Exploring Mixtures and their SeparationClass 9 Science NCERT Solutions
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Solution 1 of 15
Q1Revise, Reflect, Refine
Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? Choose the correct option.
(i)
Air-Hm, Milk-Ht, Sugar solution-Hm, Smoke-Hm
(ii)
Brass - Ht, Fog - Ht, Vinegar - Ht, Muddy water - Hm
(iii)
Copper sulfate solution-Hm, Salt solution-Hm, Milk-Hm, Bronze-Hm
(iv)
Muddy water - Ht, Milk - Ht, Blood - Ht, Brass - Hm
Solution
The correct option is (iv) Muddy water - Ht, Milk - Ht, Blood - Ht, Brass - Hm.
Explanation:
- Muddy water is a heterogeneous (Ht) mixture because the mud particles are suspended in water and are not uniformly distributed. They can be seen and will settle over time.
- Milk is a colloid, which is a type of heterogeneous (Ht) mixture. Although it appears uniform, its particles (fat globules) are dispersed and not truly dissolved. They scatter light (Tyndall effect).
- Blood is also a colloid, making it a heterogeneous (Ht) mixture. It consists of cells (red blood cells, white blood cells, platelets) suspended in plasma.
- Brass is an alloy of copper and zinc. Alloys are solid solutions and are considered homogeneous (Hm) mixtures because the constituent metals are uniformly mixed at the atomic level.
Why other options are incorrect:
- (i) Smoke is a heterogeneous mixture (solid particles in a gas), not homogeneous.
- (ii) Brass is a homogeneous mixture, not heterogeneous. Vinegar (acetic acid in water) is a homogeneous mixture, not heterogeneous. Muddy water is heterogeneous, not homogeneous.
- (iii) Milk is a heterogeneous mixture (colloid), not homogeneous.
Q2Revise, Reflect, Refine
Choose the correct options, and explain the reason for the correct and incorrect options. 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
(i) a and b (ii) b and d (iii) a and c (iv) c and d
Solution
The correct option is (iii) a and c.
Explanation:
The Tyndall effect is the scattering of a light beam by particles in a colloid or a suspension, making the path of light visible. True solutions do not show this effect because their particles are too small to scatter light.
- (a) Air and dust particles: This is a suspension. The dust particles are large enough to scatter light. Therefore, it shows the Tyndall effect.
- (b) Copper sulfate and water: This forms a true solution. The copper sulfate particles (ions) are very small (less than 1 nm) and do not scatter light. Therefore, it does not show the Tyndall effect.
- (c) Starch and water: This forms a colloid. The starch particles are intermediate in size (1-1000 nm) and are large enough to scatter light. Therefore, it shows the Tyndall effect.
- (d) Acetone and water: This forms a true solution as they are miscible liquids. The particles are of molecular size and do not scatter light. Therefore, it does not show the Tyndall effect.
Since mixtures (a) and (c) are a suspension and a colloid, respectively, they will both exhibit the Tyndall effect.
Q3Revise, Reflect, Refine
A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Utilise the words or phrases provided in the box to fill in the Table 5.2. Words and phrases may be used more than once. Words and Phrases Large-sized particles; Particles remain evenly distributed; Small-sized particles (less than 1 nm diameter); Moderate-sized particles (1-1000 nm); Settles down when left undisturbed (more than 1000 nm in diameter); Does not settle down; Scatters light; Separates by filtration; Transparent; Salt solution; Milk; Sand in water; Smoke; Heterogeneous mixture; Cannot be separated by filtration; Mud; Butter; Brass. Complete the Table 5.2. Table 5.2 Solution Suspension Colloid Properties Properties Properties Examples Examples Examples
Solution
Here is the completed Table 5.2 based on the provided words and phrases.
