentage = \frac{50 g}{250 g} \times 100 = \frac{1}{5} \times 100 = 20 % m/m

Saturated Solution: A solution in which no more solute can be dissolved at a given temperature is called a saturated solution.Effect of temperature on solubility: The solubility of a solid solute in a liquid solvent generally increases with an increase in temperature. This is because at higher temperatures, the particles have more kinetic energy, which helps in overcoming the forces of attraction between solute particles and dissolving them in the solvent.

Crystallization: Crystallization is a process of forming pure solid crystals from a saturated solution. When a hot saturated solution of a substance is cooled slowly, the solute starts separating out in the form of pure crystals as its solubility decreases with a decrease in temperature.Example: Obtaining pure copper sulfate crystals from an impure sample. Dissolve impure copper sulfate in a minimum amount of hot water, filter out insoluble impurities, and then cool the solution slowly. Pure crystals of copper sulfate will form.

Distillation: Distillation is a process used to separate a homogeneous mixture of two miscible liquids with a sufficient difference (at least 25 °C) in their boiling points. It involves heating the mixture to vaporize the liquid with the lower boiling point, and then cooling the vapor to condense it back into a liquid, which is collected separately.Conditions for use: Distillation is used when the boiling points of the two miscible liquids differ significantly (by at least 25 °C). If the difference is less than 25 °C, fractional distillation is used.

(Competency-based/Situational) The ink from a sketch pen separating into different colors indicates that the ink is not a single substance but a mixture of different colored components. The separation of these components by a solvent (water) moving up the paper is due to the differences in their solubility and their affinity for the paper and the solvent. This process is called paper chromatography. Therefore, the ink is likely a heterogeneous mixture of different colored dyes, which can be separated by chromatography.

Separating Immiscible Liquids using a Separating Funnel:

Pour the mixture of two immiscible liquids (e.g., oil and water) into a separating funnel.
Allow the mixture to stand undisturbed for some time. The liquids will separate into two distinct layers based on their densities. The denser liquid will form the lower layer, and the less dense liquid will form the upper layer.
Open the stopcock of the separating funnel carefully. Collect the lower, denser liquid in a beaker.
Close the stopcock just as the interface between the two liquids reaches the stopcock.
Then, open the stopcock again and collect the upper, less dense liquid in another beaker.
Diagram: A labelled diagram of a separating funnel showing the two layers (e.g., oil on top, water at the bottom) and the process of draining the lower layer.

Sublimation: Sublimation is the process where a solid changes directly into a gas on heating, without passing through the liquid state. On cooling, the gas changes back into a solid without becoming a liquid (deposition).

Example 1: Camphor
Example 2: Naphthalene

(Competency-based/Application) Milk is a colloid where fat globules and other proteins are dispersed in water. The dispersed particles in a colloid are large enough to scatter light (Tyndall effect), making the beam visible. A milkman adds water to milk. This addition of water, while potentially diluting the milk, doesn't make it appear as a suspension because the milk fats and proteins are still dispersed colloidally, and the mixture still exhibits the Tyndall effect to some extent if the dilution is not too extreme. However, the dilution reduces the concentration of milk solids, making the milk less rich. The deception is exploitable because the visual appearance and the Tyndall effect might not immediately indicate the dilution, especially if the milkman uses a small amount of water.

Tyndall Effect: The scattering of light by particles in a colloid or suspension is called the Tyndall effect. When a beam of light passes through a colloid, the particles scatter the light, making the path of the beam visible. For example, when a beam of sunlight passes through the canopy of a dense forest, the path of light is visible due to scattering by dust particles and water droplets in the air.Why solutions do not exhibit Tyndall effect: In true solutions, the solute particles are very small (less than 1 nm in diameter) and are uniformly dispersed. These particles are too small to scatter light effectively, hence the path of light is not visible.

Comparison of Solution, Suspension, and Colloid:

Property	Solution	Suspension	Colloid
Particle Size	Less than 1 nm	More than 1000 nm (1000 Å)	1 nm to 1000 nm (10 Å to 10000 Å)
Settling of Particles	Do not settle down	Settle down when left undisturbed	Do not settle down on standing
Visibility	Particles not visible even under microscope	Particles visible to the naked eye	Particles visible only under microscope
Filtration	Cannot be separated by filtration	Separated by filtration	Cannot be separated by filtration
Tyndall Effect	Does not exhibit	Exhibits	Exhibits

(Competency-based/Data Interpretation)From Table 5.4, at 40 °C, the solubility of potassium nitrate is 62 g per 100 g of water.This means that in 50 g of water, the mass of potassium nitrate needed to prepare a saturated solution would be half of that required for 100 g of water.Mass of

K N O_{3}

needed =

\frac{1}{2} \times 62 g = 31 g

.When a saturated solution at 60 °C is cooled to 40 °C, the solubility decreases.Mass of