Table 5.2
Solution
Suspension
Colloid
Properties
-
Small-sized particles (less than 1 nm diameter)
-
Particles remain evenly distributed
-
Does not settle down
-
Transparent
-
Cannot be separated by filtration
Properties
-
Large-sized particles (more than 1000 nm in diameter)
-
Heterogeneous mixture
-
Settles down when left undisturbed
-
Scatters light
-
Separates by filtration
Properties
-
Moderate-sized particles (1-1000 nm)
-
Heterogeneous mixture
-
Particles remain evenly distributed
-
Does not settle down
-
Scatters light
-
Cannot be separated by filtration
Examples
-
Salt solution
-
Brass
Examples
-
Sand in water
-
Mud
Examples
-
Milk
-
Smoke
-
Butter
Q4Revise, Reflect, Refine
Solve the following problems:
(i)
A cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of all-purpose flour and 5 g of sodium hydrogencarbonate. Express the concentration of each component in the mixture using an appropriate method.
(ii)
A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.
Solution
(i) Concentration of dry ingredients in a cake recipe:
Since all ingredients are solids measured by mass, the most appropriate method to express concentration is the mass by mass percentage (% m/m).
Given:
Mass of sugar = 75 g
Mass of all-purpose flour = 420 g
Mass of sodium hydrogencarbonate = 5 g
To Find:
Concentration of each component.
Formula:
Calculation:
Total mass of the mixture = Mass of sugar + Mass of flour + Mass of sodium hydrogencarbonate
Total mass = 75 g + 420 g + 5 g = 500 g
-
Concentration of sugar:
-
Concentration of all-purpose flour:
-
Concentration of sodium hydrogencarbonate:
Final Answer (i): The concentrations are: Sugar 15% m/m, all-purpose flour 84% m/m, and sodium hydrogencarbonate 1% m/m.
(ii) Quantities of copper and zinc in brass:
Given:
Total mass of brass = 120 g
Concentration of copper = 70% by mass
Brass is an alloy of copper and zinc.
To Find:
Mass of copper and mass of zinc.
Calculation:
-
Mass of copper: Mass of copper = 70% of the total mass of brass
-
Mass of zinc: Mass of zinc = Total mass of brass - Mass of copper
Final Answer (ii): The quantities present in 120 g of brass are 84 g of copper and 36 g of zinc.
Q5Revise, Reflect, Refine
The label on a cooking oil pack says one litre (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.
Solution
Yes, if cooking oil is mixed with water, it will form separate layers because oil and water are immiscible liquids.
Which substance will be on top?
To determine which liquid will be on top, we need to compare their densities.
- Density of oil: Given: Volume = 1 litre = 1000 mL, Mass = 910 g
- Density of water: The density of water is approximately 1.0 g/mL.
Since the density of oil (0.91 g/mL) is less than the density of water (1.0 g/mL), the oil will float on top of the water.
How to separate the two layers?
The two immiscible layers can be separated using a separating funnel. The process is as follows:
- Pour the mixture of oil and water into a separating funnel and let it stand undisturbed for some time.
- Two distinct layers will form, with the less dense oil on top and the denser water at the bottom.
- Place a beaker below the funnel and carefully open the stopcock to allow the lower layer (water) to drain out.
- Close the stopcock just as the last of the water layer has passed through.
- Place another beaker under the funnel and open the stopcock again to collect the upper layer (oil).
Diagram of the apparatus:
The apparatus used is a separating funnel. A diagram would show a conical glass vessel with a stopcock at the bottom and a stopper at the top, mounted on a stand. Inside, two layers would be depicted, with the oil layer above the water layer. (Refer to Figure 5.16 in the textbook for a visual representation).
Q6Revise, Reflect, Refine
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. Choose the correct option:
(i)
Both A and R are true, and R is the correct explanation of A.
(ii)
Both A and R are true, but R is not the correct explanation of A.
(iii)
A is true, but R is false.
(iv)
A is false, but R is true.
Solution
The correct option is (iii) A is true, but R is false.
Explanation:
-
Assertion (A): "Solutions do not exhibit the Tyndall effect." This statement is true. True solutions are homogeneous mixtures where the solute particles are extremely small (at the ionic or molecular level) and do not scatter a beam of light passing through them. The path of light is not visible.
-
Reason (R): "The particles in solutions are larger than 100 nm, so they cannot scatter light." This statement is false. The reason solutions do not scatter light is because their particles are extremely small, specifically less than 1 nm in diameter. Particles larger than 100 nm (found in suspensions and colloids) are the ones that do scatter light. The reason given is the opposite of the actual scientific principle.
Therefore, the assertion is true, but the reason is false.
Q7Revise, Reflect, Refine
How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method. If a mixture cannot be separated, explain why. Table 5.3 Mixture Method of separation Reason for selection Mud from muddy water Plasma from other components in the blood sample Naphthalene and sand Chalk powder and common salt Common salt and water Oil from water Pigments of the flower
Solution
Here is the completed Table 5.3 with the appropriate separation methods and reasons.
Table 5.3
Mixture
Method of separation
Reason for selection
Mud from muddy water
Coagulation followed by decantation/filtration, or Centrifugation.
Muddy water is a suspension. Fine particles may not settle easily or pass through a filter. Coagulation clumps them together for easier separation. Centrifugation uses force to separate denser mud particles from the lighter water.
Plasma from other components in the blood sample
Centrifugation
Blood is a colloid. Centrifugation spins the sample at high speed, forcing the denser components (red and white blood cells) to the bottom, leaving the lighter plasma on top.
Naphthalene and sand
Sublimation
Naphthalene is a sublimable substance (it turns directly from solid to gas upon heating), while sand is not. Heating the mixture will vaporize the naphthalene, which can then be collected by cooling (deposition), leaving the sand behind.
Chalk powder and common salt
Dissolution in water, followed by filtration and then evaporation/crystallization.
This method is based on the difference in solubility. Common salt is soluble in water, while chalk powder is insoluble. After dissolving the salt, filtration separates the chalk powder. The salt can then be recovered from the water by evaporation or crystallization.
Common salt and water
Evaporation or Distillation.
Evaporation is used if only the salt (solute) needs to be recovered. Distillation is used if both the salt and the pure water (solvent) need to be recovered, as it involves boiling the water and then condensing the vapour.
Oil from water
Using a separating funnel.
Oil and water are immiscible and have different densities. They form separate layers in a separating funnel, which allows the denser lower layer (water) to be drained off, leaving the oil behind.
Pigments of the flower
Paper chromatography.
A flower petal contains a mixture of different coloured pigments. Paper chromatography separates these components based on their different solubilities in a solvent and their differential movement up the paper.
Q8Revise, Reflect, Refine
Two miscible 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. Suggest a method to separate them. Also, draw a labelled diagram of the method suggested.
Solution
The suggested method to separate the two miscible liquids A and B is distillation.
Reason:
Distillation is used to separate a mixture of two miscible liquids that have different boiling points. The difference in their boiling points is 90 °C - 60 °C = 30 °C. Since this difference is more than 25 °C, simple distillation will be effective.
Process:
- The mixture of liquids A and B is taken in a distillation flask and heated.
- Liquid A has a lower boiling point (60 °C), so it will start to vaporize first.
- The vapours of liquid A will rise and pass into the condenser.
- A thermometer is placed at the neck of the flask to monitor the temperature, which will remain constant at 60 °C during the distillation of A.
- In the condenser, cold water circulates in the outer jacket, which cools the vapours of liquid A and condenses them back into liquid form.
- The pure liquid A is collected in a separate receiving flask.
- Liquid B, having a higher boiling point (90 °C), will remain behind in the distillation flask.
Labelled Diagram:
The diagram should show a standard distillation apparatus. The key labels are:
- Round-bottom distillation flask (containing the mixture)
- Heating source (like a Bunsen burner with a tripod stand and wire gauze)
- Thermometer (with its bulb at the level of the side tube opening)
- Condenser (with a water inlet at the bottom and a water outlet at the top)
- Receiving flask (to collect the distillate, which is pure liquid A)
(Refer to Figure 5.12 in the textbook for a detailed labelled diagram of a distillation set-up).