K N O_{3}

at 60 °C (in 100 g water) = 106 g.Mass of

K N O_{3}

at 40 °C (in 100 g water) = 62 g.Decrease in solubility = 106 g - 62 g = 44 g per 100 g of water.If the initial saturated solution was made with 100 g of water, 44 g of solid would be deposited.Since the question implies a saturated solution is cooled, but doesn't specify the amount of water, we assume it refers to the change in solubility. The question is ambiguous about the initial quantity of solution. If we interpret it as "what mass of solid will be deposited per 100 g of water when cooling from 60°C to 40°C", then the answer is 44g. However, if it implies a specific, but unstated, quantity of solution, it cannot be solved without that information. Given the context of other questions, it's most likely asking for the change in solubility. Let's re-evaluate the question to see if it refers to a specific amount of solution. The question says "If you cool a saturated solution of potassium nitrate from 60°C to 40°C". This usually implies the total amount of solute that precipitates out. Without knowing the initial mass of the solution or the initial mass of the solute and solvent, we cannot give a definitive mass.Let's assume the question is flawed or implying a typical scenario. A common interpretation in such problems is to calculate the amount deposited per 100g of water if the initial solution was made with 100g of water. In that case:Mass deposited = Solubility at 60°C - Solubility at 40°C = 106 g - 62 g = 44 g.If we assume the question is asking how much solid would be deposited from a saturated solution prepared with 50g of water at 60°C, then:Mass of

K N O_{3}

dissolved at 60°C in 50g water =

\frac{106 g}{100 g water} \times 50 g water = 53 g

.Mass of

K N O_{3}

that can remain dissolved at 40°C in 50g water =

\frac{62 g}{100 g water} \times 50 g water = 31 g

.Mass deposited = 53 g - 31 g = 22 g.Given the ambiguity, and the simpler calculation for "per 100g of water", let's go with the first interpretation if the question expects a single numerical answer. However, the phrasing "how much solid will be deposited" suggests a specific quantity. The problem is poorly worded. Let's assume it implies that the initial solution was prepared with 100g of water.

Revised Answer for Q13: When a saturated solution of potassium nitrate prepared with 100 g of water at 60 °C is cooled to 40 °C, the mass of solid deposited will be the difference in solubility at these temperatures.
Mass deposited = Solubility at 60 °C - Solubility at 40 °C
Mass deposited = 106 g - 62 g = 44 g.

(Competency-based/Scenario) The practice of adding 'fitkari' (alum) to muddy water is based on the process of coagulation. Alum acts as a coagulant. When added to water containing suspended mud particles, alum causes the fine suspended particles to clump together, forming larger, heavier aggregates. These larger aggregates (called flocs) then settle down at the bottom of the container due to gravity (sedimentation). This makes the water clearer and easier to separate from the settled impurities by decantation or filtration.

It is necessary to mention the temperature when stating the solubility of a substance because the solubility of most solid solutes in liquid solvents generally increases with an increase in temperature. Conversely, the solubility of gases in liquids generally decreases with an increase in temperature. Therefore, the amount of solute that can dissolve in a given amount of solvent is dependent on the temperature.

Alloy: An alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal, prepared by melting them together at high temperatures. Alloys are generally prepared to obtain materials with properties superior to those of the constituent elements, such as increased strength, hardness, or resistance to corrosion.

Example 1: Brass (Copper and Zinc)
Example 2: Bronze (Copper and Tin)
Example 3: Stainless Steel (Iron, Chromium, Nickel, Carbon)

(Competency-based/Application) I would use the crystallization method to prepare pure copper sulfate crystals from an impure sample.Reason: Crystallization is effective for purifying solids from homogeneous mixtures. It relies on the difference in solubility of the desired substance and impurities at different temperatures. Impurities that are less soluble or insoluble can be removed by filtration, and the desired pure substance can be obtained as crystals when the solution is cooled.Steps:

Dissolve the impure copper sulfate in a minimum amount of hot water to form a saturated solution.

Filter the hot solution to remove any insoluble impurities.

Allow the hot, saturated solution to cool slowly without disturbing it. Pure copper sulfate crystals will form as the solution cools.

Filter the crystals to separate them from the remaining solution (mother liquor).

Wash the crystals with a small amount of cold water and then dry them.

(Competency-based/Situational) When a mixture of sand and camphor is heated, the camphor will turn directly into a vapor (gas) without melting. This process is called sublimation. The sand, which does not sublime, will remain in the china dish. The camphor vapor will then rise and cool on the inner walls of the inverted funnel, where it will deposit back into solid camphor.

Process to separate components of blood: The process used to separate components of blood, such as plasma from red blood cells, is centrifugation.Principle: Centrifugation uses rapid spinning of the sample in a centrifuge. This high-speed rotation creates a centrifugal force that pushes the heavier particles towards the bottom of the container, while the lighter components remain at the top. In blood, the red blood cells are denser and heavier than plasma, so they settle at the bottom, allowing plasma to be collected from the top.

(Competency-based/Data Interpretation) To determine which of the four salts shows the least change in solubility with a change in temperature, we need to compare the difference in solubility at the extreme temperatures (10 °C and 80 °C) or the slope of their solubility curves if they were plotted.Let's look at the change from 10 °C to 80 °C:

Potassium nitrate ( $K N O_{3}$ ): Solubility change = 167 g - 21 g = 146 g
Sodium chloride (NaCl): Solubility change = 37 g - 36 g = 1 g
Potassium chloride (KCl): Solubility change = 54 g - 35 g = 19 g
Ammonium chloride ( $N H_{4} C l$ ): Solubility change = 66 g - 24 g = 42 g

Comparing these changes, Sodium chloride (NaCl) shows the least change in solubility (only 1 g) with a significant increase in temperature from 10 °C to 80 °C. This indicates that its solubility is least affected by temperature variations among the given salts.

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Exploring Mixtures and their Separation — Important Questions

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