Q9Revise, Reflect, Refine
Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?
Solution
Evaporation, crystallization, and distillation are all methods used to separate components of a homogeneous mixture, typically a solid dissolved in a liquid. Here is a comparison:
Feature
Evaporation
Crystallization
Distillation
Principle
A volatile solvent is removed as vapour by heating, leaving the non-volatile solute behind.
A pure solid is obtained in the form of crystals from its saturated solution by cooling. It is based on the difference in solubility of the substance and its impurities at different temperatures.
Separation based on the difference in boiling points of the components of a liquid mixture. The more volatile component vaporizes and is then condensed back to a liquid.
Recovery of Components
Only the solute is recovered. The solvent is lost to the atmosphere.
The solute is recovered as pure crystals. The solvent is not typically recovered.
Both the solute (if non-volatile) and the solvent can be recovered. Or, two miscible liquids can be separated.
Purity of Solute
May yield an impure solid if soluble impurities are present.
Yields a very pure solid, as impurities remain dissolved in the solution.
The non-volatile solute left behind may contain impurities. The recovered solvent (distillate) is pure.
Situations for preference:
-
Prefer Evaporation: When you need to recover a non-volatile, stable solid solute from a volatile solvent, and you do not need to recover the solvent. For example, obtaining salt from seawater on a large scale where the water is not needed.
-
Prefer Crystallization: When you need to obtain a pure solid substance from a solution that may contain impurities. It is superior to evaporation for purification because it separates the desired solid from dissolved impurities, resulting in well-defined crystals. For example, purifying copper sulfate from an impure sample.
-
Prefer Distillation: When you need to:
- Recover a pure liquid solvent from a solution containing a non-volatile solute (e.g., obtaining pure water from salt water).
- Separate a mixture of two miscible liquids with significantly different boiling points (e.g., separating acetone and water).
Q10Revise, Reflect, Refine
Blood is an example of a colloidal mixture. (i) What would happen if blood behaved like a true suspension inside the body? (ii) In a blood sample, identify the dispersed phase and the dispersion medium.
Solution
(i)
If blood behaved like a true suspension inside the body, its components would not remain uniformly distributed. The heavier particles, such as red blood cells, would settle down due to gravity when the blood flow is slow or stops. This would lead to several catastrophic problems:
- Blockages: The settled cells could clog smaller blood vessels (capillaries), cutting off the supply of oxygen and nutrients to tissues and organs.
- Failure of Transport: The primary function of blood—transporting oxygen, nutrients, and hormones uniformly throughout the body—would fail. Areas of the body would be deprived of essential substances.
- Inconsistent Composition: The composition of blood would vary in different parts of the circulatory system, making its functions unreliable. In essence, the circulatory system would cease to function correctly, leading to rapid organ failure and death.
(ii)
In a blood sample, which is a colloid:
- Dispersed Phase: This consists of the solute-like components or particles that are suspended. In blood, the dispersed phase includes the solid components: red blood cells (RBCs), white blood cells (WBCs), and platelets.
- Dispersion Medium: This is the solvent-like component in which the dispersed phase is suspended. In blood, the dispersion medium is the liquid part, called plasma, which is mostly water.
Q11Revise, Reflect, Refine
You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). The Fig. 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques.
Solution
The mixture contains sand, common salt, and naphthalene. To separate these three components, we must use techniques that exploit their different physical properties.
- Naphthalene: It is a sublimable solid.
- Common Salt: It is soluble in water.
- Sand: It is insoluble in water.
The correct sequence of separation techniques based on these properties is as follows:
-
Step 1: Sublimation. Heat the mixture gently. The naphthalene will turn directly into vapour (sublime), leaving behind the sand and salt. The naphthalene vapour can be collected and cooled on a cold surface to turn back into solid naphthalene. This corresponds to diagram 1, which shows the setup for sublimation.
-
Step 2: Dissolution and Filtration. Take the remaining mixture of sand and salt and add water to it. Stir well to dissolve the common salt. The sand will not dissolve. Then, filter the mixture. The insoluble sand will be left on the filter paper as residue. The salt solution (filtrate) will pass through. This corresponds to diagram 2, which shows filtration.
-
Step 3: Evaporation or Crystallization. Heat the filtrate (salt solution) to evaporate the water. The common salt will be left behind as a solid. This corresponds to diagram 3, which shows evaporation from a china dish.
Therefore, the correct sequence of the steps shown in the figures is 1 → 2 → 3.
Q12Revise, Reflect, Refine
Why is distillation an effective method for separating a mixture of water and acetone?
Solution
Distillation is an effective method for separating a mixture of water and acetone because there is a significant difference in their boiling points.
- The boiling point of acetone is approximately 56 °C.
- The boiling point of water is 100 °C.
The difference in their boiling points is 100 °C - 56 °C = 44 °C.
The principle of simple distillation states that it can be used to separate two miscible liquids that have a difference in their boiling points of at least 25 °C. Since the 44 °C difference is well above this threshold, the separation is efficient.
When the mixture is heated, the more volatile liquid (acetone) vaporizes at a much lower temperature than the water. These acetone vapours can then be passed through a condenser, cooled, and collected as pure liquid acetone, leaving the less volatile water behind in the distillation flask.
Q13Revise, Reflect, Refine
Answer the following questions with the help of the data given in Table 5.4. Table 5.4 Salts Temperature (°C) 10 °C 20 °C 30 °C 40 °C 60 °C 80 °C Potassium nitrate 21 32 45 62 106 167 Sodium chloride 36 36 36.3 36.5 37 37 Potassium chloride 35 35 37.4 40 46 54 Ammonium chloride 24 37 41 41 55 66
(i)
What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40 °C?
(ii)
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). What would she observe as the solution cools? Explain.
(iii)
What is the effect of a change in temperature on the solubility of salts? Also, compare the changes in the solubility of the four given salts with increasing temperature from 10 °C to 80 °C.
Solution
(i) Mass of potassium nitrate needed:
From the table, the solubility of potassium nitrate at 40 °C is 62 g. This means 62 g of potassium nitrate is needed to prepare a saturated solution in 100 g of water.
To find the mass needed for 50 g of water, we can set up a proportion:
Mass in 100 g of water = 62 g
Mass in 50 g of water = ?
Answer: 31 g of potassium nitrate would be needed.
(ii) Observation on cooling a saturated potassium chloride solution:
From the table, the solubility of potassium chloride at 80 °C is 54 g (per 100 g of water).
At room temperature (25 °C), the solubility will be between its value at 20 °C (35 g) and 30 °C (37.4 g), approximately 36 g.
Since the solubility of potassium chloride decreases significantly as the temperature drops from 80 °C to 25 °C, the solution will become supersaturated upon cooling. The excess dissolved potassium chloride, which the water can no longer hold at the lower temperature, will separate from the solution.
Observation: The student would observe the formation of solid crystals of potassium chloride at the bottom of the container as the solution cools.
(iii) Effect of temperature on solubility:
Based on the data in the table, the general effect of an increase in temperature is an increase in the solubility of these salts in water.
Comparison of solubility changes from 10 °C to 80 °C:
- Potassium nitrate: Its solubility increases dramatically, from 21 g to 167 g. This is the most significant change among the four salts.
- Sodium chloride: Its solubility shows a very slight increase, from 36 g to 37 g. Temperature has a minimal effect on its solubility.
- Potassium chloride: Its solubility shows a moderate increase, from 35 g to 54 g.
- Ammonium chloride: Its solubility shows a significant increase, from 24 g to 66 g, though less dramatic than that of potassium nitrate.
Q14Revise, Reflect, Refine
Three students, A, B and C, are preparing sugar solutions for an experiment: Student A dissolves 20 g of sugar in 80 g of water. Student B dissolves 20 g of sugar in 100 g of water. Student C dissolves 30 g of sugar in 80 g of water.
(i)
Calculate the mass percentage (% m/m) concentration of sugar in each student's solution.
(ii)
Whose solution is the most concentrated? Explain why.
Solution
(i) Calculation of mass percentage (% m/m) concentration:
Formula:
where, Mass of solution = Mass of solute + Mass of solvent.
-
Student A: Mass of solute (sugar) = 20 g Mass of solvent (water) = 80 g Mass of solution = 20 g + 80 g = 100 g
-
Student B: Mass of solute (sugar) = 20 g Mass of solvent (water) = 100 g Mass of solution = 20 g + 100 g = 120 g
-
Student C: Mass of solute (sugar) = 30 g Mass of solvent (water) = 80 g Mass of solution = 30 g + 80 g = 110 g
Final Answer (i):
- Concentration of Student A's solution = 20% m/m
- Concentration of Student B's solution = 16.67% m/m
- Concentration of Student C's solution = 27.27% m/m
(ii) The most concentrated solution:
By comparing the calculated mass percentages:
- Student A: 20%
- Student B: 16.67%
- Student C: 27.27%
Student C's solution is the most concentrated.
Explanation:
Concentration refers to the amount of solute dissolved in a given amount of solvent or solution. A higher concentration means a larger proportion of solute relative to the solvent. Student C's solution has the highest mass percentage (27.27%), which indicates it contains the greatest amount of sugar per unit mass of the solution compared to the other two solutions.
Q15Revise, Reflect, Refine
Examine Fig. 5.26.
(i)
Identify the separation technique marked as 'S'.
(ii)
Label the apparatus A, B and C.
(iii)
Which of the following mixtures can be separated by the technique identified above? Use the data given in Table 5.5. Mixtures:
(a)
water - acetone
(b)
water - salt
(c)
acetone - alcohol
(d)
sand - salt
(e) alcohol - chloroform
(f) alcohol - benzene
Table 5.5: Boiling points of some compounds
Solvent
Water
Acetone
Alcohol
Chloroform
Benzene
Temperature (°C)
100 °C
56 °C
78 °C
61 °C
80 °C
Solution
(i) Identification of the technique:
The separation technique 'S' shown in Figure 5.26 is Distillation.
(ii) Labelling the apparatus:
- A: Distillation flask (or round-bottom flask)
- B: Condenser
- C: Receiving flask (or conical flask)
(iii) Mixtures that can be separated by distillation:
The technique shown is simple distillation, which is effective for separating:
- A liquid from a non-volatile solid.
- Two miscible liquids with a difference in boiling points of at least 25 °C.
Let's analyze the mixtures using the data from Table 5.5:
-
(a) water - acetone: Boiling points: Water = 100 °C, Acetone = 56 °C. Difference = 100 - 56 = 44 °C. Since the difference is > 25 °C, this mixture can be separated by distillation.
-
(b) water - salt: Salt is a non-volatile solid dissolved in liquid water. This mixture can be separated by distillation to recover the pure water.
-
(c) acetone - alcohol: Boiling points: Acetone = 56 °C, Alcohol = 78 °C. Difference = 78 - 56 = 22 °C. Since the difference is < 25 °C, simple distillation is not very effective. Fractional distillation would be required.
-
(d) sand - salt: This is a mixture of two solids. Distillation is not an appropriate method.
-
(e) alcohol - chloroform: Boiling points: Alcohol = 78 °C, Chloroform = 61 °C. Difference = 78 - 61 = 17 °C. Since the difference is < 25 °C, simple distillation is not effective.
-
(f) alcohol - benzene: Boiling points: Alcohol = 78 °C, Benzene = 80 °C. Difference = 80 - 78 = 2 °C. The boiling points are very close, so simple distillation cannot separate them.
Final Answer: The mixtures that can be effectively separated by the simple distillation technique shown are (a) water - acetone and (b) water - salt.